Geometry Problems

Quant Funda Class (Geometry)

Plane Geometry ANGLES and PARALLELS (a) Two straight lines which meet at a point form an angle between them.

a

b

c

Acute angle : 0° < a < 90° Right angle : b = 90° Reflex angle : 180° < d < 360°

d

Obtuse angle : 90° < c < 180°

(b) Theorems : If AOB is a straight line, then a + b = 2 right angles (Adjacent angles on a straight line) If a + b = 2 right angles, then AOB is a straight line. (Adjacent angles are supplementary) The sum of all the angle at a point, each being adjacent to the next, is 4 right angles.  a + b + c + d + e = 4 right angles (Angles at a point) If two straight lines intersect, the vertically opposite angles are equal. a = b, c = d (Vertically Opposite angles)

Prepared By Harshwardhan Phatak

a

a

b

b

a e b d c

a d c b

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Parallel lines PQ and RS are cut by a transversal LM, then we have : 1. 2. 3.

The corresponding angles are equal c = b (Corresponding angles, PQ || RS) The alternate angles are equal a = b (Alternate angles, PQ || RS) The interior angles are supplementary b + d = 2 right angles (Interior angles, PQ || RS)

L P

c a d

Q

R

b

S

M

If PQ and RS are cut by transversal LM, the two lines are parallel if a = b (Alternate angles) or c = b (Corresponding angles) or b + d = 2 right angles (Interior angles are supplementary) (c) (i) Two angles whose sum is 90°, are complementary. Each one is the complement of the other. (ii) Two angles whose sum is 180º, are supplementary. Each one is the supplement of the other. TRIANGLES PROPERTIES : 1. Sum of the three interior angles is 180° 2. When one side is extended in any direction, an angle is formed with another side. This is called the exterior angle. 3. There are six exterior angles of a triangle. 4. Interior angle + corresponding exterior angle = 180°. 5. An exterior angle = Sum of the other two interior angles not adjacent to it 6. Sum of any two sides is greater than the third side. 7. Difference of any two sides is less than the third side. 8. Side opposite to the greatest angle will be the greatest and vice versa. 9. A triangle must have at least two acute angles. 10. Triangles on equal bases and between the same parallels have equal areas. 11. If a, b, c denote the sides of a triangle then (i) if c² < a² + b², Triangle is acute angled (ii) if c² = a² + b², Triangle is right angled (iii) if c² > a² + b², Triangle is obtuse angled IMPORTANT POINTS WITH RESPECT TO A TRIANGLE : a. Centroid : When a vertex of a triangle is joined to the midpoint of the opposite side, we get a median. The point of intersection of the medians is called the CENTROID of the triangle. The centroid divides any median in the ratio 2 : 1.


A

A

G

B

D

C

B

D

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E

C

Quant Funda Class (Geometry) Any median of a triangle bisects the area of the triangle. In the figure AD is the median from A. and BD = CD We have the formula : 2 x (Median)² + 2 x (½ Third side)² = Sum of squares of other two sides  2 x (AD)² + 2 x (½ BC)² = AB² + AC². Area of triangle ABD = Area of triangle ADC. If BD = DE = EC then the areas of triangles ABD, ADE, AEC are equal. D & E are called the points of trisection of BC. b. Circumcentre : The point at which the perpendicular bisectors of the sides of a triangle meet is the circumcentre of the triangle. The circumcentre S of a triangle is equidistant from the three vertices. We have SA = SB = SC = circumradius. The circle with center S and passing through A,B & C is called the circumcircle of triangle ABC. In the triangle : BSC = 2  BAC,  ASB = 2  ACB,  CSA = 2  ABC Any point on the perpendicular bisector of a line is always equidistant from the ends of the line.

A

S B

c. Incentre : The point of intersection of the angle bisectors of a triangle is called the incentre I. The perpendicular distance of I to any one side is inradius and the circle with centre I and radius equal to inradius is called the incircle of the triangle. The three sides are tangent to the incircle. Any point on the bisector of an angle is equidistant from the arms of the angle. The incentre divides the bisector of  A in the ratio (b + c) : a.

C

A

c

b I r

B

a D

.

Also  BIC = 90 + A/2,  AIB = 90 + C/2,  AIC = 90 + B/2. The bisectors of two exterior angles at B & C and the bisector of  A meet at a point called excenter I’. There can be three ex–circles of a triangle. Also I’E = Ex–radius. There can be three ex–radii to a triangle.  BI’C = 90 – A/2. NOTE : (a) Angle bisector Theorem : In the figure if AD is the angle bisector (Interior) of  BAC then 1 . AB/AC = BD/DC 2 . AB x AC – BD x DC = AD²

C

I'

A

C

B D


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Quant Funda Class (Geometry) (b) In the figure If AD is the bisector of exterior angle at A of triangle ABC, then : AB/AC = BD/DC

E

A

B

D

C A

(c) In the figure, if in triangle ABC, AN is the bisector of  BAC and AM is perpendicular to BC then  MAN = ½ ( B –  C).

C

B M

N

(d) In a triangle ABC, if BC is produced to D and AL is the bisector of  A, then  ABC +  ACD = 2 ALC (e) In triangle ABC, if side BC is produced to D and bisectors of  ABC and  ACD meet at E, then  BEC = A/2 d.

Orthocentre : The perpendiculars drawn from vertices to opposites (called altitudes) meet at a point called orthocentre of the triangle. Also

A

 BOC = 180 –  A  AOB = 180 –  C  AOC = 180 –  B

Area of triangle = ½ x Base x Altitude.

O B

C

D

NOTE : There is similarity between the topic on geometry & that on mensuration. You are requested to go through this topic only after thoroughly grasping the chapter on mensuration as formulae and facts discussed in that chapter are not repeated here. Some Important Points : 1.

2.

Isosceles triangle : In this the base angles (or any two angles are equal). The bisector of  A is perpendicular bisector of the base and is also the median to the base.

A

C

B AB=AC

Equilateral Triangle : (a) All the four points viz. centroid, circumcentre, incentre, orthocentre coincide. (b) Medians, angle bisectors, altitudes, perpendicular bisectors of sides are all represented by same straight lines. (c) Given the perimeter, equilateral triangle has the maximum area. (d) Of all the triangles that can be inscribed in a circle, the equilateral triangle has the greatest area. Points (c) and (d) give you a hint regarding the nature of symmetry.

3.

Right triangle : Median to the hypotenuse = ½ x hypotenuse = circumradius


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Quant Funda Class (Geometry) GENERAL THEOREMS ON SIMILARITY 1.

Proportionality Theorem : Intercepts made by two transversal lines (cutting lines) on three or more parallel lines are proportional. In the figure, lines X 1Y 1 & X 2Y 2 are transversals cutting the three parallel lines AB, CD, EF. Then AC, CE, BD, DF are intercepts Also, AC/BD = CE/DF

A

Y1 Y2

C

B D

E

F

X1

X2

2.

Midpoint Theorem : A triangle, the line joining the mid points of two sides is parallel to the third side and half of it.

3.

Basic Proportionality Theorem : A line parallel to any one side of a triangle divides the other two sides proportionally. If DE is parallel to BC, then (a) AD/BD = AE/EC, (b) AB/AD = AC/AE, (c) AD/DE = AB/BC and so on.

A D B

E

C

PROPERTIES OF SIMILAR TRIANGLES : (1) If two triangles are similar, then Ratio of sides = Ratio of heights = Ratio of medians = Ratio of angle bisectors = Ratio of in–radii = Ratio of circumradii

Note : In the above result, Ratio of sides refers to the Ratio of corresponding sides etc. (2) Ratio of areas = Ratio of squares of corresponding sides A (3) RIGHT TRIANGLE : o ABC is a Right Triangle with A as the Right angle. 90 AD is perpendicular to BC then 90 o (a) Triangle ABD ~ Triangle CBA & BA² = BC x BD B C D (b) Triangle ACD ~ Triangle BCA & CA² = CB x CD (c) Triangle ABD ~ Triangle CAD & DA² = DB x DC.


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Quant Funda Class (Geometry) QUADRILATERALS Properties : (a) Sum of the four interior angles = 360° (b) If a quadrilateral can be inscribed in a circle, it is called a cyclic quadrilateral. Here opposite angles are supplementary. If one side is produced, then the exterior angle = Remote interior angle. (c) The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram. (d) If a quadrilateral is circumscribed about a circle, the sums of opposite sides are always equal. PARALLELOGRAM : If in a quadrilateral, the opposite sides are parallel and equal, it is a || gm . a (a) Opposite angles are equal (b) Diagonals bisect each other (not at 90°) b (c) Sum of any two adjacent angles = 180° (d) Bisectors of four angles enclose a rectangle. (e) Each diagonal divides it into 2 equal triangles. (f) When inscribed in a circle, it becomes a rectangle (g) When circumscribed about a circle, it becomes a rhombus. (h) Diagonals divide it into 4 equal triangles. (i) Point of intersection of medians is equidistant from the four vertices. (j) The figure formed by joining the Mid–points of adjacent sides of a || gm is a ||gm . (k) Diagonals are unequal in lengths and do not bisect angles at vertices. RHOMBUS : If in a parallelogram, all sides are equal, it is a Rhombus. (a) Opposite angles are equal (b) Diagonals bisect each other at 90°. (c) Diagonals bisect angles at vertices. a (d) Sum of any two adjacent angles = 180°. (e) Figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle. (f) Diagonals are unequal. (g) Point of intersection of medians is equidistant from the four vertices.

a a a

RECTANGLE : (a) Opposite sides equal, each angle = 90° (b) Diagonals bisect each other (not at 90°). l (c) Of all rectangles of given perimeter, a square has max. area (d) When inscribed in a circle, it will have maximum area when it’s a square. (e) Figure formed by joining the midpoints of a rectangle is a rhombus. (f) If P is any point within a rectangle ABCD then PA² + PC² = PB² + PD² (g) The biggest circle that can be inscribed in a rectangle will have the diameter equal to the breadth of the rectangle. (h) When a rectangle is inscribed in a circle, the diameter of the circle is equal to the b diagonal of the rectangle.

b

b

l l


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Quant Funda Class (Geometry) SQUARE : (a) All sides equal, all angles 90°. (b) Diagonals bisect each other at 90° and are equal (c) When inscribed in a circle, diagonal = diameter of circle (d) When circumscribed about a circle, Side of square = Diameter of circle. a

a

a

a

a

a

a

a

a

a

a

a

TRAPEZIUM (TRAPEZOID) : (a) The median is half the sum of parallel sides. (b) If inscribed in a circle, it becomes an isoscles Trapezium. The oblique sides are equal, angles made by each parallel side with oblique sides are equal. Diagonals are equal. (c) Diagonals intersect proportionally and the ratio = ratio of length of parallel sides. (d) Four similar triangles are obtained by joining the mid points of adjacent sides. C

D

(e) In the figure ABCD is a trapezium, then AC² + BD² = BC² + AD² + 2 AB.CD (f) In the above figure if X & Y are mid points of diagonals, then XY = ½ (AB – CD).

A

CYCLIC QUADRILATERAL :

A

B

B

x

1.

The four vertices lie on a circle.

2.

Opposite angles are supplementary.

3.

If any one side is produced, Exterior angle = Remote interior angle

4.

If one pair of opposite sides are equal, diagonals are equal

5.

The line joining the points of intersection of the bisectors of opposite angles of a cyclic quadrilateral with the circle is the diameter of the circle.

6.

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

7.

Sum of products of opposite sides = product of diagonals

y x C

D o


x+y=180

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Quant Funda Class (Geometry) CIRCLES A (a) Tangent is perpendicular to radius. (b) Perpendicular from centre to a chord bisects the chord. P O (c) Tangents drawn from an external point are equal. B (d) Equal chords of a circle are equidistant from the centre. (e) When two circles touch, their centres & the point of contact are collinear (f) If two circles touch externally, distance between centres = sum of radii (g) If two circles touch internally, distance between centres = difference of radii (h) Circles with same centre and different radii are concentric circles. Q P (i) Points lying on the same circle are called concyclic points. (j) Only one circle can pass through three given points (k) Equal arcs subtend equal angles at the centre. O (l) Measure of an arc means measure of central angle. 60 o (m) m(minor arc) + m(major arc) = 360° (n) Angle at the centre made by an arc A = twice the angle made by the arc at any point on the remaining part of the circumference. We have  APB = ½  AOB = 30° =  AQB (o) Angle in a semicircle is a right angle. D P (p) Alternate Segment Theorem : In the fig.

B

A

B

C

if BAC is the tangent at A to a circle and if AD is any chord, then  DAC =  APD or            PAB =  PDA (Angles in alternate segment)


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SOME MORE RESULTS A

4.

(a) If triangle ABC is an acute angled triangle then AB² = BC² + AC² – 2 BC x CD

C 90o

(b) If  A is obtuse, then BC² = AB² + AC² + 2AB x AD

B

90o

C

D

A

(c) If in triangle ABC, PQ is parallel to BC such that area of triangle APQ = Area of trapezium PQBC; then AP/PB = 1/(2 – 1) 5.

B

A

D

P

Q

B

C

Three times the sum of squares of sides in a triangle is equal to four times the sum of squares of medians.

CONGRUENCE & SIMILARITY OF TRIANGLES 1.

Two triangles are said to be CONGRUENT if they are equal in all respects. (a) The three sides of one must be respectively equal to the three sides of the other. (b) The three angles of the first must be respectively equal to the three angles of the other. Thus, if  ABC and  XYZ are congruent, Z

A

B

C

X

Y

(represented as  ABC   XYZ) then AB = XY, AC = XZ, BC = YZ  B =  Y,  C =  Z.

&  A =  X,

(c) Theorems : X

A

*

Triangle ABC  Triangle XYZ if AB = XY  A =  X AC = XZ (SIDE ANGLE SIDE) S.A.S.

C

B

X

A

*

Triangle ABC  Triangle XYZ if  B =  Y  C =  Z AC = XZ (ANGLE ANGLE SIDE) A.A.S.


B

Z

Y

C Y

Z

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Quant Funda Class (Geometry) *

Triangle ABC  Triangle XYZ if  B =  Y  C =  Z BC = YZ (ANGLE SIDE ANGLE) A.S.A.

X

A

*

Triangle ABC  Triangle XYZ if AB = XY AC = XZ BC = YZ (SIDE SIDE SIDE) S.S.S.

A

X

C

B

*

Triangle ABC  Triangle XYZ if  B =  Y = 90° Hypotenuses AC = XZ, AB = XY (RIGHT ANGLE HYPOTENUSE SIDE) R.H.S.

Z

Y X

A

Z

Y

C

B

*

Z

C Y

B

A

If two sides of a triangle are equal, the angles opposite to these sides are equal. If AB = AC, then  B =  C (BASE ANGLES, ISOSCELES TRIANGLE)

C

B A

*

2.

If two angles of a triangle are equal, the sides opposite to these angles are equal. If  B =  C then AB = AC (SIDES OPPOSITE EQUAL ANGLES)

C

B

Two triangles are SIMILAR if (a) The angles of one are respectively equal to the angles of the other (b) The corresponding sides are proportional (c) Theorems : *

*

If 2 triangles are equiangular, their corresponding sides are proportional. In triangles ABC and XYZ, if  A =  X,  B =  Y,  C =  Z then AB/XY = AC/XZ = BC/YZ (EQUIANGULAR TRIANGLES) A.A.A. If 2 triangles have their corresponding sides proportional, they are equiangular. In triangles ABC and XYZ, if AB/XY = AC/XZ = BC/YZ then  A =  X,  B =  Y,  C =  Z (3 SIDES PROPORTIONAL)


X

A

C

B

A

B

Z

Y

X

C

Y

Z

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Quant Funda Class (Geometry) *

If 2 triangles have one angle of the one A equal to one angle of the other and the sides about these equal angles are proportional, then the triangles are similar. If  A =  X, then AB/XY = AC/XZ B then  B =  Y,  C =  Z (RATIO OF TWO SIDES, INCLUDED ANGLES)

(d) If triangle

3.

ABC & Triangle PQR are

similar

X

C

Z

Y

then it is denoted by  ABC ~  PQR.

If two chords AB & CD intersect externally at P then (a) PA x PB = PC x PD (b) P = ½ [m (arc AC) – m (arc BD)]

A B P D C

B

4.

If PAB is a secant and PT is a tangent then, (a) PA x PB = PT² (b) P = ½ [m (arc BXT) – m arc (AYT)]

A P

Y

X T

5.

If chords AB & CD intersect internally at P, then A

(a) PA x PB = PC x PD

D

P

x

y

C

B

(b) BPD = ½ [m (arc AXC) + m (arc BYD)] 6.

Common Tangents For the two circle with centers A & B, PQ is a direct common tangent and RS is a transverse common tangent. P

Q

(a) Length of direct common tangent =

r1

(Distance between centres)² – (r1 – r2 )²

A

B

r2

R

(b) Length of transverse common tangent A

=

(Distance between centres)² – (r1 + r2 )² r O

(c) In this figure 1. OM /O’M = r/r’ 2. OP/O’P = r/r’


P

B r' O' D

M

C

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EXERCISE

1.

Find the value of x in the figure given

2.

Find the values of x, y in the figure given

3. 4.

A chord of length 10 cm subtends an angle120° at the centre of a circle. Calculate its distance from the centre An equilateral triangle is inscribed in a circle of radius 6 cms. Find its sides.

5.

In the diagram B : C = 2 : 3. Find B, C.

6.

In the following figure, equal sides BA and CA of a triangle ABC are produced to Q and P respectively, so that AP = AQ. Prove that PB = QC.

7.

ABC is an isosceles triangle such that AB = AC. Side BA is produced to a point D such that

AB = AD. BCD = ?


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Quant Funda Class (Geometry) 8.

If X and Y are two points on equal sides AB and AC of a ABC, such that AX = AY, prove that XC

= YB.

9.

In ABC, D and E are points on AB and AC, such taht DE | | BC. Show that ABC ~ ADE.

10.

ABC, is a triangle, right angled at A, and AD is perpendicular to BC. If DAB = B, Show that

AD = BD = CD.

11. 12. 13.

In the above figure, A = 90°, AD is perpendicular to BC and B = 45°. If AB = x find AD in terms of x. In a trapezium ABCD, AB || DC, AB = 2CD. If the diagonals intersect at O, show that area of AOB = 4 × area of COD. ABCD is a trapezium with AB ||CD. If AC and BD intersect at E such that BE DE = 2 : 3 show that

AEB 4  . CED 9 Fin the length of a chord, which is at a distance of 4 cms from the centre of a cricle whose radius is 5 cms. In a circle of radius 5 cms, AB and CD are two parallel chords of length 8cms and 6 cm respectively. Calculate the distance between the chords, if they are on (i) the same side of the centre (ii) opposite sides of the centre ABC is an equilateral triangle inscribed in a circle of radius r. What is the length of the side of the triangle? O is the centre of a circle of radius 5cms. P is any point in the circle such that OP = 3 cms. A is the point travelling along the circumference. X is the distance from A to P. What is the least and greatest values of x in cms? ABCD is cyclic quadrilateral. A circle passing through A and B meets AD and BC at E and F respectively. Is EF parallel to DC? O is the centre of the circle, PA and PB are tangnets to the circle from point P. If PAB = 70, find PBA and APB. Three circles with centres A, B and C touch each other externally at P, Q and R, if AB = 5cms, BC = 7cms and CA = 6 cms, find the radius of each circle. area of the triangles

14. 15.

16. 17.

18. 19. 20.


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Quant Funda Class (Geometry) 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

31. 32. 33.

Two circles touch internally at P and a straight line ABCD meets the outer circle in A and D and the inner circle in B and C. Is APB = CPD? Two parallel lines cut two lines AB, AC at points B, D, C, E respectively. Prove that the circle ABC and ADE touch each other. P and Q are mid points of the sides AB and AC of a triangle ABC. PQ is produced to R such that PQ = QR. Is PRCD a parallelogram? ABCD is a trapezium with AB || CD. E is the mid point of AD. Prove that the line parallel to AD through E bisects BC. ABCD is a quadrilateral. P, Q, R, S are mid points of the sides AB, BC, CD, AD respectively. Is PQRS a parallelogram? In an isosceles ABC, AD is the median to the unequal side meeting BC at D. DP is the angle bisector of ADB and PQ is drawn parallel to BC meeting AC in Q. PDQ = ? ABC is a triangle with AB = AC. D is the midpoint of BC. What is the ratio of DAB to DAC? The bisectors of angle sB and C of a ABC meet the opposite sides at D and E respectively. BD and CE intersect at O. If ABC is isosceles, what kind of triangle is ODE? ABC is an isosceles triangle with B = C. The bisectors of B and C meet the opposite sides at P and Q. Is PQ || BC? In ABC, AD and BE are perpendiculars from A and B to the sides BC and AC respectively. Show that (i) ADC ~ BEC (ii) CA . CE = CB . CD The diagonal BD of a parallelogram ABCD intersects the segment AE at a point F where E is any point on the side BC. Then (DE . EF) : (FB . FA) = ? In ABC, DE is parallel to BC. If AD = 5 cm, AE = 6 cm, BC = 12 cm, AB = 15 cm. Find AC and DE. ABC and DBC are two right angled triangles with common hypotenuse BC. The sides AC and BD

AP.PC ? DP.PB ABCD is a quadrilateral inscribed in a circle and the diagonal BD bisects AD. Prove that AD . AB = CD . CB. Triangle ABC and DEF are similar. If AB and PQ are 7.5cms and 3.5 cms respectively,find the ratio of the areas of the two triangles. The medians BE and CF of the ABC intersect at P. FInd the ratio of the area of BPC to that of FPE. Three circles with equal radii touch one another externally. Prove that the triangle formed by joining their centres in an equilateral triangle. CD is a tangent at C to the circumcircle of ABC intersecting AB produced in D. Prove that DBC ~ DCA. Two circles touch each other externally at P. APC and BPD are drawn through P to meet the wto circles in A, B and C, D respectively. Prove that (i) PAB ~ PCD (ii) AB || CD Six cubes each with 12 cm edge are joined end to end. Find the surface area of the resulting cuboid. The outer dimensions of a closed box are 42 cm, 30 cm and 20 cm. If the box is made of wood of thickness 1 cm, determine the volume of wood used. Water flows at the rate of 10 metres per minute from a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth is 24 cm. The hieght of a right circular cone is 40 cm and the radius of the base is 28 cm. Find the area of crosssection of the cone formed by a plane parallel to the base of the cone and at a distance of 10 cm from the vertex of the cone. extended intersect at P, Then

34. 35. 37. 38. 39. 41.

42. 43. 44. 45.


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Quant Funda Class (Geometry) 46. 47.

48. 49.

50.

If the radii of the bottom and top ends of a bucket 24 cms high, are 5 cm and 15 cms. Find the volume of the bucket. A cylinder is circumscribed about a hemi-sphere and a cone is inscribed in the cylinder so as to have its vertex at the centre on one end and the other end as its base. Find the ratio of the volumes of the cylinder, hemispheer and the cone. A right circular cylinder and a right cone have equal bases and equal height. If their curbed surfaces are in the ratio 8:5, what is the ratio of the radius of their base to their height. A spherical ball of lead has been melted and made into smaller balls of half of the radius of the original one. How many such balls can be made? Find the ratio of the total surface area of all the smaller balls to that of the original one. If successive squares are formed by joining the mid points of the sides of a parent square and the process is repeated infinitely, what is sum of the areas of all squares thus formed?

SOLUTIONS 1.

Draw AC parallel to the other parallel lines. Then a = 180° - 128° = 52°. a + 20 + x = 180° or x = 108°

2.

2 = 90° (alternate angles). = x (corresponding angles). 1 + 2 = 3x + 10. x + 90° = 3x + 10 2x = 80° or x = 40°. y = 180 - 1 (adjacent angles). y = 180° - 40° = 140° 3.

AB = 10 cm. d represents  distance from centre of chord AB tothe center of the circle. Hence, D is AD 5 5  3  OD  OD OD 3 Circum radius of an equilateral triangle = a / 3 (a is the side of the triangle) so a = 6 3 DAC = B + C (external angle = sum of the interior opposite angles) 130 = 2x + 3x x = 26 B = 52°; C = 78° In PAB and QAC are congruent since, PA + QA; BA = CA; and PAB = QAC. Hence, PB = QC (corresponding sides of congruent triangles)

the mid point of AB. AOD = 60°. tan 60  4. 5. 6.


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8. 9. 10. 11.

In ABC, AD = AC. so, DAC = AD = AC, so, D = DCA In BDC, D + C = 180° 2x + 2y = 180° x + y = 90° BCD = 90° Altnernatively: With A as the centre we can draw a circle that passes through the three points B, C, D. Hence A is the circum centre of the triangle ABC. So C must be a right angle. Since AB = AC, B = C. In XBC and YBC, BX = CY and XBC = YCB. BC is the common side, Hence XBC and YCB are congruent. So, XC = YB D = B; E = C and A is common to both the triangles. So, ADE and ABC are similar. B = C = 45° AD = BD = CD = AC If B = 45° AD = BD = CD = AC Hence, AD = x / 2

12.

AOB and COD are similar. The ratio of the proportional sides is 1 : 2. Hence the ratio fo the areas of the triangles DOC and AOB is 1 : 4. 13.

BE : DE = 2 : 3. The triangles AEB and CED are similar. The ratio of the proportional sides is 2 : 3

Area (AEB) 4 therefore, Area (CED)  9 14. 17.

6 cms.

15. 1 cm, 7 cms. 16. R = a / 3 or a = R3 x = 2 is the smallest value of x, when O, P and A are in a straight line

{A and B lie on the same side of O}. x = 8 is the largest value, when O, P and A lie on a straight line and P and A lie on the opposite sides of O.


Page 16 of 54


Both ABCD and ABFE are cyclic quadrilaterals

ABF + AEF = 180° (Sum of opposite angles = 180°). ABF + ADC = 180°. AEF - ADC = 0°. AEF = ADC. These are corresponding angles. Since they are equal EF || DC. 19.

PA = PB (tangents from external point). So PBA = PAB = 70°. BPA = 180° - 140° = 40° 20.

Let x1, x2 and x3 be radii of circles with centres A, B, C resp. Then x1 + x2 = 5, x2 + x3 = 7, x3 + x1 = 8, 2(x1+ x2 + x3) = 20 x1 + x2 + x3 = 10,  x1 = 3, x2 = 2, x3 = 5. 21.

Join the mid point of the Chord AB i.e. E and the point P. E would also be the mid point of the chord BC. Triangles AE and DEP are congruent. So AP = DP. Similarly triangles BPE and CPE are congruent. So BP = CP. AB = CD (Since AE - BE = DE - CE). Hence the triangles ABP and CDP are congruent. so BPA = CPD.


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Since the 2 circles have a common pt. A through which both pass, they touch internally at A. 23.

Since P and Q are mid points of AB and AC resp. PQ = 1/2 BC  Bc = 2 PQ = PR and PR || BC. So, PRCB is parallelogram. 24.

Since EF || AB, but AB || CD therefore, EF || CD. Since EG || CD, Also GF || AB,

AE AG  AD AC

BF AG BF AE 1    . So, F is the mid pt. of BC . So, CB CA CB AD 2

25.

Consider triangles ABC and ACD. PQ || AC; RS || AC. Hence, PQ || RS. Similarly, from triangles ABD and BCD we get SP || QR. So the pairs of opposite sides are equal. Hence, PQRS is a parallelogram.


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AB = AC. AD is the median. So ADB = 90°, PDA = 45° ( DP is angle bisector). Now, DQ is also angle bisector ABC is isosceles triangle PDQ = 90° 27.

ABD = ACD. AB = AC and BD = DC So triangle ABD and ADC are congruent. So BAD = CAD or AD bisects A 28.

29.

If ABC is an isoceles triangle with AB = AC. Then the point D divides AC in the ratio AB/BC and E divides AB in the ratio AC/BC. Since D, E divide sides in the same ratio we find that ED is parallel to BC. So  EDB and  DEC are congruent. Therefore BDE = CED. OED is isosceles triangle. Proof Similar to the last question.

30.

(i) For triangle ADC and BEC the angles are all equal. B = A . E = D = 90°. C is common to both. Hence, ADC - BEC (ii) From (i)

CB CE   CB.CD  CA.CE . CA CD


Page 19 of 54


’s AFD and EFB are similar. So

FA DF  or FA . FB = DF . EF EF FD

32.

Since DE || BC 

AD AE DE 15 5    AC  6   18 cm and DE   12  4cm. AB AC BC 5 15

33.

PD AP  PC PB

Triangles ABP and DCP are similar so, 34. 35.

AD . AB = CD . CB. AB 2

since ABC ~ DEF



DE

AB AC BC 10    DE DF EF 7

2



AC 2 DF

2

8.2 10  DE 7



BC 2 EF

2



Area ( ABC) 100  Area ( DEF) 49

DE = 5.74

36.

ABC ~ PQR

AB 2 PQ 2

2



2

Area ( ABC)  7.5   15  225      Area ( PQR)  3.5   7  49


Page 20 of 54


Point P divides the median in the ratio 2 : 1. So, s) so FPE ~ CPB. Hence,

CP BP 2   . And FPE = BPC (vertically opp. PF PE 1

Area(FPE) PF 2 1   Reqd. ratio = 4 : 1 Area ( CPB) CP 2 4

38.

In a circle AP . PB = DP . PC [Refer Properties]. Since AP = CP we have PB = DP AB = AP + PB = CP + PD = CD 39.

If A, B, C are the centers of 3 triangles each of chose radii is r units. Then AB = AC = BC = 3r units. Hence, ABC is an equilateral triangle. 40.

In the circle DC is a tangent at C. BA extended meets the tanget at D. 1 = 2 (Alternate segment Theorem). 4 is common to both the triangles. Since 2 angles are equal the third angle must also be equal by AAA property triangles BDC and CDA are similar.

41.


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42. 43. 44.

(i) PAB ||| PCD. Since BPA = CPB (vertically opp. angles). XPB = A = YPB = C similar YPB = C. B = D. Alternate segment Theorem (ii) D = B. CD || AB If l, b, h are length, breadth, height of cuboid. l = 12 × 6 = 72 cm. b = 12 cm. h = 12 cm Hence surface area = 2 (lb + bh + lh) = 2(864 + 144 + 864 = 2 × 1872 = 3744 cm2. Outside Volume of box = 42 × 30 × 20 cm3 = 25200 cm3. Inner volume of the box = 40 × 28 × 18 cm3 Volume of wood = 25200 - 20160 = 5040 cm3 Volume of water flowing every minute = 1000 cm ×  × (.5)2 cubic cms = 250 cu cms.

1 2 1 1 400  24 2 2 min utes  12.8 minutes Volume  r h  .(20) .24  3200 cm . Time    3 3 3 250 45.

BE | | CD 

AB AE BE 10    7 cm Area of cross section =  (r)2 = 54 cm2. So, BE  28  AC AD CD 40

46.

AB BE AB 5 1    . AB = 12. CSA (curved surface area) of cone =  r l = 15 × 39  = . AC CD AB  24 15 3 585. CSA of cone with radius BE =  × 5 × 13 = 65.  Lateral Surface area of bucket = 520. Total Surface area = 520 + base area of bucket = 520 + 25 = 545 cm2. 47. All have same base radius r. Also, height of cylinder = Height of cone = radius of hemisphere = r.

48.

2 3 1 2 2 3 2 3 1 3 Vcylinder : Vhemisphere : Vcone  r h : r : r h  r : r : r = 3 : 2 :1 3 3 3 3 For cylinder let radius and the height be R, H respectively. For cone they are r, h respectively. Curved Surface Area for cylinder : cone = 8 : 5  2 RH : r r 2  h 2  8 : 5 But R = r and H = h 

2 RH R R2  H 2



8 R 3 H2 16  .    3H  4 R or 2 2 5 H 4 25 R H 3

49.

4r 2 4 4 3 4 r 4 r 3 TSA big     r       Original volume . Volume of new spheres . 3 3 2 3 8 TSA small r2 1 4 4


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 Required ratio = 1 : 4. Twice the area of the parent square (by infinite GP sum formula)

Exercise 1.

O is a point within the triangel ABC. Areas of triangles AOB, BOC and COA are equal. Then O is the (a) incentre (b) orthocentre (c) circumcentre (d) centroid of the triangle

2.

ABCD is a rectangle and P is any point within it. If AB = 20 cms, BC = 15 cms, PD2 + PB2 = 344.5 sq. cms, then OP where O is the point of intersection of the diagonals, is (a) 2.5 cm (b) 3 cm (c) 4 cm (d) 3.5 cm

3.

Of all the triangles with the same baes length and equal areas, the one which has the least perimeter is (a) equilateral (b) right angled (c) isosceles (d) none of these

4.

In the given figure, P is the midpoint of BC and E is the midpoint of AB, AD, BC and EF are parallel to each other. If the area of the triangle ABP is 4 sq. units, the area of the triangle ECP is

(a) 2 sq. units 5.

(b) 3 sq. units

(c) 3/2 sq. units

(d) 1 sq. units

The incircle of the triangle ABC touches BC, CA, AB at D, E, F respectively. The angle EDF is (a)

1 (B  C) 2

(b)

1 (C  A) 2

(c)

1 ( A  B) 2

(d) none of these.

6.

The perpendiculars from B and C to the sides AC and AB repectively of the triangle ABC meet at O. Then AC2 + BO2 is (a) AB2 + OC2 (b) AB2 + AO2 (c) BC2 + AO2 (d) BC2 + BO2

7.

If AB and DC of a quadrilateral ABCD are produced to intersect at E and if E = 90°, then AC2 + BD2 is equal to (a) AD2 + BC2 (b) AB2 + CD2 2 2 2 2 (c) 1/2 (AB + BC + CD + DA ) (d) none of these

8.

ABCD is a trapezium in which AB is parallel to DC and AD = BC. The quadrilateral formed by joining the midpoints of adjacent sides of this trapezium is (a) a rhombus (b) a square (c) a rectangle (d) another trapezium


Page 23 of 54


Which of the following statemens is true? (a) All the angles of a parallelogram can be acute (b) If the diagonals of a quadrilateral are at right angles, it is a rhombus (c) If the diagonals of a quadrilateral are at right angles, the figure formed by joining the midpoints of adjacent sides is a rectangle. (d) If one pair of opposite sides are parallel and the other pair of opposite sides are equal in a quadrilateral, the quadrilateral is a parallelogram.

10.

CD is a median of the triangle ABC. BE and AF are drawn perpendicular to CD. Then which of the following is not true? (a) Angle ACD = Angle BCD (b) BE = AF (c) ADC = Area of BDC (d) Angle DBE = Angle BCD A

r

e

a

o

f

11.

AB, CD, EF are three equidistant parallel lines. PRL and QSM are any two transversals cutting them at P, R, L and Q, S, M respectively. Then (a) 2 RS = P + LM (b) PR = QS (c) PR = SM (d) QS = SM

12.

The bisectors of the angles of a parallelogram enclose (a) another parallelogram (b) a rhombus (c) a rectangle (d) a trapezium

13.

In the quadrilateral ABCD, the bisectors of the angles A and C meet on the diagonal BD. Then the bisectors of the angles B and D. (a) are parallel (b) are perpendicular (c) meet on AC (d) none of these

14.

The median AD of the triangle ABC meets BC at D. The internal bisectors of angles ADB and ADC meet AB and AC at E and F respectively. Then EF (a) is perpendicular to AD (b) is parallel to BC (c) divides AD in the ratio AB : AC (d) none of these

15.

ABCD is a parallelogram. P is a point on AB such that AP : PB = 3 : 2. q is a point on CD such that CQ : QD = 7 : 3. If PQ meets AC at R, then AR : AC is (a) 5 : 11 (b) 6 : 13 (c) 4 : 7 (d) 2 : 5

16.

The radius of the incircle of a triangle with sides 15 cms, 15 cms and 18 cms is (a) 4 cm (b) 5 cm (c) 6 cm (d) 8 cm

17.

ABCD is a trapezium with AB parallel to DC. AC and BD intersect in E. Through E. a line is drawn parallel to AB meeting the sides AD and BC at F and G. Then E (a) is one of the points of trisection of FG (b) divides FG in the ratio 2 : 3 (c) bisects FG (d) none of these

18.

Which of the following statements is not true? Two triangles are similar if (a) the sides of one are respectively parallel to the sides of the other (b) the side sof one are respectively perpendicular to the sides of the other


Page 24 of 54

Quant Funda Class (Geometry) (c) two of the sides of one are proportional to the sides of the other and an angle of one is equal to an angle of the other. (d) the three side sof one are proportional to the three sides of the other. 19.

Two circles with centres A and B intersect at P and Q. Then (a) AB is the perpendicular bisector of PQ (b) PQ is the perpendicualr bisector of AB (c) the figure formed by joining the midpoints of AP, PB, BQ, QA is a rectangle (d) area of triangle APQ = triangle PBQ

20.

Two circles intersect at A and B. P is a point on the common chord AB extended. PS and PT are tangents from P to the two circles. Then (a) PS + PT (b) SP2 = PA . PB (c) Triangles SPA and PTA are similar (d) Both (a) and (b)

21.

A square is inscribed in a circle of radius a. Another square is inscribed inside the square with its vertices at the midpoints of the sides of the new square. The length of a side of the new square is (a) 2 a (b) 1/2 a (c) a (d) 2/3 a

22.

A square and an equilateral triangle are inscribed in a circle of radius r. Then their side sare in the ratio (a) 1 : 3 (b) 2 : 3 (c) 1 : 3 (d) 2 : 3

23.

AB, AC are two chords of a circle of radius r. If 2 AB = AC and p, q are respectively the perpendicular distances of AB and AC from the centre then, 4p2 - q2 is equal to (a) r2 (b) 3r2 (c) 3r2 (d) 4r2

24.

ABCD is a cyclic quadrilateral whose diagonals intersect at E. If BEC = 80°, DBC = 60° and BCD= 40° which of the following statements is true? (a) BC = bisects Angle ADC (b) AB = BC (c) DA = DC (d) AC bisects angle BCD

25.

Two circles intersect at A and B. Through A straight line is drawn terminating on the circumferences of the circles. The greatest possible length of the line is (a) twice the distance between the centres (b) three times the distance between the centres (c) when it is parallel to the line joining the centres (d) none of these

26.

Chords AB, BC, CD subtend angles of 80°, 60° and 70° respectively at the centre O of a circle. Then the acute angle between AC and BD is (a) 60° (b) 75° (c) 65° (d) 80°

27.

PQ, RS are two chords of a circle which intersect at L within the circle. PR and SQ meet outside thecircle at M. If angle M is 30° and PRS is 50° the angle PLS is (a) 110° (b) 100° (c) 80 ° (d) 70°


Page 25 of 54


Two circles intersect at A and B and through A, two straight lines PAQ and XAY are drawn terminated by the circumferences. Then angle PBQ and angle XBY (a) are equal (b) are complementary (c) are supplementary (d) have no relation between them

29.

O is the centre of a circle of radius r. AOB is a diameter and circles are drawn on OA and OB as diameters. If a circle is drawn to touch these three circles, its radius is (a)

2r 3

(b)

r 2

(c)

r 4

(d)

r 3

30.

The distance between the centres of two circles with radii 9 cms and 4 cms is 13 cms. The length of a direct common tangnet between them is (a) 10 cms (b) 12 cms (c) 11 cms (d) 10.5 cms

31.

Two concentric circles have radii 13 cms and 5 cms.If AB, the chord of the bigger circle, touches the smaller circle, the length of AB is (a) 24 cms (b) 18 cms (c) 12 cms (d) 8 cms

32.

Two circles intersect each other at O and P. AB is a common tangent to the circles. Then the angles subtended by the segment AB at O and P are (a) complementary (b) supplementary (c) equal (d) none of these

33.

Two circles intersect each other at P and Q. From a point C on any one of the circles, lines CP and CQ are drawn meeting the other circle at A and B. Then the tangent at C (a) is parallel to PQ (b) is parallel to AB (c) is perpendicular to CS, where S is the centre of the circle PCQ. (d) both (b) and (c)

34.

Tangents at A, B on a circle intersect at C. D is any point on the minor arc AB. If ACB = 40°, then ADB is (a) 140° (b) 130° (c) 110° (d) 120°

35.

P is any point on a circle whose centre is O. A chord which is parallel to the tangent at P bisects OP. IF the length of the chord is 12 cms. the radius of the circle is (a) 4 cms (b) 4 3 cm (c) 3 cms (d) 2 5 cms

ANSWER BOX 1.(d) 11.(a) 21.(c) 31.(a)

2.(c) 12.(c) 22.(b) 32.(b)

3.(c) 13.(c) 23.(c) 33.(d)

4.(a) 14.(b) 24.(d) 34.(c)

5.(a) 15.(b) 25.(a) 35.(b)


6.(c) 16.(a) 26.(b)

7.(a) 17.(c) 27.(d)

8.(a) 18.(c) 28.(a)

9.(c) 19.(c) 29.(d)

10.(a) 20.(d) 30.(b)

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SOLVED EXAMPLES Part 1 1.

The interior angles of a polygon are in a ratio of an A.P. whose first term is 2 and the common difference is 1. Find the area of the polygon if the number of sides is equal to five. Sol. Let the angles be 2 X, 3 X, 4 X, 5 X, and 6 X  Given (2 X + 3 X + 4 X + 5 X + 6 X) = (n – 2) 180° where n = 5 Solving it we get : X = 27  Angles are 54°, 81°, 108°, 135°, and 162° Answer. 2.

The biggest possible circle is inscribed in a rectangle of length 16 cms and breadth 6 cms. Find its area. Sol. Diameter of the circle = Breadth of the rectangle  Required area =  x (6/2)² = 9  cms² Ans. 3.

Determine whether the following statements are True or False. If False, give the True statement. (a) If the diagonals of a quadrilateral bisect each other then definitely it is a square. (b) The sum of two complementary angles is 180°. (c) In a cyclic quadrilateral, the opposite angles are supplementary. (d) If a triangle is inscribed in a circle and it has the maximum area possible, then it must be an isosceles triangle. (e) The angle in a semi–circle is a right angle. (f)The tangents at any point of a circle and the radius through that point are perpendicular to each other. (g) The straight line joining the mid–points of any two sides of a triangle is parallel to the third and one–third of it. Sol. (a) False : If the diagonals of a quadrilateral bisect each other, the quadrilateral could be a square or a rhombus or a rectangle or a ||gram. (b) False : The sum of two complementary angles is 90°. (d) False : If a triangle is inscribed in a circle and it has the maximum area possible than it is an equilateral triangle. (g) False : The straight line joining the mid points of any two sides of a triangle is parallel to the third side and half of it. ALL OTHER STATEMENTS ARE TRUE. 4.

(a) The complement of an angle exceeds the angle itself by 60°. Find the angle. (b) The supplement of an angle is one–fourth of the angle itself. Find the angle. Sol. (a) Let the angle be X  Its complement = (90 – X) Given that : (90 – X) – X = 60°  X = 15° (b) Let the angle be X  Its supplement = X/4 Given that : X/4 + X = 180°  X = 144° Ans. 5. In a triangle ABC, AB² + AC² = 200 cms². Median AD = 10 cms. Find BC. Sol. Use the formula for the length of MEDIAN  2 (AD)² + 2 (BD)² = AB² + AC²  2 x 100 + 2 BD² = 200  BD² = 0  BD = 0  The triangle does not exist under the given conditions Ans. 6.

The sides of a triangle are in the ratio 5 : 4 : 3. Find the sides of this triangle, given that another similar triangle of sides 30, 24, 18 cms has an area 92 times the area of the triangle. Sol. Use the property of similar triangles. Let the required sides be 5 X, 4 X, and 3 X respectively  Area of bigger triangle : Area of smaller triangle = 92 : 1 = 30² : 25 X²  X² = 22  X = 1.68. Thus the req lengths of sides are 5 x 1.68, 4 x 1.68, 3 x 1.68 cms. 7. Find the area enclosed when three circles of radius 10 cms each touch each other. Sol. Req Area = (Area of the equilateral triangle ABC) – (Area of 3 sectors) = 3/4 x 20² – 3 x ( x 10² x 60° / 360°) = 100 3 – 50  = 16.12 cms² Ans.


Page 27 of 54


A number of triangles are drawn with perimeter 20 cms. Find the type of triangle which will have the maximum area. Also find its area. Sol. The maximum area shall be of an equilateral triangle (due to symmetry)  Sum of the three sides = 20 = 3 x Side  Side = 20/3 = 6.66 cms  Area = 3/4 x 6.66² = 19.2 cms² Ans. 9.

A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a tower casts a shadow 40 m long on the ground. Determine the height of the tower. Sol. Use similar triangles property :  Height of Tower : Height of stick = Shadow of tower : Shadow of stick  X : 12 cms = 40 m : 8 cms  X = 60 m Ans. 10. ABCD is a cyclic quadrilateral whose side AB is a diameter of the circle through A, B, C, D. If angle ADC = 140°, find angle BAC. Sol. Opposite angles of a cyclic quadrilateral are supplementary  angle ABC = 180° – 140° = 40° Then we know that angle ACB = 90° becasuse it is the angle contained in the half–circle  Angle BAC = 180° – 90° – 40° = 50° Ans. 11. How many common tangents can be drawn through to two circles of radius 2.8 cms and 2 cms respectively under each of the following conditions : (1) the centres of the two circles are 2 cms apart (2) the centres of the two circles are 4.8 cms apart (3) the centres of the two circles are 6.8 cms apart (4) the centres of the two circles are 0.6 cms apart Sol. Difference of radius = 2.8 – 2 = 0.8 cms Sum of radius = 2.8 + 2 = 4.8 cms (1) Since distance between centres is less than the sum, they do not touch externally but cross each other (intersect) at two points. This is also evident from the fact that the distance between centres is more than 0.8 cms which indicates internal touching. Thus the number of direct common tangents possible are 2. No transverse common tangent is possible. (2) Number of direct common tangents possible is 2. Number of transverse common tangents possible is 1 (3) Number of direct common tangents possible is 2. Number of transverse common tangents possible is 2 (4) The smaller circle lies wholly in the bigger one  No tangent is possible. Ans. 12. With the vertices of the triangle ABC as centres, three circles are made each touching the other externally. If the sides of the traingle are 4 cm, 6 cm, and 8 cm resp, find the radii of the circles. Sol. Let the radii of the three circles be R1, R2, R3 respectively. (Three equations / Three unknowns  Can be solved)  R1 + R2 = 6, R2 + R3 = 8, R3 + R1 = 4  R1 = 1 cm, R2 = 5 cms, R3 = 3 cms Ans. 13. Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal. Sol. Let the square’s side be A  Its diagonal is A 2  Area of the first equilateral triangle = 3/4 x A² Area of the second equilateral triangle = 3/4 x (A 2)²  Required ratio = 1 : 2 Ans. 14. The perimeters of two similar triangles are 30 cms and 20 cms resp. If one side of the first triangle is 15 cms, then find out the corresponding side of the other triangle. Sol. For similar triangles, Ratio of perimeters = Ratio of corresponding sides  30 / 20 = 15 / X  X = 10 cms Ans.


Page 28 of 54

Quant Funda Class (Geometry) 15. Two isosceles triangles have equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding heights. Sol. Area = ½ Base x Altitude For similar triangles : Ratio of areas = Ratio of squares of corresponding heights  9/16 = height 1² / height 2²  height 1 : height 2 = 3 : 4 Ans. 16. A sector of a circle of radius 12 cms has an angle of 120°. It is rolled up so that two bounding radii are joined together to form a cone. Find the volume of the cone. Sol. Arc of the sector = 120/180 x  x 12 cms = 8  cms  The circumference of the base of cone = 8  cms Let R cms be the base of the cone  2  R = 8   R = 4 cms Given L = 12 cms  Height of the cone = (L² – R²) = (144 – 16) = 82 cms  Volume of the cone = 189.5 cc Ans. 17. A rectangular sheet of iron foil is 44 cm long and 20 cm wide. A cylinder is made out of it by rolling the foil. Find the volume of the cylinder. Sol. The cylinder obtained from the foil has perimeter 44 cms and height 20 cms  2  R = 44  R = 22/ cms  Volume of the cylinder = 3080 cc Ans. 18. The exterior angle of a regular polygon is one–third of its interior angle. How many sides has the polygon ? Sol. Each exterior angle of a regular polygon of n sides = 360°/n Each interior angle of a regular polygon of n sides = (2n – 4) x 90°/n Using the given informations, we get 360° = 1 x (2n – 4) x 90° n 3 n 12 = 2n – 4  2n = 16  n=8 Hence the given polygon has 8 sides Ans. 19. Each interior angle of a regular polygon is 150°. Find the number of sides of the polygon. Sol. Since interior angle is 150°  Exterior angle = 180° – 150° = 30°  Sum of the exterior angles of a polygon is 4 rt. angles  Number of sides = 360°/External angle = 360°/30 = 12 Ans. 20. Two regular polygons are such that the ratio between their number of sides is 1 : 2 and the ratio of measures of their interior angles is 3 : 4. Find the number of sides of each polygon. Sol. Let the number of sides of two polygons be X and 2X respectively. Since the ratio of measures of their interior angles is 3 : 4. 2X – 4 x 90° X = 3/4  X = 5 on solving. 2(2X) – 4 x 90° 2X Hence the number of sides of two polygons are 5 and 10 resp. Ans.

{ {

}

}

SOLVED EXAMPLES part 2 Ex 1.

(a) In figure (a), find X and Y. (b) In figure (b) AB is parallel to EF.  ABC = 65°,  CDE = 50°,  DEF = 30°, find  BCD. B

D 65o

y

50o

x o

40

F

o

C

40

30o

fig (a)


A

fig (b)

E

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Quant Funda Class (Geometry) Sol.

(a)

X = 40° (vert. opp. angles) Ans. Y = 180° – 40° – X (adj. angles on st.line) = 180° – 40° – 40° = 100° (b) Draw CG || AB  CG || EF F Draw DH || EF  CG || DH G s = 30° (alt. angles, DH || EF) D B  r = 50° – 30° = 20° y x X = r = 20° (alt. angles, DH || CG) r s 30 65 C Y = 65° (alt. angles, CG || AB)  BCD = X + Y = 20° + 65° = 85° Ans. H E A o

o

C

Ex 2.

Given: the figure below.

To find: a, b, c.

50 o b

A

Sol.

Ex 3.

70 o

B

a + 36° + 70° = 180° (sum of triangle)  a = 180° – 36° – 70° = 74° Ans b = 36° + 70° (Ext. angle of triangle) = 106° Ans. c = a – 50° (Ext. angle of triangle) = 74° – 50° = 24° Ans. In the figure, AXD, BXE and CXF are straight lines, find in degrees: F

(a)

(b) the angle a, if b = c = d = e = f = x

E

f e

the angle X, if X = c + d, x

A

X

a b

B

Sol.

c 36 o

D a

d

D

c

C

(c) the value of a + b + c + d + e + f (a) X =  CXD (vert. opp angles) c + d +  CXD = 180° (sum of angles)  c + d + X = 180°  X + X = 180° (because given X = c + d)  2X = 180°  X = 90° Ans. (b) X =  CXD (vert. opp. angles)  X + c + d = 180° (sum of angles)  3X = 180° (given c = d = X)  X = 60° e = f = 60° (given e = f = x) FXE = 180° – 60° – 60° = 60°  EXD = 180° – 60° – 60° (adj. angles on stt. line) = 60°  AXB = 60° = b a = 180° – 60° – 60° = 60° Ans. (c) a + b + c + d + e + f = 3 x 180° – (360°/2) = 540° – 180° = 360° Ans. A

Ex 4.

In the figure, AD, BE and CF are the altitudes of the triangle ABC intersecting at G. Find the sum of  BAC and  BGC.

F G B

Sol.

E

b r

s

C

 BAC = 90° – b (ext. angles of triangle)  BGC = 180° – r – s   BAC +  BGC = (90° – b) + (180° – r – s) = 270° – (b + r + s) = 270° – 90° = 180° Ans.


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Quant Funda Class (Geometry) Ex 5. In the figure, if AC : CB = 1 : 3, AP = 4.5 and BQ = 7.5, PA || RC || QB, find CR. A

C

B

7.5 4.5

P

R

Q

3

A

Sol. Join AQ cutting CR at K. CK/7.5 = 1/4  CK = 7.5 / 4 KR /4.5 = 3/4  KR = 13.5 /4  CR = CK + KR = 7.5/4 + 13.5/4 = 5.25 Ans.

1

B

C

7.5

K

4.5

P

Q

R

F

Ex 6. In the figure, circle ADB, centre O, AC = CB = 12 cm, CD = 8 cm, OD  AB. Find the radius r of the circle. D Sol. OD = r cm (same radii) OC = r – 8 cm 12 8 12 A C r² = 12² + (r – 8)² (Pythagoras’ Theorem) r  r² = 144 + r² – 16 r + 64 O  r = 13 cm Ans.

Ex 7. AB and CD are parallel chords of a circle with centre O, MN = 3 cm, AB = 4 cm, CD = 10 cm. To find : the radius, r cm, of the circle. Sol. Let ON = X cm AM = MB = ½ (4) cm ( from centre bisects chord) = 2 cm CN = ND = ½ (10) cm ( from centre bisects chord) = 5 cm A In Triangle AMO, (1) r² = 2² + (3 + X)² (Pythagoras’ theorem) C In Triangle CNO (2) r² = 5² + X² (Pythagoras’ theorem)  2² + (3 + X)² = 5² + X² (1) = (2)  X = 2 cm Put X = 2 cm in (2)  r² = 5² + 2² = 29  r = 29 cm. Ans.


B

M r r

N x

B D

O

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Quant Funda Class (Geometry) Ex 8. Given : In the figure, diameter AB produced and chord QR produced meet at P.  QPA = 28° and  QAR = 32° To find : X and r.

A o

32

x

1

B o

28

r

x

Q

R

Sol. AB is a diameter (Given)   AQB = 90° ( in semi–circle) X = X1 (angles in same seg.)  32° + X1 + 28° + X + 90° + 180° (sum of  of triangle) 2X = 30° X = 15° Ans. r = 28° + X1 (Ext.  of triangle) = 28° + 15° = 43° Ans.

R r

Ex 9. Given:In the figure, ABCD is a cyclic quad. To prove: r + s = 180° – 2c Sol. c = c1 (Vert. opp.  s) b = c + s (Ext. ) d = c1 + r (Ext. ) But b + d = 180° (Opp. s, cyclic quad.)  c + s + c1 + r = 180°  r + s + 2c = 180°  r + s = 180° – 2c. Ex 10. In the figure ABCD is a cyclic quadrilateral. If AB is a diameter, BC = CD, and ABD = 40°, find DBC. Sol. In ABD,  d = 90° ( in semicircle)  a = 180° – d – b (sum of triangle) a = 180° – 90° – 40° = 50° In cyclic quadrilateral ABCD, c + a = 180° (Opp.  of a cyclic quad.)  c = 180° – a = 180° – 50° = 130° But DCB is an isosceles triangle, with BC = CD.  X = X1 (base s, isos triangle)  X1 + X + C = 180° (sum of  of triangle)  X1 + X = 180° – c = 180° – 130° = 50°  X1 = 1/2. 50° = 25°   DBC = 25° Ans.

P

D c1 d

C

b

A

s

B

S

D C

x d

c x1

A

40o b

a

Ex 11. In the figure ABC is a triangle inscribed in a circle, centre O. If  ABE = 25°,  BAC = 50°, find the angles a1, a2, a3, a4, b1 and b2. Sol. b1 = 2  BAC ( at centre is twice  at circumference) = 2 x 50° = 100°  a1 = b1 = 100° (vert. opp. ) b2 = 50° + 25° (Ext.  of triangle) = 75° E A o a2 = b1 – b2 (Ext.  of triangle) b 50 = 100° – 75° = 25° a Also BOC is isosceles with OB = OC, same radii, ob  a3 = a4 and a3 + a4 + b1 = 180° 25o a  a3 + a4 = 180° – b1 = 180° – 100° = 80°  a3 = a4 = 40° B  a1 = 100°, a2 = 25°, a3 = 40°, a4 = 40° b1 = 100° and b2 = 75°.

2

a2

B

C

a4

1

1

3


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Quant Funda Class (Geometry) Ex 12. Given : In the figure, 2 circles, centres Y and Z touch each other externally at A. Another circle, centre X, touches the other 2 circles internally at B and C. XY = 6 cm, YZ = 9 cm, ZX = 7 cm. To find : The radii of the circles. Sol. Let X, Y, Z be the radii of the circles, centres X, Y, Z resp. YAZ, XYB, XZC are straight lines (Contact of circles) XY = X – Y = 6 .... (1) XZ = X – Z = 7 .... (2) YZ = Y + Z = 9 .... (3)  (1) + (2) + (3) X 2X = 22 Y A Z  X = 11, Y = 5, Z = 4 The radius of the circle, centre X, is 11 cm. C B The radius of the circle, centre Y, is 5 cm. The radius of the circle, centre Z, is 4 cm. Ans.

Ex 13. Given : In the figure, X is a point on diameter AB of the circle with centre O, such that AX = 9 cm, XB = 5 cm. To find : the radius of the circle, centre Y, which touches the diameter at X and touches the circle, centre O, internally at Z. Sol. Let YX = YZ = r (Same radii) Z OYZ is a straight line (Contact of circles) r YX  AB (Tangent  to radius) Y AX = 9, XB = 5 (Given) O  AB = 14, 7 2 X 5 B A OB = OZ = 7 (same radii) OX = 7 – 5 = 2 In triangle OXY, OY = 7 – r YX = r, OX = 2  OY² = YX² + OX² (Pythagoras’ Theorem) (7 – r)² = r² + 2²  49 – 14r + r² = r² + 4  14 r = 45  r = 33/14 cm Ans. Ex 14. Given : In the figure, chord AC and BD of a circle are produced to meet at P, PA = 10 cm, PB = 8 cm, PC = 5 cm, AC = 6 cm. To find : BD, PD. Sol. In Triangles ACP and BDP a = a1 ( in same seg.) A p=p (common) a D  ACP =  BDP (3rd  of triangle)  Triangle ACP ~ triangle BDP (A.A.A.) K p P BD/BP = AC/AP (corr. sides of ~ triangles) a1 C  BD/8 = 6/10 BD = 4.8 cm Ans. PD / BP = PC / AP (Corr. sides) B PD/8 = 5/10 PD = 40/10 = 4 cm Ans. Ex.15. Determine the length of the mirror required to form a complete image of a man 6 feet tall. Given that the distance of the eye from the topmost part of the head is 5 cms. Sol. For all such questions, the required length of the mirror is always HALF the man’s height  3 feet answer Ans.


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Quant Funda Class (Geometry) Ex.16. The interior and the exterior angles of an isosceles triangle are in the ratio 1 : 4. Find all the interior and exterior angles of the triangle. Also find out the lengths of all the sides of the triangle if the perimeter of the triangle is 20 cms. Sol. Let the interior angle and the exterior angle be X and 4 X  X + 4 X = 180°  X = 36° Thus there are two possible cases : CASE I : If we take 36° to be one of the equal angles, the 3 internal angles would be 36°, 36°, and (180 – 2 x 36 = 108°) resp. The exterior angles will be 144°, 144°, 72° respectively (in the same order).Thus, we use the Sine rule : Sin A / a = Sin B / b = Sin C / c We know that a + b + c = 20, A = 36°, B = 36°, C = 108° Also since triangle is isosceles, a = b Thus we can easily solve the Sine rule equations to get a = 5.54 cms, b = 5.54 cms, c = 8.92 cms NOTE : To solve by the Sine rule, you need SINE INVERSE functions. This is generally NOT asked in the MBA tests but you must have an idea of the procedure. CASE II : If we take 36° to be the unequal angle, the 3 angles work out to be 36°, 72°, 72°. The exterior angle will be 144°, 108°, 108° resp (in the same order). Now to get the edge lengths we have b = c, a + b + c = 20 cms. Again using the same procedure of Sine rule, we get a = 7.64 cms, b = c = 6.18 cms NOTE : WE HAVE SOLVED THE PROBLEM TAKING THE RATIO 1 : 4 INTO CONSIDERATION THROUGHOUT. Ex.17. A regular polygon of 8 sides is inscribed in a circle of radius 20 cms. Find its area and also the value of the angle a side of the polygon subtends at the centre of the circle. Also find the value of an internal angle of this polygon. Sol. Sum of all interior angles = (n – 2) 180° Put n = 8  Sum of angles = 1080°  Each internal angle = 135° Thus now if we join each vertex of this 8–sided polygon to the centre of the circle, the central angle subtended by each side would be 360°/8 = 45° Now to find the area : Area of the polygon = Sum of areas of the eight isosceles triangles = Sum of areas of 16 right angled triangles Area of a right angled triangle = ½ Base x Height = ½ B x P = ½ (20 x Cos 67.5°) x (20 x Sin 67.5°)  Area of right triangle = 70.7 sq cms  Area of the Polygon = 16 x 70.7 = 1131.2 cms² Ans. Ex.18. In a triangle ABC, AD is drawn perpendicular to BC. Let If a = 25 cms, P = 12 cms, BD = 9 cms. Find b and c. Sol. We can get the sides using the Pythagoras theorem  b = (12² + 16²) = 20 cms, c = (9² + 12²) = 15 cms Ans.

P denote the length AD.

Ex.19. Two circles of diameter 10 cms and 18 cms touch each other internally. Find the distance betewen their centres. Find the same if they touch externally. Sol. When they touch each other externally, the distance between their centres = 10 + 18 = 28 cms When they touch each other internally, the distance between their centres = 18 – 10 = 8 cms Ans. Ex.20. In a trapezium the distance between the parallel sides is 10 cms and the segment joining the midpoints of the oblique sides is 5. Find the area of the trapezium. Sol. Area of a trapezium = ½(Sum of parallel sides)x(Distance between them) We know that : Sum of parallel sides / 2 = Length of the segment joining midpoints of the oblique sides  Required Area = ½ (5 x 2) x 10 = 50 cms² Ans.


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Quant Funda Class (Geometry) Ex.21. M and N are points on the sides PQ and PR respectively of a  PQR. For each of the following cases state whether MN is parallel to QR : (a) PM = 4, QM = 4.5, PN = 4, NR = 4.5 (b) PQ = 1.28, PR = 2.56, PM = 0.16, PN = 0.32 Sol. (a) The triangle PQR is isosceles  MN || QR by converse of Proportionality Theorem (b) Again by Converse of Proportionality theorem, MN || QR Ans. Ex.22. P and Q are the points on the sides AB and AC respectively of a  ABC.If AP = 2 cm, PB = 4 cm, AQ = 3 cm, QC = 6 cm; Prove that BC = 3 PQ. Sol. The two triangles ABC and APQ are similar. Thus BC : PQ = AB : AP  BC = PQ x (AB / AP) = PQ x (AP + PB) / AP On substituting the values, we get BC = PQ x (2 + 4)/2 = 3 PQ Ans. Ex.23. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle. Sol. Let us assume that the two given chords are on the same side of the centre (Even if you take them to be on the opposite sides of the centre, the final answer shall be the same as one distance shall come negative)  Given : PQ = OQ – OP = 3 cms Now we have the right angled triangles OAQ and OCP Let radius = R In triangle OAQ, OQ² = R² – AQ²  OQ² = R² – 2.5² In triangle OCP, OP² = R² – CP²  OP² = R² – 5.5² On subtracting, we get : OQ² – OP² = 24  (OQ – OP) (OQ + OP) = 24  3 x (OQ + OP) = 24  OQ + OP = 8  On solving for OP and OQ, we get OQ = 5.5 cms, OP = 2.5 cms  R² = OQ² + AQ² = 5.5² + 2.5²  R = 6.04 cms Ans. NOTE : In this problem, we have used the property that a perpendicular from the centre to a chord bisects the chord. Ex.24. In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cms. Find the length of the chord BC. Sol. Considering the triangle ABP, we get : (5 – X)² + Y² = 36 Considering the triangle OPB, we get : X² + Y² = 25 Solving these : we get Y = 4.8 cms  Req answer = 2 Y = 9.6 cms. Ex.25. AB and CD are two chords of a circle such that AB = 10 cm, CD = 24 cm, and AB is parallel to CD. The distance between AB and CD is 17 cms. Find the radius of the circle. Sol. Question is similar to Q 43 : with the difference that the two chords will finally work out to be on different sides of the centre. However, even if you draw the diagram assuming the chords on the same side of the centre, the final answer would definitely be the same. Final Answer: Radius = 13 cms. Ex.26. ABCD is a cyclic quadrilateral whose diagonals intersect at P. If angle DBC = 70° and angle BAC = 30°, find angle BCD. Sol. We use the property that the angles subtended by an arc at any point on the circumference are equal  Arc DC subtends angle CBD and CAD at the circumference  Angle CBD = Angle CAD = 70°  Angle DAB = Angle DAC + Angle BAC = 70° + 30° = 100° Since opposite angles of a cyclic quadrilateral are supplementary, we get Angle DAB + Angle BCD = 180°  Angle BCD = 80°


Page 35 of 54


Questions from Previous CATs and other METs 27.

Find the area of the shaded region in the diagram below.

Sol.

(1) 6  2 2  (2) (6  2 ) (3) 9( - 1) (4) 9  Shaded area = Area of big semicircle - (Area of 2 small semi-circles + area of triangle)

 r12  2r22 1    bh   18  9  9  9(  1) . Hence, (3).  2  2 2  ABCD is quadrilateral whose diagonals intersect at E. Which one of the following is not sufficient to prove AB || DC? (1) ABD = BDC (2) AE : EC = BE : ED (3) AB : BE = DC : DE (4) ADC = BCD and AD = CB 

28.

Sol.

(1) ABD = BDC AB|| DC (2) AE : EC = BE : ED  In ABE, and CDE, we have corresponding sides proportional and their included angles equal (vertically opposite)  They are similar triangles ABD = BDC  AB || DC (3) AB : BE = DC : DE. We take the same two triangles as in (2), but as we do not know whether the included angles ABD and BDC are equal or not, we cannot conclude that AB || DC. (4) ADC = BCD and AD = BC. We extend DA and CB to meet at O. As ADC = BCD, ODC is isosceles  OD = OC. As AD = BC, OA = OB,

 OA OB   , If a line (seg AB) divides any two sides of a triangle ( ODC) in equal ratio   AD BC  then it is parallel to the third side (seg DC). Hence, AB || DC. Hence, (3). 29.

A solid metal sphere is melted and smaller spheres of equal radii are formed. 10% of the volume of the sphere is lost in the process. The smaller spheres have a radius which is 1/9th the larger sphere. If 10 litres of paint were needed to paint the larger sphere, how many litres are needed to paint all the smaller spheres? (1) 90 (2) 81 (3) 900 (4) 810


Page 36 of 54


Volume available on melting 

4 ( R)3 (0.9) 3

[10% loss]

4 ( R)3 (0.9)  (0.9)93 . Paint required is proportional to total surface Number of smaller spheres  3 3 4 R   3 9 area i.e., 4  r2. Total surface area of the bigger sphere = 4  R2 Total surface area of all small 2

(8.1)4R R 2  10  81 litres. Hence, (2). sphere  9 (0.9)4   (8.1)4R .  Paint required  4R 2 9 Circles shown in the figure are of radius = 3 cm each. AB and CD are tangents to both the circles. CD = 10 cm. What is the length of AB? 2

3

30.

(1) Cannot be determined (3) 6 cm

(2) 8 cm (4) 10 cm

Sol.

PC = PA = QD = QB = 3 cm Join PQ. Since PC  CD and QD  CD. PCDQ is a rectangle.  PQ = 10 cm.  PAO = QBO = 90°; AOP = QOB (vertically opp.) AP = BQ.  AOP  BOQ (SAA test of congruency)  AO = OB and PO = OQ. Since PQ = 10 cm. PO = OQ = 5. Since AP = 3, PO = 5 and  AOP is right angled triangle.

AO  52  32  16  4 .  AB = 2AO = 2 × 4 = 8. Hence, (2). 31.

An Eskimo constructs an igloo such that his son’s head touches the roof when he stands 30 cm away from the centre. The thickness of the walls is 8 cm. If the son’s height is 40 cm, find the amount to be spent by the Eskimo to plaster the walls of his dwelling from inside as well as outside at the rate of Rs. 0.5 per sq. cm. (1) 11728 (2) 2500 (3) 3364 (4) 5864


Page 37 of 54


Pythagoras triplet: r = 50 cm. Now, surface area of hemisphere 

4r 2  2  r2 2

Exterior surface area of igloo = 2  (58)2 Interior surface area of igloo = 2  (50)2 Total surface area = 5864 × 2  = 11728 cm2. Cost of plastering = 11728 × 0.5 = Rs. 5864. (4). 32.

A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the wooden toy. (1) 104 cm3 (2) 162 cm3 (3) 427 cm3 (4) 266 cm2

Sol.

Radius of hemisphere = 4.2 cm. Height of the whole toy = 10.2 cm. Height of hemisphere = radius = 4.2 cm Hight of cone ABC = 10.2 - 4.2 = 6 cm. Total volume of wooden toy = Vol of cone + Vol of hemisphere

33.

Sol.

1 2 1 22 2 22  r 2 h  r 3    4.2  4.2  6    4.2  4.2  4.2 Hence, (4). 3 3 3 7 3 7 A cylinder is filled to half of its volume. It is then tilted so that the level of water coincides with one edge of its bottom and top edge of the opposite side. What is the amount of water spilled in this process? (1) 0 cc (2) 24 cc (3) 16 cc (4) Cannot be determined Let the volume of the cylinder be ‘v’

Initial position. In this positon 


the volume of water is half that

Page 38 of 54


of cylinder. Thus, in both, the cylinder can hold same volume. No water will be spilled. Hence, (1). A poor farmer has to fence his 100 m2 rectangular garden. He has barbed wire of 30 m only. Therefore, he decides to use the compound wall of this house as the 4th fence. Assuming compound wall is very long, find the dimensions of the garden if he can exactly fence his garden using all the stock of barbed wire.

1 (2) 20 × 5 (3) 25 × 4 (4) None of these 3 If l = length and b = breadth of the field. 2 l + b = 30  b = 30 - 2 l. l b = 100 l(30 - 2l) = 100  30 l - 2 l2 = 100 or 2 l2 - 30 l + 100 = 0 l = 10 or l = 5.  Possible dimension are 10 × 10 or 20 × 5. Hence, (2).  ABC is an isosceles triangle (AB = AC) where  (AC) = 10 cm. If  (AO) = 5 cm, find the third side of the  ABC. (1) 12  8

Sol.

35.

Sol.

(1) 3 7 (2) 2 7 (3) 5 7 O is the centroid, and it divides AD in the ratio 2 :1.

(4) 6 7 2

 AO 

2 3 1  AD  AD  AO  7.5 Apollonius theorem: 2  ( AD) 2  2   BC   AB 2  AC 2 3 2 2   2

1  2  (7.5)  2   BC   10 2  10 2 BC = 57. Hence, (3). 2  2

Alternatively. O is the centroid.  AD 

3 AO  7.5 . Since  ABC is isossceles, AD is perpendicular 2

BC 2  AB 2  AD 2 BC  5 7 . Hence, (3). 4 The ratio of the radius of the front tyres of a tractor to its rear tyres is 3 : 5. The rear wheel has a diametr of 1.8 m. Find the total revolutions made by its 4 tyres when the tractor as travelled 2 km. (1) 1946 (2) 2112 (3) 1714 (4) 1887 Rear wheel circumference =  d = 1.8  m. Distance travelled = 2 km  No. of revolution of real 2 bisector to BC.  BD 

36.

Sol.

37.

Sol.

5 wheels   353.8  589.7 . Total revolution of 4 wheels = 2(353.8 + 589.7) = 1887. Hence, (4). 3 A ladder 8 m long is rested against a buidling such that it reaches a point 8 m below the top of that building. At the foot of the ledder, the angle of elevation of the top of the building is 60°. Find the height of the buidling. (1) 10 m (2) 15 m (3) 12 m (4) 11 m AB = height of the building, CD = ladder. CBA = 30° (complimentary angles).


Page 39 of 54


 BDC is isosceles since BD = CD. BCD = 30°. Hence, DCA = 60° -  BCD = 30°.

AD 1  sin 30  AD  8   4 m.  AB = (AD + DB) = 12 m. Hence, (3). CD 2 ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A 

38.

and C respectively. If AC = 5 cm and AD 

Sol.

3 5 cm, find the length of CE. 2

(1) 2 5 cm (2) 2.5 cm (3) 5 cm (4) 4 2 cm 2 2 Since AD and CE are two medians. BD = DC and AE = EB. AB = AC - BC2 = 25 - BC2 .....(1)

2

2 3 5    BD 2  45  BC AB  AD  BD    2  4 4   2

2

2

2 Equating (1) and (2), 25  BC 

.....(2).

45 1  BC 2 4 4

55 3 55  BC2  BC2  . From (1), 4 4 3 2

55 20 Also, BE2 + BC2 = CE2   1 AB   BC2  CE 2 AB  25   2  3 3 5 55  CE 2  CE = 25 cm. Hence, (1). ¼AB2 + BC2 = CE2   3 3 A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top which is oen is 5 cm. It is filled with water upto the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, 1/4 th of water flows out. Find the number of lead shots dropped in the vessel. (1) 50 (2) 100 (3) 150 (4) 200 2

39.

Sol.

4 1 200 3 3 Volume of water = Volume of cone  (5) 2 8  cm . Volume of the lead shot  (0.5) 3 3 3 1 200   Water flown out of vessel  4 3  100 . Hence, (2). Number of lead shots  Volume of one lead shot 4 1 1 1     3 2 2 2


Page 40 of 54


Sol. 41.

The radius OA of the circle in the figure is:

(1) 9 (2) 7 (3) 5 (4) 3 Since BD is diameter of circle: C is a riht angle. Hence, by Pythogoras theorem, BD = 10.  Radius = 5. Hence, (3). A spherical ball is melted and another of radius 1/3rd of the original cast. The remaining is used to form a cylinder whose base radius is 2/3rd the radius of the original ball. What will be the ratio of height of the cylinder to its radius. (1) 2

Sol.

2 9

(2) 2

Volume of ball of radius ' r ' 

8 9

(3) 3

6 7

(4) 3

12 27

4 3 4 r3 r . Volume fo small ball formed      Volume remaining 3 3 27

26 4 2 26 4 26 h 26 8   r 3  h  r    2 . (2). of original volume.    r h  27 9 27 3 9 r 9 9 What is mDCA + mABD in the adjoining figure? for cylinder 

42.

Sol. 43.

(1) 100° (2) 70° (3) 50° (4) 40° In  DBC, 2y - x + x + 90 + 30 = 180  y = 30°. In  ABC, 70 + 30 + 30 + 60 - x = 180  x = 10°.  x + y = 30 + 10 = 40°. Hence, (4). A cone with height 6 cm and base radius 4.5 cm has its base painted with blue colour. What fraction of his surface area is painted? (1)

Sol.

3 8

(2)

3 5

(3)

3 4

(4)

1 2

Slant height  (6) 2  ( 4.5) 2  7.5 . Total surface area =  r l +  r2 =  r(4.5 + 7.5) = 12  r Area of base = Area painted =  r2 = 4.5  r.  Required fraction 


4.5r 3  . Hence, (1). 12r 8

Page 41 of 54


Sol. 45.

What is the radius of a spherical ball in inches which is formed by melting a cylinder of base diameter 8 inches and height 160 inches, if the conversion wastage results in a 10% weight loss? (1) 6 (2) 8 (3) 12 (4) 16

4 3 9 Volume of cylinder =  × 42 × 160. Wastage results in 10% loss  r    4  4  160  3 10 3 r =4×4×4×9×3  r = 12. Hence, (3). Let ‘O’ be an incentre and circumcentre of a  ABC and length of OB is 10 cm. Find the area of the triangle.

75 25 100 3 3 sq. cm 3 sq. cm. sq. cm. (2) (3) (4) 75 3 sq. cm. 4 4 4 As the incentre and the circumcentre are located at a common point the triangle is an equilateral triangle. Hence, the radius of the incircle is 5 cm and the side of an equilateral triangle will be 10 3. (1)

Sol.

 Area of the traingle  46. Sol.

sq. cm. Hence, (4). In the above Q. find the area of the circumcircle which is not enclosed by the triangle. (1) 7 5  (2) 100 - 753 (3) 75 3 - 25 (4) 25 - 75 3 As the incentre and the circumcentre are located at a common point the triangle is an equilateral triangle. Hence, the radius of the incircle is 5 cm and the side of an equilateral triangle will be 10 3.  Area of the traingle 

47.

Sol. 48.

Sol.

3 (10 3 ) 2  75 3 sq. cm.  Area of the circumcircle = (10)2 = 100  4

3 (10 3 ) 2  75 3 sq. cm.  Area of the circumcircle = (10)2 = 100  4

sq. cm. Hence, (2). A rectangular cerpet has an area of 60 sq. ft. The sum of its diagonal and the longer side is five times the shorter side. What is the perimeter of the rectangle? (1) 30 (2) 17 (3) 34 (4) 98

l 2  b 2  l  5b ; Solving l = 12, b = 5; 2(l + b) = 34. Hence, (3). In the figure below, BD is the diameter of a circle with area 1386 kms. Find the cost of leveling the shaded region, given that it costs Rs. 49 per sq. ft. of levelling. (All curved surfaces are semicircles on the diametrs, AB, AD, BC, CD.) lb = 60;

(1) 475000 (2) 49000 (3) 22638 Length of diametr = BD = 42 kms. AB = 14, BC = 28

(4) 45276

14  14 7  7    147  22  49   22638 . (3). Area of the shaded region     2  147 . Cost  2 2 7  


Page 42 of 54


Sol.

50.

Sol. 51.

Sol.

In the following figure are seen two circles, who have a common point A. C is the centre of the larger circle. BD = 9 cm, EF = 5 cm. What are the diameter of the two circles?

(1) 50, 41 (2) 49, 40 (3) 25, 16 (4) 43, 32 Since C is the center of the bigger circle. Let AC = R. Join DE and EA. m  DEA = 90° [Diameter subtends 90° at any point on the circle]. CE = R - 5 and CD = R - 9. We know that (CE)2 = (CD)(CA) [From Similar triangles].  (R - 5)2 = R(R - 9).  R2 - 10R + 25 = R2 - 9R  R = 25.  Diameter of the larger circle would be 50 and that of the smaller circle would be 50 - 9 = 41. Hence, (1). Find the value of LDM in the following figure:

(1) 54° (2) 36° (3) 34° (4) 40° Let LDM = x;MAD = 180 - 32 - x = 148 - x; BCD = 180 - (148 - x) = 32 + x 40 + 32 + x + x = 180.  x = 54. Hence, (1). Find the value of x in the following figure?

(1) a + b + c (2) b - a - c (3) 180 - c (4) None of above In CDB b + y + z = 180°  y + z = 180° - b. In  ADB  a + x + y + c + z = 180°  a + c + x + 180° - b = 180°  x = b - a - c. Hence, (2).


Page 43 of 54


Sol. 53.

Sol. 54.

Sol. 55.

Sol. 56.

Sol.

In the figure L1 is parallel to L2. If AB = CD, then

(1) ABE is an equilateral triangle (2) AC = BC (3) AD = BC (4) AC = AD AB || CD and AB = CD. Then ABCD is a parallelogram and AC = BD. Hence, (2). Line BC divides  ADE into 2 sections, one of them an isosceles triangle (AB = AC). Angle DBC is equal to 105°. What is the sum of the measure of angles D and E?

(1) 100° (2) 125° (3) 150° (4) 175° AB = AC B = C = 75° A = 30°, A + D + E = 180°D + E = 150°. Hence, (3). The radius OA of the circle in the figure is:

(1) 9 (2) 7 (3) 5 (4) 3 Since BD is diameter of circle C is a right angle. Hence by Phythogoras theorem, BD = 10  Radius = 5. Hence, (3). A cone is cut 1 cm from the top, parallel to its base. The base area of the cut out piece is 20 cm2 and the base area of the cone is 80 cm2. Find what height from the base of the cone, was the portion cut? (1) 7 cms (2) 1 cm (3) 3.7 cms (4) Can’t get Ans. (2). Diagonals of a cyclic quadrialteral ABCD intersect in P and the area of  APB is 24 cm2 if AB = 8 cm and CD = 5 cm, calculate the area of the  DPC. (1) 24 cm2 (2) 75/8 cm2 (3) 48 cm 2 (4) 75 cm2 A1 S12 82 24(52 ) 75    A    APB is similar to  DPC . Hence, (2). 2 A2 S 22 52 8 82


Page 44 of 54


Sol.

5 concentric circles are drawn whose circumferences are in the ratio 2 : 3 : 7 : 9 : 13. If a line is drawn joining the centre to a point on the outermost circle., what is the ratio of the length of the line to the radius of the 2nd innermost circle? (1) 13 : 3 (2) 13 : 5 (3) 11 : 3 (4) Can’t get Radii are in the ratio 2 : 3 : 7 : 9 : 13  Radii are 2r, 3r, 7r, 9r, 13r

13r  13 : 3 . Hence, (1). 3 A cricket pitch has to be flattened by 2 rollers, of equal length but with the bigger roller having a radius 7/3 times that of the smaller roller. If the bigger roller takes 6 minutes to do the job, how much time will the smaller roller take? (1) 14 min (2) 16 1/3 min (4) 2 4/7 min (4) 21 min Let bigger roller have radius ‘r’, length ‘l’  Area covered per roll = 2  r l. For smaller roller,  Desired ratio 

58.

Sol.

59.

Sol. 60.

3 7  2  r  Smaller roller takes 7/3 times time taken by bigger roller.  6   14 minutes. 7 3 A square PXCM is rotated on a plane anticlockwise about its centre such that ‘C’ moves to the position origianally occupied by ‘X’. By how many degree will the square have rotated if PXCM are the four vertices named in anticlockwise manner? (1) 90° (2) 180° (3) 270° (4) 315° When it rotates 90° ‘C’ moves to ‘M’. When it rotates 180° ‘C’ moves to ‘P’. When it rotates 270°, ‘C’ moves to ‘X’. Hence, (3). O is the origin and AB is the diameter of the semicircle that is shown shaded. Find the area of the semicircle.

(1) 21.13

(2) 169

(3) 204.28

(4) 84.5

Sol.

OA = 12; OB = 5

1  13  169  21.13 . Hence, (1).  AB = 13  Area        2 8 2

61.

From a cube of side 9 cms, how many cubes can be formed of size 45 mm.? (1) 4 (2) 8 (3) 16 (4) 32

Sol.

Number of cubes 

2

62.

9  9  9  10  10  10  8 cubes. Hence, (2). 45  45  45 In a trapezium, the diagonals intersect at point O. The ratio of the length of one of the diagonals from one vertex of the trapezium to thepoint O to its entire length is 2:5. Find the ratio of its parallel sides (smaller side : larger side) (1) 2 : 5 (2) 2 : 3 (3) 2 : 7 (4) 5 : 7


Page 45 of 54


Given DO : DB = 2 : 5

DO 2 DO 2 2 DO 2 AB    .  AOB is similar to  COD    By dividing . DB 5 DB  DO 5  2 3 OB 3 DC Hence, (2). If the side of a cube is twice the radius of a sphere, which of the following is true? I. The cube completely fits into the sphere. II. The sphree completely fits into the cube. (1) I only (2) II only (3) Both I and II (4) Neither I nor II Since

63.

Sol.

64.

Sol.

If a = 2r, then the sphere completely fits into the cube. Hence, (2).

A cylinder of base radius 5 cm and height 58 cm is inclined in such a manner that the smallest vertical distance from the ground to its top surface is 29 cms. Find the angle of inclinations. (1) 15° (2) 30° (3) 45° (4) 60° Midpoint theorem

29 1   x = 30°. Hence, (2). 58 2 A point on the largest side of a triangle is equidistant from the vertices of that side. A line passing through this point, and parallel to the smallest side intersects the longest and second longest sides in two points which are 17 cms apart. Find the length of the smallest side. (1) 8.5 cm (2) 34 cm (3) 51 cm (4) Cant’ get Midpoint theorem sin x 

65.

Sol.

 ADE   ABC 

AD DE 1    BC = 17 × 2 = 34 cms. Hence, (2). AB BC 2


Page 46 of 54


A sphere is melted and half the liquid is used to form 11 identical cubes. Whereas the remaining half is used to form 7 equal smaller sphere. The ratio of the side of the cube to the radius of the new small spheres is: (1) 1 : 33 (2) 2 : 33 (3) 2 33 (4) 33 : 2 : 32

Sol.

Volume 

3 1/ 3 a 8 2 21 3 3 a  8      3 Hence, (2). a  14r1  r = 14 r1 ....(ii) 3 r1 3 r1  3  4 3 The difference in the circumferences of 2 circles is a multiple of 6. Which of the following statement is/are true? i. If the larger circle has a radius that is a multiple of 3, then the samller circle cannot have a radius that is a multiple of 3. ii. If the smaller circle has a radius that is a multiple of 3, then the larger circle cannot have a raidus that is a multipel of 3. iii. The smaller circle cannot have a radius that is half the radius of the larger circle. (1) i and ii only (2) ii and iii only (3) iii only (4) None of these Let radii be r1 and r2 (r1 > r2)  2  r1 - 2p r2 = 2 (r1 - r2). We already have a factor 2, so r1 - r2 is a multiple of 3 i. is False, If r1 is a multiple of 3, r2 should also be a multiple of 3. ii. is False, If r2 is a multiple of 3, r1 should also be a multiple of 3. iii. is False, r1 = 2r2  r1 - r2 = 2r2 - r2 = r2 and if r2 = 3 or multiple of 3 then the condition is satisfied. Hence, (4). A palm tree swings with the breeze in such a manner that the angle covered by its trunk is 14°. If the topmost point of the tree covers a distance of 33 meters, find the length of the tree. (1) 20 mts. (2) 99 mts. (3) 135 mts. (4) 125 mts. 3

67.

Sol.

68.

Sol.

4 3 2 3 3 7 21 3 2 4 r  r  11a 3  r 3  11a 3    a ...(i). Similarly,, r 3  7  r13 3 3 2 22 4 3 3 3

 Length of the arc  2    r 

 where ‘r’ will be the length of the tree and  will be the angle 360

69.

22 14 33  360  7 r  33 r   135 metres Hence, (3). 7 360 14  2  22 A spherical balloon is blown until its diameter reaches 10 ft. Then it is continued to be blown until the diameter reaches 12 ft. By what proportion has the 2nd volume increaesd as compared to the first volume? (1) 60% (2) 60.8% (3) 72.8% (4) 80.6%

Sol.

Volume of a sphere  (radius)3 Radii are in ratio 6 : 5  Ratio of volumes 

made by its trunk  2 

666  1.728 555

 % increase = (1.728 - 1) × 100 = 72.8% Hence, (3).


Page 47 of 54


Sol. 71.

Sol.

72.

In the diagram below, what is BOD?

Given ODC = 30°, AOC = 50°. ‘O’ is the centre of the circle. (1) 10° (2) 45° (3) 85° (4) 112.5° DOC = 180 - 50 = 130°. DCA = 180 - (30 + 130) = 20°. OBC = 20° BOC = 180 - 40 = 140° BOD = 140 - 130 = 10°. Hence, (1). In the figure alongside are two concentric circles, with a tangent to the inner circle forming a chord to the outer one. The length of the chord is 8 cm. If the radii of both circles are integers, what is the radius of the inner circle?

(1) 3 cm. (2) 4 cm. (3) 5 cm. (4) 6 cm. The radius of the inner circle makes 90° angle with the tangent and cuts it in half. The only rightangled triangle with 4 as one side has 3 and 5 as the other sides. So the smaller circle’s radius is 3 cm. (As all the sides are integres). Hence, (1). The figure below is an equilateral triangle with incircle and circumcircle. Find the (area of A) : (area of B) : (area of C).

(1)

3 3   1 4  3 3 : : 27 27 27

(3) 3 3 -  :  : 4 - 3 3 Sol.

a - altitude of the triangle.

(2)

3 3    4  3 3 : : 27 9 27

(4) Data Insufficient

a 2a ;  radius of incircle and circumcircle respectively.. 3 3

Side of an equilateral triangle 

2 3

a . Area of equilateral triangle 

3 2 a . Area of the circumcircle 3

1 2 4  a 2 . Area of the incircle  a . Area of C = 1/3 (area of circumcircle - area of the triangle) 9 9 1 4 3 3 2  4  3 3 2   a 2  a  a Area of A = 1/3 (area of triangle - area of incircle) 3 9 9 27 


Page 48 of 54


73.

Sol. 74.

Sol.

1  3 3 2 1 2  3 3   2   a  a   a  Area of A : Area of B : Arae of C  3 3   :  : 4  3 3 . 3 9 9 27 27 9 27  What is the arae of the largest triangle that can be fitted into a rectangle of length l and breadth b? (1) lb/3 (2) lb/2 (3) 2lb/3 (4) 3lb/4 The maximum area of the triangle = lb/2 . Hence, (2). What is the ratio of the area of an equilateral triangle to that of a circle circumscribing it? (1) 1 : 3 3 (2) 3 3 : 4 (3) 3 3 : 4 (4) Depends on the side of the triangle

Let side of triangle = a; so area of triangle 

2 3 a 3 2 a a . Radius of circle   . Area of 3 2 3 4

a 2 3  : i.e., 33 : 4. Hence, (3). . So, Ratio  3 4 3 A circular park has sum its area and perimeter equal to 8. The diagonal of the park equals: (1) 4 (2) 8 (3) 2 (4)  A circular park has sum of its Area + Perimeter = 8  r2 + 2  r = 8 where r = radius r2 + 2r - 8 = 0  r = 2 or r = -4 but radius cannot be negative  r = 2  d = 4 Hence, (1). Two circles x and y with centre A and B intersect at C and D. Area of circle X is 4 times area of circle Y. Then AB = ?

circle  75. Sol. 76.

5 r 2 ACB = 90°. Angle at the point of intersection to the centre of the circles. BC = r. AC = 2r (as area

(1) 5r Sol.

77.

Sol.

(2) 5 r

(3) 3r

(4)

of X = 4 area of Y)  AB  r 2  4r 2  5r . Hence, (2). From a circular paper a man makes two conical caps. The surface area of the two are in the ratio 2 : 1. He then covers the face of the caps with other circular pieces of paper. The ratio of the area of these pieces is: (1) 1 : 2 (2) 4 : 1 (3) 1 : 1 (4) 2 : 1

Surface Area A 2 1l1   Surface Area B 1  2l2 Where r1, r2  radii of bases l1, l2  slant heights. Since both are from r1 2 r 2 4 same circular piece of paper l1 = l2  r  1  Ratio of area of bases  12  . Hence, (2). 2 r2 1


Page 49 of 54


OA = R. What is the ratio of area of circle X and Y?

(1) 2 : 1

(2) 4 : 1 R

Sol.

OA = R  OC 

79.

What is the area of  PSR?

(1) 20

2

( RtOAC )  OB 

(3) 3 : 1

(4) 8 : 1

Area of circle A R 2 4 R   . ( RtOCB)  2 Area of circle B 1 2 R   2

(2) 24

(3) 40

(4) None of these

Sol.

PR = 10, RS = 8. PS  10 2  82  6 . Area = 1/2 × 8 × 6 = 24 sq. cm. Hence, (2).

80.

AG 

3 AC . What is the ratio of the area of ECG to that of square ABCD? 8

(1) 3 : 32 (2) 15 : 128 (3) 5 : 64 (4) None of these

Sol.

3 2 x . For squares AHGE and ABCD, diagonals are in the ratio 8 3 : 8  sides are in ratio 3 : 8. Let sides be 3x and 8x  Area of ABCD = 8x × 8x = 64x2 ......(i) In  ECG, with EG as the base, the length ED will be the height  base = 3x, height = 8x - 3x = 5x Let AB = x.  AC = 2 x  AG 

 Area 

1 15 x 2  3x  5 x  .....(ii)  Ratio of Areas of  ECG to ABCD is 2 2

15 x 2 : 64 x 2  15 : 128 . Hence, (2). 2


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Sol.

82.

Sol.

83.

Sol. 84.

Sol. 89.

What is the angle in degrees made by a sector the ratio of whose area with the area of the semicircle is equal to 1 : 10? (1) 36 (2) 18 (3) 24 (4) 9 We know that area of the sector/area of the circle = angle of the sector/360. Thus we get 1/10 = angle of the sector / 180 (as the second curve is a semi-circle). Thus the angle of the sector = 18 degrees. Hence, (2). The length of the spiral string below can be

(1) 12  a (2) 14  a (3) 18  a (4) 20  a Consider arc ABC. ABC > 2  a and ABC > 2  (2a). Greater than circle of radius ‘a’ less than circle of radius ‘2a’.  2  a < ABC < 2  a. Similarly, 2  (2a) < CDE < 2  (3a) 2  (3a) < EFG < 2  (4a). Adding, 2  (6a) < ABC + CDE + EFG < 2  (9a) Only one choice 14  a lies between 12  a and 18  a. Hence, (2). The sum of the perpendicular sides in a right angled triangle is 18. Find the length of the hypotenuse when the area of the triangle is maximum. (1) 10 2 (2) 9 2 (3) 164 (4) None of these Maximum area at a = b = 9. Hypotenuse = 9 2. Hence, (2). Find 1 + 2, if x = 60°.

(1) 60° (2) 120° (3) 150° (4) 90° For a cyclic quadrilateral opposite angles add up to 180.  180 - (1 + 2) + 180 - 60 = 180  1 + 2 = 180 - 60 = 120° Hence, (2) The heights of two right triangles are in the ratio 1 : 2 the bases are in the ratio 2 : 1 and the ratio of smaller to the longer side of the triangles other than the hypotenuses are in the ratio 2 : 1 then: (1) the hypotenuse are equal (2) the areas are equal (3) the triangles are congruent (4) all of these


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Given,

90.

PQ = 2AB, BC = 2QR, and PQ = 2QR  BC = 2AB  PQ = BC and AB = QR   ABC  RQP. Hence, (4). The perimeter of a right angled triangle is four times the shortest side. The ratio of other two sides is: (1) 5 : 6 (2) 3 : 4 (3) 4 : 5 (4) 2 : 3

Sol.

We know that a + b + x = 4x {x < a < b}  a + b = 2x .....(I) and a2 + x2 = b2 i.e., a2 - b2 = -x2 3x(a - b) = -x 2  a  b 

 (a + b) (a - b) = -x 2

x  x2 ba 3 3x

.....(II)

4x 5x , and b  . Hence, the required ratio = 4/5. Hence, (3). 3 3 A circle passes through the vertex of an equilateral triangle and is tangent to the opposite side at the midpoint. What is the ratio in which the circle cuts the other side? Solving (I) and (II), we get a 

91.

(1)

1 1

DC2

a a a2  4a  x  = CE × CA     x  a  4 4 2

(2)

3 2

(3)

2 3

(4)

3 1

2

Sol.

3a ax 3 . Hence, the required ratio is , which is . Hence, (4). 4 x 1 ABCD is a rectangle such that AC + AB = 5AD and AC - AD = 8. What is the area of the rectangle? (1) 60 (2) 64 (3) 65 (4) 72 Therefore, a  x 

92.


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93.

AC + AB = 5AD. AC - AD = 8  AC = 8 + AD  8 + AD + AB = 5AD. 4AD = 8 + AB  AB = 4AD - 8. Now, AC2 = AB2 + BC2 = AB2 + AD2. (8 + AD2) = (4AD - 8)2 + AD2. 64 + AD2 + 16AD = 16AD2 + 64 - 64AD + AD2 0 = 16AD2 - 80AD  0 = AD - 5  AD = 5  AB = 12  AC = 13  Area = AB × AD = 12 × 5 = 60. Hence, (1). In the figure, the area of parallelogram ABCD is 24. What is the area of  BFE, if CE = 2BC and AB = BF?

(1) 24 Sol.

(2) 72

(3) 48

24 AB = BF. CE = 2BC. Area of parallelogram ABCD = base × height. AB × h1 = 24  AB  h 1 Area of BFE 



94.

Sol.

(4) 36

1 1  h2  BF   h2  AB 2 2

h h 1 h 1 24 x h1  h2   12 2 . BHC and BGE are similar triangles.    1   2 3 2 h1 h1 3x h2 h2 3 h1

 Area of BDE = 12 × 3 = 36. Hence, (2). PQ = PR and QS = TR. Which of the following is not necessarily true?

(1) SPT + PST = SPT + PTS (2) QPS = RPT (3)  PRS   PQT (4)  PST is an equilateral triangle. Clearly (1) and (2) are true PRS and PQT by SAS test of congruence. Hence, (3) is true. PST is an isosceles triangle but not necessarily an equilateral triangle. Hence, (4).


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The medians of a triangle ABC intersect the sides AB and AC at D and E. CE and BD interesect at M. What is the ratio of the area of ADME to the area of ABC? (1)

Sol.

2 3

(2)

1 3

Area of BDC = Area of BEC 

(3)

1 4

(4)

1 2

1 A(ABC ) (Median divides triangle into equal areas). 2

1 Also, in BEC, area of BEM = 1/3 area of BEC (Same altitude and EM  EC ). 3 11  1  Hence, A(BEM )   AABC    A(ABC )  . Thus, A ( 3 2  6 

96.

ADME)

1  1  1  = A(ABC) - A(BEM) - A(BDC)  A(ABC )   AABC    AABC    AABC  6  2  3  Hence, (2). Given the triangle ABC and its side a, b, c then the length of the median mb which is a median on side b is (1) (a2 + b + 2 - c2)1/2 (3) (2a2 + 2c2 - 2b2)1/2

1 (2a 2  2c 2  b 2 )1 / 2 2 (4) (a2 + c2 - 2b2)1/2 (2)

Sol.

Take a point D such that AD = CD, thus ABCD is parallelogram since the diagonals bisect each other. And we know in parallelogram sum of squares of diagonal is equl to sum of square of sides.

1 Thus (2mb)2 + b2 = 2(a2 + c2), thus mb  (2a 2  2c 2  b 2 )1 / 2 . Hence, (2). 2


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Geometry Problems

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