Geometry Problems

Prepared ByHarshwardhan Phatak Page 1 of 54

Quant Funda Class (Geometry)

a dcb

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

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

(b) Theorems :

If AOB is a straight line, thena + b = 2 right angles(Adjacent angles on a straight line)

If a + b = 2 right angles, thenAOB is a straight line.(Adjacent angles are supplementary)

The sum of all the angle at a point, eachbeing 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, thevertically opposite angles are equal.a = b, c = d(Vertically Opposite angles)

ba

ba

adbc

deab

c

Plane Geometry



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

1 . The corresponding angles are equal c = b(Corresponding angles, PQ || RS)

2. The alternate angles are equal a = b(Alternate angles, PQ || RS)

3. The interior angles are supplementaryb + d = 2 right angles(Interior angles, PQ || RS)

If PQ and RS are cut by transversal LM, thetwo lines are parallel ifa = 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 180, are supplementary. Each one is the supplement of the other.

TRIANGLESPROPERTIES :

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 c < a + b, Triangle is acute angled (ii) if c = a + b, Triangle is right angled(iii) if c > a + b, Triangle is obtuse angled

IMPORTANT POINTS WITH RESPECT TO A TRIANGLE :

a. Centroid :When a vertex of a triangle is joined to the midpointof the opposite side, we get a median. The point ofintersection of the medians is called the CENTROIDof the triangle. The centroid divides any median inthe ratio 2 : 1.

P Q

R S

L

M

ca d

b

A

B CD E

G

A

B CD


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 = CDWe have the formula :

2 x (Median) + 2 x ( Third side) = Sum of squares of other two sides 2 x (AD) + 2 x ( BC) = AB + AC.Area of triangle ABD = Area of triangle ADC. If BD = DE = EC then the areas of trianglesABD, ADE, AEC are equal. D & E are called the points of trisection of BC.

b. Circumcentre :The point at which the perpendicular bisectors ofthe sides of a triangle meet is the circumcentre ofthe triangle. The circumcentre S of a triangle isequidistant 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 ABCAny point on the perpendicular bisector of a line isalways equidistant from the ends of the line.

c. Incentre :The point of intersection of the angle bisectors ofa triangle is called the incentre I. The perpendiculardistance of I to any one side is inradius and thecircle with centre I and radius equal to inradius iscalled the incircle of the triangle. The three sidesare tangent to the incircle. Any point on the bisectorof an angle is equidistant from the arms of theangle. The incentre divides the bisector of A inthe ratio (b + c) : a.

Also BIC = 90 + A/2, AIB = 90 + C/2, AIC = 90 + B/2.The bisectors of two exterior angles at B & C andthe bisector of A meet at a point called excenterI. There can be three excircles of a triangle.Also IE = Exradius. There can be three exradiito a triangle. BIC = 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 = AD

A

B C

S

.I'

B

I

C

c b

a D

r

A

A

B CD


Quant Funda Class (Geometry)(b) In the figure

If AD is the bisector of exterior angle at A of triangleABC, then : AB/AC = BD/DC

(c) In the figure, if in triangle ABC, AN is thebisector of BAC and AM is perpendicular toBC then MAN = ( B C).

(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 calledorthocentre of the triangle.

Also BOC = 180 A AOB = 180 C AOC = 180 B

Area of triangle = x Base x Altitude.

NOTE : There is similarity between the topic on geometry & that on mensuration. You arerequested to go through this topic only after thoroughly grasping the chapter on mensuration asformulae and facts discussed in that chapter are not repeated here.

Some Important Points :

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

2. 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 = x hypotenuse = circumradius

AE

B DC

A

B CM N

A

B CO

D

CB

A

AB=AC



A

B C

D E

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X2

Y1B

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CA

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Y2

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CB D90o

90o

GENERAL THEOREMS ON SIMILARITY

1. Proportionality Theorem :Intercepts made by two transversal lines(cutting lines) on three or more parallellines are proportional.In the figure, lines X1Y1 & X2Y2 aretransversals cutting the three parallel linesAB, CD, EF. Then AC, CE, BD, DF areintercepts Also, AC/BD = CE/DF

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 atriangle divides the other two sidesproportionally. 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.

PROPERTIES OF SIMILAR TRIANGLES :

(1) If two triangles are similar, thenRatio of sides = Ratio of heights = Ratio of medians = Ratio of angle bisectors = Ratio of inradii = Ratio of circumradiiNote : 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(3) RIGHT TRIANGLE :

ABC is a Right Triangle with A as the Right angle.AD is perpendicular to BC then(a) Triangle ABD ~ Triangle CBA & BA = BC x BD(b) Triangle ACD ~ Triangle BCA & CA = CB x CD(c) Triangle ABD ~ Triangle CAD & DA = DB x DC.



QUADRILATERALSProperties :

(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 interiorangle.

(c) The figure formed by joining the mid points of the adjacent sides of a quadrilateral is aparallelogram.

(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) Opposite angles are equal(b) Diagonals bisect each other (not at 90)(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 Midpoints 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.(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.

RECTANGLE :(a) Opposite sides equal, each angle = 90(b) Diagonals bisect each other (not at 90).(c) Of all rectangles of given perimeter, a square has max. area(d) When inscribed in a circle, it will have maximum area when its 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 PA + PC = PB + PD(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 thediagonal of the rectangle.

a

b

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b

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b



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.

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.(e) In the figure ABCD is a trapezium,

then AC + BD = BC + AD + 2 AB.CD

(f) In the above figure if X & Y are mid points of diagonals,then XY = (AB CD).

CYCLIC QUADRILATERAL :

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 cyclicquadrilateral 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

a

aa

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a

a

aa

a

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D C

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xC

yD

BA

x+y=180o



CIRCLES

(a) Tangent is perpendicular to radius.(b) Perpendicular from centre to a chord

bisects the chord.(c) Tangents drawn from an external point

are equal.(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.(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.(l) Measure of an arc means measure of central angle.(m) m(minor arc) + m(major arc) = 360(n) Angle at the centre made by an arc

= twice the angle made by the arc at any point on theremaining part of the circumference.We have APB = AOB = 30 = AQB

(o) Angle in a semicircle is a right angle.(p) Alternate Segment Theorem : In the fig.

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)

P

B

A

O

P Q

O

A B

60o

B A C

P D



SOME MORE RESULTSA

B CD

90o90o

A B

C

DA

B C

P Q

4. (a) If triangle ABC is an acute angled triangle thenAB = BC + AC 2 BC x CD

(b) If A is obtuse, thenBC = AB + AC + 2AB x AD

(c) If in triangle ABC, PQ is parallel to BCsuch that area of triangle APQ = Area oftrapezium PQBC; then AP/PB = 1/(2 1)

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

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,

(represented as ABC XYZ) then AB = XY, AC = XZ, BC = YZ & A = X, B = Y, C = Z.

(c) Theorems :

* Triangle ABC Triangle XYZif AB = XY A = X AC = XZ(SIDE ANGLE SIDE) S.A.S.

* Triangle ABC Triangle XYZif B = Y C = Z AC = XZ(ANGLE ANGLE SIDE) A.A.S.

X Y

ZA

B C

A

B

X

Y ZC

A

B C

X

Y Z



* Triangle ABC Triangle XYZif B = Y C = Z BC = YZ(ANGLE SIDE ANGLE) A.S.A.

* Triangle ABC Triangle XYZif AB = XY AC = XZ BC = YZ(SIDE SIDE SIDE) S.S.S.

* Triangle ABC Triangle XYZif B = Y = 90Hypotenuses AC = XZ,AB = XY(RIGHT ANGLE HYPOTENUSE SIDE) R.H.S.

* 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)

* 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)

2. 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, theircorresponding sides are proportional.In triangles ABC and XYZ,if A = X, B = Y, C = Zthen AB/XY = AC/XZ = BC/YZ(EQUIANGULAR TRIANGLES) A.A.A.

* If 2 triangles have their correspondingsides proportional, they are equiangular.In triangles ABC and XYZ,if AB/XY = AC/XZ = BC/YZthen A = X, B = Y, C = Z(3 SIDES PROPORTIONAL)

XA

B C Y Z

A

B C

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Y Z

XA

B C Y Z

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B C

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B C

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B C

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Y Z

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B C

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Y Z

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DP

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AB

T

X Y

PA

ByxD

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r1B r2

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C

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Or P r'

O' M

B

D

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

(d) If triangle ABC & Triangle PQR are similar then it is denoted by ABC ~ PQR.

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

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

5. If chords AB & CD intersect internally at P, then

(a) PA x PB = PC x PD

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

6 . Common TangentsFor the two circle with centers A & B, PQ is a direct common tangent and RS is a transversecommon tangent.

(a) Length of direct common tangent

= (Distance between centres) (r r )1 2

(b) Length of transverse common tangent

= (Distance between centres) (r + r )1 2

(c) In this figure1. OM /OM = r/r2. OP/OP = r/r



EXERCISE

1. Find the value of x in the figure given

2. Find the values of x, y in the figure given

3. A chord of length 10 cm subtends an angle120 at the centre of a circle. Calculate its distance fromthe centre

4. 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 = ?


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. In the above figure, A = 90, AD is perpendicular to BC and B = 45. If AB = x find AD in termsof x.

12. In a trapezium ABCD, AB || DC, AB = 2CD. If the diagonals intersect at O, show that area of AOB= 4 area of COD.

13. ABCD is a trapezium with AB ||CD. If AC and BD intersect at E such that BE DE = 2 : 3 show thatarea of the triangles 9

4

CEDAEB

.

14. Fin the length of a chord, which is at a distance of 4 cms from the centre of a cricle whose radius is 5cms.

15. 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

16. ABC is an equilateral triangle inscribed in a circle of radius r. What is the length of the side of thetriangle?

17. O is the centre of a circle of radius 5cms. P is any point in the circle such that OP = 3 cms. A is thepoint travelling along the circumference. X is the distance from A to P. What is the least and greatestvalues of x in cms?

18. ABCD is cyclic quadrilateral. A circle passing through A and B meets AD and BC at E and Frespectively. Is EF parallel to DC?

19. O is the centre of the circle, PA and PB are tangnets to the circle from point P. If PAB = 70, findPBA and APB.

20. 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.


Quant Funda Class (Geometry)21. 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?22. 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.23. 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?24. 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.25. ABCD is a quadrilateral. P, Q, R, S are mid points of the sides AB, BC, CD, AD respectively. Is

PQRS a parallelogram?26. 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 = ?27. ABC is a triangle with AB = AC. D is the midpoint of BC. What is the ratio of DAB to DAC?28. 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?29. 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?30. 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

31. The diagonal BD of a parallelogram ABCD intersects the segment AE at a point F where E is anypoint on the side BC. Then (DE . EF) : (FB . FA) = ?

32. In ABC, DE is parallel to BC. If AD = 5 cm, AE = 6 cm, BC = 12 cm, AB = 15 cm. Find AC and DE.33. ABC and DBC are two right angled triangles with common hypotenuse BC. The sides AC and BD

extended intersect at P, Then ?.

. PBDPPCAP

34. ABCD is a quadrilateral inscribed in a circle and the diagonal BD bisects AD. Prove that AD . AB =CD . CB.

35. Triangle ABC and DEF are similar. If AB and PQ are 7.5cms and 3.5 cms respectively,find the ratioof the areas of the two triangles.

37. The medians BE and CF of the ABC intersect at P. FInd the ratio of the area of BPC to that ofFPE.

38. Three circles with equal radii touch one another externally. Prove that the triangle formed by joiningtheir centres in an equilateral triangle.

39. CD is a tangent at C to the circumcircle of ABC intersecting AB produced in D. Prove that DBC~ DCA.

41. Two circles touch each other externally at P. APC and BPD are drawn through P to meet the wtocircles in A, B and C, D respectively. Prove that(i) PAB ~ PCD (ii) AB || CD

42. Six cubes each with 12 cm edge are joined end to end. Find the surface area of the resulting cuboid.43. 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.44. 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.45. The hieght of a right circular cone is 40 cm and the radius of the base is 28 cm. Find the area of cross-

section of the cone formed by a plane parallel to the base of the cone and at a distance of 10 cm fromthe vertex of the cone.


Quant Funda Class (Geometry)46. 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.47. 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 thecylinder, hemispheer and the cone.

48. A right circular cylinder and a right cone have equal bases and equal height. If their curbed surfacesare in the ratio 8:5, what is the ratio of the radius of their base to their height.

49. A spherical ball of lead has been melted and made into smaller balls of half of the radius of the originalone. How many such balls can be made? Find the ratio of the total surface area of all the smaller ballsto that of the original one.

50. If successive squares are formed by joining the mid points of the sides of a parent square and theprocess 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 orx = 108

2.

2 = 90 (alternate angles). = x (corresponding angles). 1 + 2 = 3x + 10. x + 90 = 3x + 102x = 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

the mid point of AB. AOD = 60. 355360tan OD

ODODAD

4. Circum radius of an equilateral triangle = a / 3 (a is the side of the triangle) so a = 6 35. DAC = B + C (external angle = sum of the interior opposite angles) 130 = 2x + 3x

x = 26 B = 52; C = 786. In PAB and QAC are congruent since, PA + QA; BA = CA; and PAB = QAC. Hence,

PB = QC (corresponding sides of congruent triangles)


Quant Funda Class (Geometry)7.

In ABC, AD = AC. so, DAC = AD = AC, so, D = DCA In BDC, D + C = 1802x + 2y = 180 x + y = 90 BCD = 90Altnernatively: 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.

8. Since AB = AC, B = C. In XBC and YBC, BX = CY and XBC = YCB. BC is the commonside, Hence XBC and YCB are congruent. So, XC = YB

9. D = B; E = C and A is common to both the triangles. So, ADE and ABC are similar.10. B = C = 45 AD = BD = CD = AC11. If B = 45 AD = BD = CD = AC Hence, AD = x / 212.

AOB and COD are similar. The ratio of the proportional sides is 1 : 2. Hence the ratio fo the areasof 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

therefore, 94

CED)(AreaAEB)(Area

14. 6 cms. 15. 1 cm, 7 cms. 16. R = a / 3 or a = R317. 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 straightline and P and A lie on the opposite sides of O.


Quant Funda Class (Geometry)18. 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 = 4020.

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 chordBC. 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 arecongruent. so BPA = CPD.



22.

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, ACAG

ADAE

Also GF || AB, CAAG

CBBF

. So, 21

ADAE

CBBF

. So, F is the mid pt. of BC

25.

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



26.

AB = AC. AD is the median. So ADB = 90, PDA = 45 ( DP is angle bisector). Now, DQ isalso 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.

If ABC is an isoceles triangle with AB = AC. Then the point D divides AC in the ratio AB/BC and Edivides AB in the ratio AC/BC. Since D, E divide sides in the same ratio we find that ED is parallel toBC. So EDB and DEC are congruent. Therefore BDE = CED. OED is isosceles triangle.

29. 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 commonto both. Hence, ADC - BEC

(ii) From (i) CECACDCBCDCE

CACB

.. .



31.

s AFD and EFB are similar. SoFDDF

EFFA

or FA . FB = DF . EF

32.

Since DE || BC cm185156 AC

BCDE

ACAE

ABAD

and .412155

cmDE 33.

Triangles ABP and DCP are similar so,PBAP

PCPD

34. AD . AB = CD . CB.

35. since ABC ~ DEF 49100

DEF)(AreaABC)(Area

2

2

2

2

2

2

EFBC

DFAC

DEAB

710

EFBC

DFAC

DEAB

7102.8

DE DE = 5.74

36.

ABC ~ PQR 49225

715

5.35.7

PQR)(AreaABC)(Area 22

2

2

PQAB



37.

Point P divides the median in the ratio 2 : 1. So, 12

PEBP

PFCP

. And FPE = BPC (vertically opp.

s) so FPE ~ CPB. Hence, 41

CPPF

CPB)(AreaFPE)Area(

2

2

Reqd. ratio = 4 : 1

38.

In a circle AP . PB = DP . PC [Refer Properties]. Since AP = CP we have PB = DPAB = 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 segmentTheorem). 4 is common to both the triangles. Since 2 angles are equal the third angle must also beequal by AAA property triangles BDC and CDA are similar.

41.


Quant Funda Class (Geometry)(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

42. If l, b, h are length, breadth, height of cuboid. l = 12 6 = 72 cm. b = 12 cm. h = 12 cmHence surface area = 2 (lb + bh + lh) = 2(864 + 144 + 864 = 2 1872 = 3744 cm2.

43. Outside Volume of box = 42 30 20 cm3 = 25200 cm3. Inner volume of the box = 40 28 18 cm3Volume of wood = 25200 - 20160 = 5040 cm3

44. Volume of water flowing every minute = 1000 cm (.5)2 cubic cms = 250 cu cms.Volume 222 cm320024.)20.(3

131 hr

. Time minutes12.8utesmin25024400

31

45.

BE | | CD CDBE

ADAE

ACAB

So, cm7401028 BE

Area of cross section = (r)2 = 54 cm2.

46.

CDBE

ACAB

. 31

155

24AB

AB. AB = 12. CSA (curved surface area) of cone = r l = 15 39 =

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.

Vcylinder : Vhemisphere : Vcone hrrhr232

31

:32

: 33331

:32

: rrr = 3 : 2 :1

48. For cylinder let radius and the height be R, H respectively. For cone they are r, h respectively. CurvedSurface Area for cylinder : cone = 8 : 5 5:8hrr:RH2 22 But R = r and H = h

582

22

HRR

RHRH

HRH 43

2516

22

2 or 4

3HR

.

49. Original volume 3r34

. Volume of new spheres 834

234 33 rr

. 14

44

4smallTSAbigTSA

2

2

r

r


Quant Funda Class (Geometry) Required ratio = 1 : 4.

50. Twice the area of the parent square (by infinite GP sum formula)

Exercise1. 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 = 344.5sq. 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 perimeteris(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 parallelto each other. If the area of the triangle ABP is 4 sq. units, the area of the triangle ECP is

(a) 2 sq. units (b) 3 sq. units (c) 3/2 sq. units (d) 1 sq. units

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

(a) )(21 CB (b) )(2

1 AC (c) )(21 BA (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(c) 1/2 (AB2 + BC2 + CD2 + DA2) (d) none of these

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


Quant Funda Class (Geometry)9. 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 thefollowing is not true?(a) Angle ACD = Angle BCD (b) BE = AF(c) A r e a o f ADC = Area of BDC (d) Angle DBE = Angle BCD

11. AB, CD, EF are three equidistant parallel lines. PRL and QSM are any two transversals cutting themat 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 thebisectors 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 ADCmeet 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 thatCQ : 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 drawnparallel 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


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 aretangents 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 itsvertices 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 perpendiculardistances 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 andBCD= 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 circumferencesof 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. Thenthe 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 outsidethecircle at M. If angle M is 30 and PRS is 50 the angle PLS is(a) 110 (b) 100 (c) 80 (d) 70


Quant Funda Class (Geometry)28. 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 asdiameters. If a circle is drawn to touch these three circles, its radius is

(a) 32r (b) 2

r (c) 4r (d) 3

r

30. The distance between the centres of two circles with radii 9 cms and 4 cms is 13 cms. The length ofa 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 thesmaller 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 anglessubtended 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 CQare 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, thenADB 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) 2.(c) 3.(c) 4.(a) 5.(a) 6.(c) 7.(a) 8.(a) 9.(c) 10.(a)11.(a) 12.(c) 13.(c) 14.(b) 15.(b) 16.(a) 17.(c) 18.(c) 19.(c) 20.(d)21.(c) 22.(b) 23.(c) 24.(d) 25.(a) 26.(b) 27.(d) 28.(a) 29.(d) 30.(b)31.(a) 32.(b) 33.(d) 34.(c) 35.(b)



SOLVED EXAMPLES Part 11. 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 = 5Solving 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 breadth6 cms. Find its area.

Sol. Diameter of the circle = Breadth of the rectangle Required area = x (6/2) = 9 cms 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 semicircle 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 midpoints of any two sides of a triangle is parallel to

the third and onethird 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 onefourth 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/4Given that : X/4 + X = 180 X = 144 Ans.

5. In a triangle ABC, AB + AC = 200 cms. Median AD = 10 cms. Find BC.Sol. Use the formula for the length of MEDIAN 2 (AD) + 2 (BD) = AB + AC 2 x 100 + 2 BD

= 200 BD = 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 92 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 = 92 : 1 = 30 : 25 X X = 22 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 20 3 x ( x 10 x 60 / 360) = 100 3 50 = 16.12 cms Ans.


Quant Funda Class (Geometry)8. 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 6.66 = 19.2 cms Ans.

9. A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same timea 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 = 40Then we know that angle ACB = 90 becasuse it is the angle contained in the halfcircle Angle BAC = 180 90 40 = 50 Ans.

11. How many common tangents can be drawn through to two circles of radius 2.8 cmsand 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 distancebetween centres is more than 0.8 cms which indicates internal touching. Thus the number ofdirect 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 thecircles.

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 theequilateral triangle described on its diagonal.

Sol. Let the squares side be A Its diagonal is A 2 Area of the first equilateral triangle = 3/4 x AArea of the second equilateral triangle = 3/4 x (A 2) Required ratio = 1 : 2 Ans.

14. The perimeters of two similar triangles are 30 cms and 20 cms resp. If one side of thefirst 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.


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 = Base x Altitude For similar triangles :

Ratio of areas = Ratio of squares of corresponding heights 9/16 = height 1 / height 2 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 twobounding 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 cmsLet R cms be the base of the cone 2 R = 8 R = 4 cms Given L = 12 cms Height of the cone = (L R) = (144 16) = 82 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 outof 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 onethird of its interior angle. How many sideshas the polygon ?

Sol. Each exterior angle of a regular polygon of n sides = 360/nEach interior angle of a regular polygon of n sides = (2n 4) x 90/nUsing the given informations, we get360

=1 x (2n 4) x 90

n 3 n12 = 2n 4 2n = 16 n = 8Hence 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 andthe 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} 2 XHence 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.

65o

50o

30o

D

E

C

B

A

40o40o

yx

fig (a) fig (b)

F


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

Draw DH || EF CG || DHs = 30 (alt. angles, DH || EF) r = 50 30 = 20X = r = 20 (alt. angles, DH || CG)Y = 65 (alt. angles, CG || AB) BCD = X + Y = 20 + 65 = 85 Ans.

Ex 2. Given: the figure below. To find: a, b, c.

Sol. a + 36 + 70 = 180 (sum of triangle) a = 180 36 70 = 74 Ansb = 36 + 70 (Ext. angle of triangle) = 106 Ans.c = a 50 (Ext. angle of triangle) = 74 50 = 24 Ans.

Ex 3. In the figure, AXD, BXE and CXF are straight lines, find in degrees:

(a) the angle X, if X = c + d,

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

(c) the value of a + b + c + d + e + f Sol. (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 = 60e = 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.

Ex 4. In the figure, AD, BE and CF are the altitudes of the triangleABC intersecting at G. Find the sum of BAC and BGC.

Sol. 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.

30o

F

D

C65o

B

EA

xysr

G

H

A

C

B

ca

b D

36o 70o

50o

E

DAXa

b cd

f e

x

B C

F

EF

srb

G

B C

A


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.

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

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.

Sol. OD = r cm (same radii)OC = r 8 cmr = 12 + (r 8) (Pythagoras Theorem) r = 144 + r 16 r + 64 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 cmAM = MB = (4) cm ( from centre bisects chord) = 2 cmCN = ND = (10) cm ( from centre bisects chord) = 5 cmIn Triangle AMO,(1) r = 2 + (3 + X) (Pythagoras theorem)In Triangle CNO(2) r = 5 + X (Pythagoras theorem) 2 + (3 + X) = 5 + X (1) = (2) X = 2 cmPut X = 2 cm in (2) r = 5 + 2 = 29 r = 29 cm. Ans.

A CB

P R Q

7.54.5

A C

P R

4.5

B

Q

7.5K

13

D

O

CA12 8

r

12

A M B

D

OC

r

rxN

F

B



E

25o

50o

a1

b2

b1

a4

a2

CA

B

a3

o

Ex 8. Given : In the figure, diameter AB produced andchord QR produced meet at P. QPA = 28 and QAR = 32 To find : X and r.

Sol. AB is a diameter (Given) AQB = 90 ( in semicircle)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.

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 = 50In cyclic quadrilateral ABCD,c + a = 180 (Opp. of a cyclic quad.) c = 180 a = 180 50 = 130But 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.

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) = 75a2 = b1 b2 (Ext. of triangle)= 100 75 = 25Also BOC is isosceles with OB = OC, same radii, a3 = a4 and a3 + a4 + b1 = 180 a3 + a4 = 180 b1 = 180 100 = 80 a3 = a4 = 40 a1 = 100, a2 = 25, a3 = 40, a4 = 40b1 = 100 and b2 = 75.

BP

RQ

A32o

r 28o

x

x 1

A

D

B S

Cd

b

c1

r

s

R

dx

c

a b

x140o

DC

A B



X

Y Z

CB

A

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)2X = 22 X = 11, Y = 5, Z = 4The radius of the circle, centre X, is 11 cm.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 thatAX = 9 cm, XB = 5 cm. To find : the radius of the circle, centre Y, which touches the diameter at X andtouches the circle, centre O, internally at Z.

Sol. Let YX = YZ = r (Same radii)OYZ is a straight line (Contact of circles)YX AB (Tangent to radius)AX = 9, XB = 5 (Given) AB = 14,OB = OZ = 7 (same radii)OX = 7 5 = 2In triangle OXY, OY = 7 rYX = r, OX = 2 OY = YX + OX (Pythagoras Theorem)(7 r) = r + 2 49 14r + r = r + 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 BDPa = a1 ( in same seg.)p = p (common) ACP = BDP (3rd of triangle) Triangle ACP

~

triangle BDP (A.A.A.)BD/BP = AC/AP (corr. sides of

~

triangles) BD/8 = 6/10 BD = 4.8 cm Ans.PD / BP = PC / AP (Corr. sides)PD/8 = 5/10PD = 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 mans height 3 feet answer Ans.

A 7 2O

X 5 B

Yr

Z

D

P

B

C

a

p

a 1

A

K


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 triangleif 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 = 36Thus 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 sameorder).Thus, we use the Sine rule : Sin A / a = Sin B / b = Sin C / cWe know that a + b + c = 20, A = 36, B = 36, C = 108Also since triangle is isosceles, a = bThus we can easily solve the Sine rule equations to get a = 5.54 cms, b = 5.54 cms, c = 8.92cmsNOTE : To solve by the Sine rule, you need SINE INVERSE functions. This is generally NOT askedin 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. Theexterior angle will be 144, 108, 108 resp (in the same order). Now to get the edge lengths we haveb = c, a + b + c = 20 cms. Again using the same procedure of Sine rule, we get a = 7.64 cms, b= c = 6.18 cmsNOTE : WE HAVE SOLVED THE PROBLEM TAKING THE RATIO 1 : 4 INTO CONSIDERATIONTHROUGHOUT.

Ex.17. A regular polygon of 8 sides is inscribed in a circle of radius 20 cms. Find its area and alsothe value of the angle a side of the polygon subtends at the centre of the circle. Also find thevalue 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 = 135Thus now if we join each vertex of this 8sided polygon to the centre ofthe circle, the central angle subtended by each side would be 360/8 = 45Now to find the area :Area of the polygon = Sum of areas of the eight isosceles triangles= Sum of areas of 16 right angled trianglesArea of a right angled triangle = Base x Height = B x P= (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 cms Ans.

Ex.18. In a triangle ABC, AD is drawn perpendicular to BC. Let P denote the length AD.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 = (12 + 16) = 20 cms, c = (9 + 12) = 15 cms Ans.

Ex.19. Two circles of diameter 10 cms and 18 cms touch each other internally. Find the distancebetewen their centres. Find the same if they touch externally.

Sol. When they touch each other externally, the distance between their centres= 10 + 18 = 28 cmsWhen 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 themidpoints of the oblique sides is 5. Find the area of the trapezium.

Sol. Area of a trapezium = (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 = (5 x 2) x 10 = 50 cms Ans.


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) / APOn 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. Ifthe 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 youtake them to be on the opposite sides of the centre, the final answer shall be the same as one distanceshall come negative) Given : PQ = OQ OP = 3 cmsNow we have the right angled triangles OAQ and OCPLet radius = RIn triangle OAQ, OQ = R AQ OQ = R 2.5In triangle OCP, OP = R CP OP = R 5.5On subtracting, we get : OQ OP = 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 R = OQ + AQ = 5.5 + 2.5 R = 6.04 cms Ans.NOTE : In this problem, we have used the property that a perpendicular from the centre to a chordbisects 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) + Y = 36Considering the triangle OPB, we get : X + Y = 25Solving 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 parallelto 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 outto be on different sides of the centre. However, even if you draw the diagram assuming the chordson 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 andangle BAC = 30, find angle BCD.

Sol. We use the property that the angles subtended by an arc at any point on the circumferenceare equal Arc DC subtends angle CBD and CAD at the circumference Angle CBD = Angle CAD = 70 Angle DAB = Angle DAC + Angle BAC = 70 + 30 = 100Since opposite angles of a cyclic quadrilateral are supplementary, we getAngle DAB + Angle BCD = 180 Angle BCD = 80



Questions from Previous CATs and other METs27. Find the area of the shaded region in the diagram below.

(1) 226 (2) )26( (3) 9( - 1) (4) 9 Sol. Shaded area = Area of big semicircle - (Area of 2 small semi-circles + area of triangle)

)1(9991821

22

2

22

21

bhrr

. Hence, (3).

28. ABCD is quadrilateral whose diagonals intersect at E. Which one of the following is notsufficient to prove AB || DC?(1) ABD = BDC (2) AE : EC = BE : ED(3) AB : BE = DC : DE (4) ADC = BCD and AD = CB

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 whetherthe 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,

If a line (seg AB) divides any two sides of a triangle ( ODC) in equal ratio

BCOB

ADOA

,

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 is1/9th the larger sphere. If 10 litres of paint were needed to paint the larger sphere, howmany litres are needed to paint all the smaller spheres?(1) 90 (2) 81 (3) 900 (4) 810



Sol. Volume available on melting )9.0()(34 3R [10% loss]

Number of smaller spheres 33

3

9)9.0(

934

)9.0()(34

R

R. Paint required is proportional to total surface

area i.e., 4 r2. Total surface area of the bigger sphere = 4 R2 Total surface area of all small

sphere 22

3 4)1.8(9

4)9.0(9 RR

. Paint required 8110

44)1.8(

2

2

R

R litres. Hence, (2).

30. Circles shown in the figure are of radius = 3 cm each. AB and CD are tangents to both thecircles. CD = 10 cm. What is the length of AB?

(1) Cannot be determined (2) 8 cm(3) 6 cm (4) 10 cm

Sol.

PC = PA = QD = QB = 3 cmJoin 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.

41635 22 AO . AB = 2AO = 2 4 = 8. Hence, (2).

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



Sol. Pythagoras triplet: r = 50 cm. Now, surface area of hemisphere 22

r22

4 r

Exterior surface area of igloo = 2 (58)2 Interior surface area of igloo = 2 (50)2Total 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 theradius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volumeof 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

3232

31

rhr 2.42.42.4722

3262.42.4

722

31 Hence, (4).

33. A cylinder is filled to half of its volume. It is then tilted so that the level of water coincideswith one edge of its bottom and top edge of the opposite side. What is the amount of waterspilled in this process?(1) 0 cc (2) 24 cc (3) 16 cc (4) Cannot be determined

Sol. Let the volume of the cylinder be v

Initial position. In this positon the volume of water is half that


Quant Funda Class (Geometry)of cylinder. Thus, in both, the cylinder can hold same volume. No water will be spilled. Hence, (1).

34. 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. Assumingcompound wall is very long, find the dimensions of the garden if he can exactly fence hisgarden using all the stock of barbed wire.

(1) 31812 (2) 20 5 (3) 25 4 (4) None of these

Sol. If l = length and b = breadth of the field. 2 l + b = 30 b = 30 - 2 l. l b = 100l(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).

35. ABC is an isosceles triangle (AB = AC) where (AC) = 10 cm. If (AO) = 5 cm, find thethird side of the ABC.

(1) 3 7 (2) 2 7 (3) 57 (4) 67Sol. O is the centroid, and it divides AD in the ratio 2 :1.

ADAO32 5.7

23 AOAD

Apollonius theorem: 222

2

212)(2 ACABBCAD

222

2 1010212)5.7(2

BC

BC = 57. Hence, (3).

Alternatively. O is the centroid. 5.723 AOAD

. Since ABC is isossceles, AD is perpendicular

bisector to BC. 222

2

4ADABBCBD 75BC . Hence, (3).

36. The ratio of the radius of the front tyres of a tractor to its rear tyres is 3 : 5. The rear wheelhas a diametr of 1.8 m. Find the total revolutions made by its 4 tyres when the tractor astravelled 2 km.(1) 1946 (2) 2112 (3) 1714 (4) 1887

Sol. Rear wheel circumference = d = 1.8 m. Distance travelled = 2 km No. of revolution of realwheels 7.5898.3533

5 . Total revolution of 4 wheels = 2(353.8 + 589.7) = 1887. Hence, (4).

37. A ladder 8 m long is rested against a buidling such that it reaches a point 8 m below the topof that building. At the foot of the ledder, the angle of elevation of the top of the building is60. Find the height of the buidling.(1) 10 m (2) 15 m (3) 12 m (4) 11 m

Sol. AB = height of the building, CD = ladder. CBA = 30 (complimentary angles).



BDC is isosceles since BD = CD. BCD = 30. Hence, DCA = 60 - BCD = 30.m.4

218AD30sin

CDAD AB = (AD + DB) = 12 m. Hence, (3).

38. ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A

and C respectively. If AC = 5 cm and2

53AD cm, find the length of CE.

(1) 52 cm (2) 2.5 cm (3) 5 cm (4) 24 cmSol. Since AD and CE are two medians. BD = DC and AE = EB. AB2 = AC2 - BC2 = 25 - BC2 .....(1)

4BC

445BD

253BDADAB

22

2222

.....(2).

Equating (1) and (2), 22 BC41

445BC25

355BCBC

43

455 22

. From (1),

320

35525AB2 Also, BE2 + BC2 = CE2 22

2CEBCAB

21

AB2 + BC2 = CE2 2CE3

5535 CE = 25 cm. Hence, (1).

39. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top whichis oen is 5 cm. It is filled with water upto the brim. When lead shots, each of which is asphere of radius 0.5 cm, are dropped into the vessel, 1/4 th of water flows out. Find thenumber of lead shots dropped in the vessel.(1) 50 (2) 100 (3) 150 (4) 200

Sol. Volume of water = Volume of cone 32 cm3

2008)5(31 . Volume of the lead shot 3)5.0(3

4

Number of lead shotsshotleadoneofVolume

vesselofoutflownWater 10021

21

21

34

3200

41

. Hence, (2).


Quant Funda Class (Geometry)40. The radius OA of the circle in the figure is:

(1) 9 (2) 7 (3) 5 (4) 3Sol. Since BD is diameter of circle: C is a riht angle. Hence, by Pythogoras theorem, BD = 10.

Radius = 5. Hence, (3).41. 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 willbe the ratio of height of the cylinder to its radius.

(1) 922 (2) 9

82 (3) 763 (4) 27

123

Sol. Volume of ball of radius 334

'' rr . Volume fo small ball formed

2734 3r Volume remaining

for cylinder 2726

of original volume. 32 34

2726

94

rhr rh926

982

926

r

h. (2).

42. What is mDCA + mABD in the adjoining figure?

(1) 100 (2) 70 (3) 50 (4) 40Sol. 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).43. 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) 83 (2) 5

3 (3) 43 (4) 2

1

Sol. Slant height 5.7)5.4()6( 22 . Total surface area = r l + r2 = r(4.5 + 7.5) = 12 r

Area of base = Area painted = r2 = 4.5 r. Required fraction 83

125.4 r

r. Hence, (1).


Quant Funda Class (Geometry)44. 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% weightloss?(1) 6 (2) 8 (3) 12 (4) 16

Sol. Volume of cylinder = 42 160. Wastage results in 10% loss 10916044

34 3 r

r3 = 4 4 4 9 3 r = 12. Hence, (3).45. Let O be an incentre and circumcentre of a ABC and length of OB is 10 cm. Find the

area of the triangle.

(1) cmsq.3475 (2) cm.sq.34

25 (3) cm.sq.4

3100 (4) 75 3 sq. cm.Sol. 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 375)310(43 2

sq. cm. Area of the circumcircle = (10)2 = 100 sq. cm. Hence, (4).

46. In the above Q. find the area of the circumcircle which is not enclosed by the triangle.(1) 75 (2) 100 - 753 (3) 75 3 - 25 (4) 25 - 75 3

Sol. As the incentre and the circumcentre are located at a common point the triangle is an equilateraltriangle. Hence, the radius of the incircle is 5 cm and the side of an equilateral triangle will be 10 3.

Area of the traingle 375)310(43 2

sq. cm. Area of the circumcircle = (10)2 = 100 sq. cm. Hence, (2).

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

Sol. lb = 60; blbl 522 ; Solving l = 12, b = 5; 2(l + b) = 34. Hence, (3).48. 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 curvedsurfaces are semicircles on the diametrs, AB, AD, BC, CD.)

(1) 475000 (2) 49000 (3) 22638 (4) 45276Sol. Length of diametr = BD = 42 kms. AB = 14, BC = 28

Area of the shaded region

1472277

21414

. Cost 2263874922147

. (3).


Quant Funda Class (Geometry)49. 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, 32Sol. 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).

50. Find the value of LDM in the following figure:

(1) 54 (2) 36 (3) 34 (4) 40Sol. Let LDM = x;MAD = 180 - 32 - x = 148 - x; BCD = 180 - (148 - x) = 32 + x

40 + 32 + x + x = 180. x = 54. Hence, (1).51. Find the value of x in the following figure?

(1) a + b + c (2) b - a - c (3) 180 - c (4) None of aboveSol. 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).


Quant Funda Class (Geometry)52. 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

Sol. AB || CD and AB = CD. Then ABCD is a parallelogram and AC = BD. Hence, (2).53. 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) 175Sol. AB = AC B = C = 75 A = 30, A + D + E = 180D + E = 150. Hence, (3).54. The radius OA of the circle in the figure is:

(1) 9 (2) 7 (3) 5 (4) 3Sol. Since BD is diameter of circle C is a right angle. Hence by Phythogoras theorem, BD = 10

Radius = 5. Hence, (3).55. 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, wasthe portion cut?(1) 7 cms (2) 1 cm (3) 3.7 cms (4) Cant get

Sol. Ans. (2).56. 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 cm2 (4) 75 cm2

Sol. APB is similar to DPC 22

22

21

2

1

58

SS

AA

875

8)5(24

2

2

2 A . Hence, (2).


Quant Funda Class (Geometry)57. 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 thelength of the line to the radius of the 2nd innermost circle?(1) 13 : 3 (2) 13 : 5 (3) 11 : 3 (4) Cant get

Sol. Radii are in the ratio 2 : 3 : 7 : 9 : 13 Radii are 2r, 3r, 7r, 9r, 13r Desired ratio 3:133

13 r. Hence, (1).

58. A cricket pitch has to be flattened by 2 rollers, of equal length but with the bigger rollerhaving a radius 7/3 times that of the smaller roller. If the bigger roller takes 6 minutes to dothe 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

Sol. Let bigger roller have radius r, length l Area covered per roll = 2 r l. For smaller roller,r

732 Smaller roller takes 7/3 times time taken by bigger roller. 14

376 minutes.

59. A square PXCM is rotated on a plane anticlockwise about its centre such that C moves tothe position origianally occupied by X. By how many degree will the square have rotatedif PXCM are the four vertices named in anticlockwise manner?(1) 90 (2) 180 (3) 270 (4) 315

Sol. 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).

60. O is the origin and AB is the diameter of the semicircle that is shown shaded. Find the areaof the semicircle.

(1) 21.13 (2) 169 (3) 204.28 (4) 84.5

Sol. OA = 12; OB = 5 AB = 13 Area

13.218

1692

1321 2

. Hence, (1).

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 8454545101010999

cubes. Hence, (2).

62. In a trapezium, the diagonals intersect at point O. The ratio of the length of one of thediagonals from one vertex of the trapezium to thepoint O to its entire length is 2:5. Find theratio of its parallel sides (smaller side : larger side)(1) 2 : 5 (2) 2 : 3 (3) 2 : 7 (4) 5 : 7


Quant Funda Class (Geometry)Sol. Given DO : DB = 2 : 5

Since 52

DBDO

By dividing 32

252 DODB

DO. AOB is similar to COD

DCAB

OBDO

32

.

Hence, (2).63. 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

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

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

Sol. Midpoint theorem

21

5829

sin x x = 30. Hence, (2).65. 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 andsecond longest sides in two points which are 17 cms apart. Find the length of the smallestside.(1) 8.5 cm (2) 34 cm (3) 51 cm (4) Cant get

Sol. Midpoint theorem

ADE ABC 21

BCDE

ABAD BC = 17 2 = 34 cms. Hence, (2).


Quant Funda Class (Geometry)66. 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 tothe radius of the new small spheres is:(1) 1 : 33 (2) 2 : 33 (3) 2 33 (4) 33 : 2 : 32

Sol. Volume 334

r 33 1132

ar 333421

227

2311 aar

...(i). Similarly,, 313 347

32

rr

r3 = 14 r13 ....(ii) 313 14421

ra 38

31

3

r

a

3

3/1

1 32

38

r

a Hence, (2).

67. The difference in the circumferences of 2 circles is a multiple of 6. Which of the followingstatement 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

Sol. Let radii be r1 and r2 (r1 > r2) 2 r1 - 2p r2 = 2 (r1 - r2). We already have a factor 2, so r1 - r2 isa multiple of 3i. 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 issatisfied. Hence, (4).

68. A palm tree swings with the breeze in such a manner that the angle covered by its trunk is14. If the topmost point of the tree covers a distance of 33 meters, find the length of thetree.(1) 20 mts. (2) 99 mts. (3) 135 mts. (4) 125 mts.

Sol. Length of the arc 3602 r where r will be the length of the tree and will be the angle

made by its trunk 3336014

7222 r 135

22214736033

r metres Hence, (3).69. 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 ascompared 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 728.1555666

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


Quant Funda Class (Geometry)70. 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

Sol. DOC = 180 - 50 = 130. DCA = 180 - (30 + 130) = 20. OBC = 20 BOC = 180 - 40= 140 BOD = 140 - 130 = 10. Hence, (1).

71. In the figure alongside are two concentric circles, with a tangent to the inner circle forminga chord to the outer one. The length of the chord is 8 cm. If the radii of both circles areintegers, what is the radius of the inner circle?

(1) 3 cm. (2) 4 cm. (3) 5 cm. (4) 6 cm.Sol. The radius of the inner circle makes 90 angle with the tangent and cuts it in half. The only right-

angled triangle with 4 as one side has 3 and 5 as the other sides. So the smaller circles radius is 3 cm.(As all the sides are integres). Hence, (1).

72. The figure below is an equilateral triangle with incircle and circumcircle.Find the (area of A) : (area of B) : (area of C).

(1)27

334:

271

:2733 (2)

27334

:9

:2733

(3) 3 3 - : : 4 - 3 3 (4) Data Insufficient

Sol. a - altitude of the triangle. 32

;3

aa radius of incircle and circumcircle respectively..

Side of an equilateral triangle a3

2 . Area of equilateral triangle 233

a . Area of the circumcircle

2

94

a . Area of the incircle 291

a . Area of C = 1/3 (area of circumcircle - area of the triangle)

222

27334

933

94

31

aaa

Area of A = 1/3 (area of triangle - area of incircle)



222

2733

91

933

31

aaa

Area of A : Area of B : Arae of C

27334

:9

:2733 .

73. What is the arae of the largest triangle that can be fitted into a rectangle of length l andbreadth b?(1) lb/3 (2) lb/2 (3) 2lb/3 (4) 3lb/4

Sol. The maximum area of the triangle = lb/2 . Hence, (2).74. 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

Sol. Let side of triangle = a; so area of triangle 243

a . Radius of circle32

332 a

a . Area of

circle3

2a . So, Ratio3

:43

i.e., 33 : 4. Hence, (3).75. 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) Sol. 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).76. 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 = ?

(1) 5r (2) 5 r (3) 3r (4) r25

Sol. ACB = 90. Angle at the point of intersection to the centre of the circles. BC = r. AC = 2r (as areaof X = 4 area of Y) rrrAB 54 22 . Hence, (2).

77. From a circular paper a man makes two conical caps. The surface area of the two are in theratio 2 : 1. He then covers the face of the caps with other circular pieces of paper. The ratioof the area of these pieces is:(1) 1 : 2 (2) 4 : 1 (3)

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