Geometry & Mensuration

Area of a n- sided regular figureB C

In a regular figure, circum-circle and in-circle will be concentric.Let us look at the decagon drawn

A D

76

8

5

4alongside. There are 10 isosceles trianglesand each of equal area.In a n-sided regular figure there will be ni l t i l d h f l

EJ8

910

4

31 2isosceles triangles and each of equal

area.In any triangle the equal sides areCircum- radius ( R) and the ┴r to the

FI

H G OCircum- radius ( R) and the ┴r to theunequal side is the in-radius (r).Let ∆ OAB be one such triangle. Let OCbe the ┴r to AB. The vertical angle in

OO

be the r to AB. The vertical angle ineach of the triangle will be 2л/n .Therefore Ľ OBC = л/2–л/n.R = a/2 cosec л/n :: r = a/2 cot л/n

R r R

7

∆ OAB = ¼{na2 cot л/n} (л/2–л/n)A a/2 a/2 BC

(л/2–л/n)

h f d h ∆ Q

RGeometry - Circles - Concepts at a glance

In the given fig. a circle and a right ∆ PQR ,right angled at P are drawn In such a waythat PQ and QR are tangents at points P and B

A respectively. The circle intersects PR at Bsuch that BR=4cm. If AR = 10cm then findAQ.

A

▪Using tangent secant relation(RA)2 = (RB)(RP)

100= 4 (RP) ::: RP = 25cm.

AS tgt. segments from an external point to a ┐AS tgt. segments from an external point to a

circle are equal , QA = QP = x (say)

QR = 10 + x ::: PR = 25cm.

┴ Q (Ľ Q 0)

PQ

But RP is ┴r to PQ (ĽRPQ = 900)

(QR)2 =(PQ)2 + (PR)2

(10+x)2 = x2 + 625

20x = 525

x = 525/20 = 105/4 =26 ¼ cm. 22

h f A d C

Geometry - Circles - Concepts at a glance

In the given fig. AB and CD are 2perpendicular chords which intersect at P.PA = 12cm, PC = 4cm, PD = 18cm .Find the A

radius of the circle.As PA . PB = PC . PD12 PB = 72PB 6 ┐

OLPB = 6 cm.

AB = 18cmFrom O draw ┴r to both AB and CD.But ┴r from O bisects AB

CD

P┐

┐

┐

M

L

But ┴r from O bisects AB.AL = LB = 9 cm ::: LP = 3cm :: OM = 3 cmCD = 22cmCM = MD = 11cm B

M

Join OD.Using Pythagoras theoremOD2 = OM2 + MD2OD2 = 32 + 112

= 130OD = √130 cm.

23

In the given fig. ABED is a cyclicGeometry - Circles - Concepts at a glance

quadrilateral. When AB and DE reproduced they meet at C. ĽACD = 45˚and ĽADC = 90˚ .AB= 6cm , BC = 4cm.Find DE.In the cyclic quadrilateral ADBE,ĽADC = 90˚ implies ĽAEB = 90˚

A

45

Hence ĽEBC = 90˚In ΔBEC, ĽBEC = 45˚Cos45˚ = BC/EC

B

451/√2 =4 / EC

Hence EC = 4 √2 cm.InΔ ADC

ED C┐ 45

InΔ ADCCos 45˚= DC/AC

1/√2 =DC/ 10

C √2DC = 5√2 cm.DE = √2 cm.

26

ACB is a right angled triangle Angle

Geometry - Circles - Concepts at a glance

AACB is a right angled triangle, AngleC = 90. A cone of greatest volume iscut out of this triangle. Find theapproximate volume of the cone.

A

pp

6

CB8C 8  

28

GEOMETRY & MENSURATIONGEOMETRY & MENSURATIONA

6

CB8  

29

GEOMETRY & • The slant heightof the cone willMENSURATION of the cone willbe the ┴r from Cto AB.

• BD will be theslant height ofthe right circularthe right circularcone that can becut from thet i l

AE

triangle.• BD = 4.8 cms• EF is the arc• EF is the arc.• EF=90/360{2*л*4.8)• EF=2.4л

6

EF 2.4л

30C B8 F

GEOMETRY & MENSURATIONGEOMETRY & MENSURATION

G Circumference of the base ofthe cone= 2.4 лRadius of the base ofthe cone= 2.4 л /2 л = 1.2GH BD f t i lGH = BD of triangle

= 4.8 cms.GJ=√{(4 8)2 - (1 2)2}= √ 21 6

J

GJ=√{(4.8) - (1.2) }= √ 21.6Volume of cone= 1/3 л (1.2) 2 √ 21.6

1.2

31

J H( )

= 2.23 л

Amy, Bob and Chip are three bees that startfrom the same flower and fly away in threedifferent directions in search of nectar. Aftersome time Chip is as far from the mid point Dof the straight line joining Amy and Bob asof the straight line joining Amy and Bob asAmy is from D. Amy and Bob are 300 m apartwhile Bob is 180m from Chip. At this instantA d di t i l hi h i i dAmy sends a distress signal which is receivedsimultaneously by the other two bees. If a beecan fly at a speed of 3 m/sec, then whichamong Bob and Chip can reach Amy quickerand by how many seconds?1) Chip by 20 sec 2) Bob by 20 sec1) Chip – by 20 sec 2) Bob – by 20 sec3) Bob – by 18 sec 4) Chip – by 18 sec5) C t b d t i d5) Cannot be determined.

32

CAs AD = BD = CD =As AD BD CD150 cms, D becomesCircum centre andangle ACB is 900angle ACB is 90 .AB = √3002 – 1802 =240

BDA 150 150 C will reach A in80 secondsB will reach A inB will reach A in100 secondsChip – by 20

dseconds

33

Barles Cabbage, an avid mathematician, had hisson’s room constructed such that the floor of theroom was an equilateral triangle in shape insteadroom was an equilateral triangle in shape insteadof the usual rectangular shape, thus making theroom in the shape of right prism with anequilateral triangle as the base One day the sonequilateral triangle as the base. One day the sonbrought home a dragonfly and tied it to one end ofa string and then tied the other end of the string toone of the corners in his room. The next day heo e o t e co e s s oo e e t day euntied the other end of the string from the cornerof the room and tied it to a point exactly at thecentre of the floor of the room. If the maximumpossible volume in which the dragonfly can fly inthe first case is V1 and in the second case is V2,then find the ratio of V2 : V1. Assume that thel th f th t i i li ibl d t thlength of the string is negligible compared to thedimensions of the room.

(1) 4 (2) 5 (3) 6 (4) 7 (5) 12

36

(1) 4 (2) 5 (3) 6 (4) 7 (5) 12

GEOMETRY &MENSURATION

• The chords P

PQ & PS of a circle of

P

radius 5 is 6cm each.

Q S

Find the length of

o

the chord QS.

37

GEOMETRY &MENSURATION

P

Join OP. OP will be

┴r bisector.P

3.6 R

┴r bisector.Let the point ofintersection of OP

QS

4.8

1.46

& QS be R.Let PR = x :: OR =(5-x)

o

4

62 - x2 = 52 - (5 - x)2

36–X2=25–(25–10X+X2)36 X2 = 10X X236 – X2 = 10X – X2

X = 3.6. QR2 = 62 – 3.62 = 4.82

38

QR = 6 3.6 = 4.8QR = 4.8 and QS = 9.6

GEOMETRY &MENSURATION• Anita was walking along the edge of a wide

road. As she walked along, she observedth t th l f l ti f th t fthat the angle of elevation of the top of apole, standing exactly on the other edge ofthe road was 30° at two distinct points andthe road, was 30 at two distinct points andwas 45° at exactly one point. If Anitacovered a distance of exactly 20√ 2 mybetween the two points from where the anglewas 30°, then find the height of the pole.

• (1) 10 √3m (2) 15m (3) 20m (4) 10√ 2m(5) 10m

39

GEOMETRY &MENSURATION

P

• Let TH be the pole & TH = h• Let Q be the point where

angle of elevation of theT tower is 300

• In TQH, Ĺ TQH = 300

• tan 300 = 1/√3h√2

R H

tan 30 = /√3• Hence PH = h√3.• Let R be the point where

angle of elevation of the

h

h

h√

450 angle of elevation of thetower is 450

• In TRH, Ĺ TRH = 450

h√2

Q 300

• tan 450 = 1. Hence RH = h• In QRH, Ĺ QRH = 900

• QR2 = (h√3) 2 – h = 2h2

h

40

30 Q ( √ )• PH = h√2 = 10√2• h = 10

GEOMETRY &MENSURATION

• Ajay and Amar inherited a rectangular field ofarea 1000 sq m If they decide to split thearea 1000 sq.m. If they decide to split theland into two halves by dividing it with asingle straight line, then find theg g ,approximate minimum cost that Ajay wouldhave to incur to fence his half of the land ath f R 1 A h Ajthe rate of Re.1 per metre. Assume that Ajay

bears the complete cost of fencing thecommon boundarycommon boundary.

• (1) Rs. 67 (2) Rs.90 (3) Rs. 95 (4) Rs. 112 (5) Rs 81(5) Rs. 81

41

GEOMETRY &MENSURATIONA BE

D F C• For minimum expenses minimum perimeter is to be arrived at.

The division should take place along the breadth side.Perimeter to be fenced is l +2b.

• Lb = 1000.• l+2b to be minimum l=2b.• 2b2 = 1000.

• b=√500.• l +2b = 4 √500 = Rs 90/

42

• l +2b = 4 √500 = Rs.90/-

GEOMETRY &MENSURATION

• In a circle of radius 6 cm a chord PQ isIn a circle of radius 6 cm, a chord PQ isconstructed at a distance of 2 cm from thecentre A square ABCD is constructed suchcentre. A square ABCD is constructed suchthat A and B lie on the minor arc PQ and Cand D lie on PQ What is the approximateand D lie on PQ. What is the approximatelength of AB?

• (1) 1 8cm (2) 2 4cm (3) 3 1cm (4) 3 4cm• (1) 1.8cm (2) 2.4cm (3) 3.1cm (4) 3.4cm(5) 3.7cm

43

GEOMETRY &MENSURATIONGEOMETRY &MENSURATION• Let OF be ┴r from O to

ABAB.• CE = ED = AF = FD = x• AB = BD = FE = 2x

BAF

AB = BD = FE = 2x• OFB, • OF=(2+2x):FB=x,OB= 6

ECD QP ( ) ,

• 62 = (2+2x) 2 +x2

• 5x2 +8x – 32 = 0O

CD Q

• x = -8+√(64+640)/10• x = 8(√11–1)/10=1.85• 2X = 3.7

44

GEOMETRY &MENSURATIONGEOMETRY &MENSURATIONIn the abovefigure, given thatA BGFE figure, given thatFJ = JK = KH, AE =EF = FG = GB, DH= HI =IC and AB =

A BGFE

HI IC and AB2AD = 3DH = 4FG,find the ratio ofthe area of theJ the area of theshaded region tothat of rectangleABCD.K ABCD.1) 3:16 2)1:183) 3:32 4)1:24D CH I

45

D CH I

GEOMETRY &MENSURATIONGEOMETRY &MENSURATIONLet AB = 12AE=EF=FG=GB=3A BGFE AE=EF=FG=GB=3DH=HI=IC=4Let AD = 6

A BGFE

ABCD=72 Sq. UnitsLet N be a point onDC such that HN = 3

JDC such that HN = 3Area of FHNG = 18Area of FHG = 9KArea of GJK = 3

GJKABCDD CH IN = 1

46

ABCDD CH IN 24

GEOMETRY &MENSURATIONGEOMETRY &MENSURATIONThree circles, each of radius r, aredrawn such that each circle touchesthe other two circles externally. Thecommon tangent drawn to the first andcommon tangent drawn to the first andsecond circles, that drawn to thesecond and third circles and that drawnsecond and third circles and that drawnto the third and first circles, together,form and equilateral triangle, enclosingthe three circles completely. Find the inradius of the triangle.

√ √1) 2r/√3 2) r/√3 3)r(√3 +1)/√3 4)r(√3 -1)/√347

GEOMETRY &MENSURATIONGEOMETRY &MENSURATIONLet AC be the commontangent of circles &A gLet AB be the commontangent of circles &Let BC be the common

A D E C

Let BC be the commontangent of circles &ABC is an Equilateral

O1 O2

Join O1 & O2. Let O1D & O2D be ┴r to AC.DE = 2r.Ĺ DAO1=300=ĹDAO2

From DAO1, DA = r√3 = ECAC 2 (√3+1) B AC = 2r(√3+1).

48

B

GEOMETRY &MENSURATIONGEOMETRY &MENSURATIONFrom DAO1, DA = r√3

√A From DAO2, EC = r√3 AC = 2r(√3+1).In an Equilateral ,

A D E C

Altitude is √3/2{a}In radius is 1/3{altitude}

O1 O2

3{ }Therefore in radius of

ABC = 1/2√3 {side}= 1/2√3 {2r(√3+1)}

= {r (√3+1)/√3}B { /√3}49

B

GEOMETRY &MENSURATIONGEOMETRY &MENSURATION• A St. line passing through Ap g g

(-3,2) and B (6,5) intersects X-axis at a point P. Find The lengthp gof PA.

B (6,5)

A (-3,2)

50

Let AD be a ┴r toPC. Let BC be a ┴r to PCPAD ||| PBC ĹAPD ĹBPC & ĹPDA ĹPCBPAD |||r PBC. ĹAPD=ĹBPC & ĹPDA= ĹPCB

PD/PC= AD/BC = 2/5.

PD DC 2 3PD: DC= 2:3.DC = 9 PD = 6PA = √ 62 + 22 = √40PA = √ 62 + 22 = √40.

B (6,5)

A (-3,2)

D (-3 0) C (6 0)

2

5

P

51

(-9,0)D (-3,0) C (6,0)3

P

Geometry & Mensuration• Two identical circles with centres A & B intersect as shown• Two identical circles with centres A & B intersect as shown.

The area of the biggest circles that can be drawn in thecommon region is 64 π cm2, while the circumference of thesmallest circle that can be drawn circum circle being thesmallest circle that can be drawn circum circle being thecommon region is 64 π cm. What’s the circumference ofcircle with centre A.(1) 136 (2) 68 (3) 72 (4) 144• (1) 136 π (2) 68 π (3) 72 π (4) 144 π

A B

52

F

atio

nns

ura

A BC D E

& M

enet

ry &

G

• CE = 16 cm :: FD = 32 cm Geo

me

CE = 16 cm :: FD = 32 cm• Let AC = x . AE = 16 + x = AF . AD = 8 + x• From ADF , AF2 = AD2 + FD2 Implies (16 + x) 2 = (8 + x) 2 + 322

2 2

G

• 256+32x + x2 = 64 +16x+ x 2 + 1024 :: 16x = 832 ::x = 52• AF = 68 . Circumference = 136 π

53

Geometry & Mensuration• A thin copper wire is wound uniformly and spirally in a

single layer from the bottom to the top around acylindrical iron rod. Circumference of iron rod = 2cm.H i ht 56 N f t 45 L th f i ?Height = 56cm. No of turns = 45. Length of wire = ?

• (1) 96 (2) 102 (3) 106 (4) 116

AA656

B C90

54

AIMCAT 11 ‐ Geometry & Mensuration• A rectangle ABCD of length ℓ and breadth b is partitioned as shown in

the figure below. If ℓ = 15 cm and b = 10 cm, find the area of the shadedregion (in sq cm)region (in sq.cm).

• (1) 75 (2) 70 (3) 110.5 (4) 72 (5) Cannot be determined

• The area of theThe area of theshaded portionsis half of the areaof the rectangle.

• Answer is

• 75 Sq.cms.

57

AIMCAT 18 - Geometry & Permutations & Combinations

Th i f fThe circumference of acircle is divided into 26equal parts by marking26 equidistant points onit. Now, using thesepoints as vertices,points as vertices,triangles are drawn suchthat the circumcentreof each of thoseof each of thosetriangles lies on one ofits sides. How many

For each diameter theresuch triangles can bedrawn?(1)156 (2)676 (3)182 (4)650 (5)312

For each diameter therewill be 24 such trianglesand therefore totally therewill be 512 triangles.( ) ( ) ( ) ( ) ( )

59

g

AIMCAT 18 – Geometry & Mensurationy

• The figure PQRSTUVW is a cube. If the lengths of PV, WU and TR are equal to the sides of a triangle the radius of the circle inscribed in thatequal to the sides of a triangle, the radius of the circle inscribed in that triangle equals

• (1)  the edge of the cube.• (2)√2/√ 3 times the edge of the cube( ) /√ 3 g• (3)1/√6 times the edge of the cube• (4)1/ √3 times the edge of the cube• (5)  None of these

√• PV = WU = TR = √2 a• In radius of a equilateral Triangle = side /2√3• = √2 a /2√3 = a / √2 √3 = a /√6• = /2√3 = / √2 √3 = /√6• 1/√6 times the edge of the cube

60

Geometry & Mensuration 

∆ ABC is such that one of its side is double the other and let the angle opposite to those sides differ by /3

A

opposite to those sides differ by angle of π/3, then the triangle is(1) Right angled triangle (2) Equilateral  triangle

900 Mα + π/3

(3) Isosceles triangle (4) Scalene  triangle.m / sin α = 2m / sin (α + π/3)sin(α + π/ ) = 2 sin α 2M

600α 300

B csin(α + π/3) = 2 sin α2sin α = ½ sin α + √3/2 cos α3/2 sin α = √3/2 cos α

tan α = 1/

2MB

tan α = 1/√3α = 300.

It is an right angled triangle

62

Geometry & MensurationABC is a right angled triangleWith ĹB =90. Let D be a pointon AC such that ĹDBC=30 Eon AC such that ĹDBC=30. Eis a point on BC such that ED= EC. If BD=√3 BE, find ĹECD.

A

Cos 300=[3X2+X2‐K2]/2√3X2

√3/2 =[3X2+X2‐K2]/2√3X2

6X2 8X2 2K26X2 = 8X2 – 2K2

X2 = K2 :: X = Kǿ = 1200

DǿĹDBC=60ĹECD=60

30k

60

B C63

E30

x k60

Geometry & Mensuration

Area of ▲PAS = Area of ▲QBR

8*8* √3 /

QP6

= 8*8* √3 / 2 * √3 *√3 *2= 16/√3Area of rectangle = 48Area of PQBRSA

4 4Area of PQBRSA = 48 - 32 /√3= [48√3 – 32] /√3Area of PQBRSA/OTHER AREA

√

W UA600

B= [48√3 – 32] /32= [3√3 –2] /2

4 4

RS 6

66