TC | CAT Geometry Geometry 4 - Solutions the radius of the circle is. (a) x16 / 4S (b) x 2.5 (c) x 2 (d) 3 2 x Solution: Figure not drawn to scale OA = radius OA = √𝑥2+2𝑥2

TC | CAT Geometry

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1. The sides of a triangle ABC are AB = 12, BC = 18

and AC = 10. There is a point D, on BC such that

both in-circles of triangles ABD and ACD touch

AD at a common point E, as shown below.

What is the length of CD?

Solution:

The tangents drawn from an external point are

always equal in length.

DE = DG = DF = x (say)

FC = HC = y (say)

AI = AE = AH = z (say)

BG = BI = k (say)

Now AB + BC + CA = k + z + k + z+ x + y + z + y =

2(k + x + y +z)

⇒ 40 = 2(k + x + y + z) ⇒ x + y + z + k = 20

z + k = AB = 12 ⇒ x + y = CD = 20 – (z + k) = 20 –

12= 8

2. Two circles with radii ‘a’ & ‘b’ respectively touch

each other externally. Let ‘c’ be the radius of a

circle that touches these two circles as well as a

common tangent to these two circles. The a, b, c

are in

(a) GP (b) H.P

(c) 1 1 1

c a b (d)

1 1 1

c a b

Solution:

There are three DCTs and they are AB, BC and AC. The

figure also says that AB + BC = AC

AB = DCT = √(𝑎 + 𝑐)2 − (𝑎 − 𝑐)2 (formula for

DCT), a+c is the distance between the centers of two

circles and a and c are the radii of two respective

circles. Similarly write the expressions for BC and AC,

and after simplification we get, AB = 2√𝑎𝑐

BC = 2√𝑏𝑐 AC = 2√𝑎𝑏

2√𝑎𝑏 = 2√𝑎𝑐 + 2√𝑏𝑐

Divide on either sides with √𝑎𝑏𝑐

Hence, answer is Option d)

3. OPQ is a quadrant of a circle and semicircles are

drawn on OP and OQ as diameter. Then area of

shaded portion A : B equals.

(a) 1 : 2 (b) 2 : 1 (c) 3 : 2 (d) 1 : 1

Solution: Since it involves ratio, let us assume values. Take 1-1- squares. Whenever such kind of figures are given, look for a square and wen no values are given, assume the values, preferably 1.

Geometry – 4 - Solutions

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The above square can be pictured as below diagrams

The above figure is a combination of the below two diagrams.

There is a square and a quarter of a circle, so the left

over area will be 1- 𝜋

4. Thus to summarize, the figure

will now become:

1 - 𝜋

4 + A + 1 - 𝜋/4 = 1 (Area of the square)

A + 1 - 𝜋

2 = 0

A = 𝜋

2 – 1

Next,

2 quarter circles + 1 square + B = area of bigger

quarter circle. 𝜋

4 +

𝜋

4 + B = 𝜋 (Area of bigger quarter,

since its radius is 2 units)

B = 𝜋

2 – 1

Ratio of the areas = 1: 1

4. In ABC , let AB = 20, AC = 11, BC = 13. The

diameter of semicircle inscribed in ABC, (whose

diameter lies on AB and side AC & BC are

tangents o the semi-circle) is equal to

(a) 13 (b) 11

(c) 17 (d) 19

Solution:

½ 13 (r) + ½ 11 (r) = √𝑠(𝑠 − 1)(𝑠 − 𝑏)(𝑠 − 𝑐)

24

2 r = 66

12 r = 66

d = 2r = 11

5. If cot2 𝜃 - (1 + √3) cot 𝜃 + √3 = 0, then what is

the value of 𝜃?

Solution:

Let cot 𝜃 = x

Then x2 – (1 + √3)𝑥 + √3 = 0

∴ x2 – x - √3x + √3 = 0

∴ x (x – 1) - √3(x – 1) = 0



TC | CAT Geometry


∴ (x – 1) (x - √3) = 0

∴ x = 1 or x = √3

∴ 𝜃 = 𝜋

4 or 𝜃 =

𝜋

6

6. ABC is a right angled at B. if BD = 8cm, BE =

6cm, AE = 10cm & CD = 10cm. Then find the area

(in sq. cm) of (quadrilateral BDOE)

(a) 24 (b) 28

(c) 32 (d) 34

Solution:

Join O and B

In triangle AOB, there are two more triangles, AEO

and BEO, both these triangles have the same height.

Hence if you see the ratio of their areas, they will be

in the ratio of their bases.

𝐴𝐸𝑂

𝐵𝐸𝑂=

10

6=

5

3

Let AEO be 5x and BEO be 3x

Similarly,

Look at triangle BOC, there are two more triangles

BOD and COD, and both these also have the same

height.

𝐵𝑂𝐷

𝐶𝑂𝐷=

8

10=

4

5

Let BOD be 4y and COD be 5y

Quadrilateral BDOE = BOE + BOD = 3x+4y

Now, area of triangle ABD = ½ * 8 * 16 = 5x+3x+4y

= 64 = 8x+4y

Similarly, area of triangle BEC = ½ * 6* 18 =

5y+4y+3x

= 54 = 9y + 3x.

Solve for x and y using the above two equations,

then find the value of 3x+4y which is the answer.

The correct answer is Option d) 34

7. If 𝛼, 𝛽, 𝛾 and 𝛿 are four angles of a cyclic

quadrilateral, then the value of

cos 𝛼 + cos 𝛽 + cos 𝛾 + cos 𝛿 is:

Solution:

In a cyclic Quadrilateral , the sum of the

opposite angles is 1800. Assuming that 𝛼 is

opposite to 𝛾, and 𝛽 is opposite to 𝛿.

𝛼 + 𝛾 = 180

𝛽 + 𝛿 = 180

𝑆𝑜 𝛼 = 180 – 𝛾

𝛽 = 180 - 𝛿

Hence the question becomes

cos (180 – 𝛾) + cos (180 – 𝛿) + cos 𝛾 + cos 𝛿

cos (180 – 𝛾) will be – cos 𝛾, because this angle

will be in the second quadrant, and cos value

will be negative in second quadrant, so will cos

(180 – 𝛿), as it becomes – cos 𝛿, hence the

expression’s value will be 0.

8. In the given figure, ABC is a triangle with AB = 5,

BC = 6 and CA = 7. Squares are drawn on each

side, as in the figure below. Find the area of

hexagon DEFGHI

(a) 90 + 24√3 (b) 90 + 24√6



TC | CAT Geometry


(c) 100 + 24√6 (d) None of these

Solution:

Carefully observe this figure, and the area of this

hexagon will be equal to area of 3 squares (ABGH,

ACDI, BCFE ) and 4 triangles (ABC, AHI, BGF, CDE).

We can find the area of the triangle ABC as the length

of the sides are given.

Area of triangle ABC is 6√6 = ½ * AB * AC * Sin A

= ½* 5*7 * Sin A = 6√6

Sin A = 12√6

35

We can also say, angles BAC+ HAD+ HAI+CAI = 3600

(Sum of all the angles), HAD and CAI = 900 each, thus

HAI = 180 – BAC

For triangle HAI, the area will be ½ * HA *AI * sin (HAI),

but Sin (HAI) = Sin (180-BAC)= Sin (180-A) = Sin A

Thus Sin (HAI) = Sin A. Thus the area of triangle HAI is

½ * 5 * 7 * 12√6

35 = 6 √6

Use the above concept and find out the areas of all

the triangles, we can also find the areas of all the

squares, sum it up to get the answer. Answer will be

option d), none of these.

Alternate method:

We can easily find the area of the three squares and

if we do that, we will get a value which is greater than

the given options, letting us know that the answer is

option d)

9. In the figure, ABCD is a parallelogram with ∠ CAD

= 2 ∠CAB. The bisector of ∠CAD meets CD at E. If

AD = 12 and DE = 8, find AC

(a) 12 (b) 15

(c) 16 (d) 18

Solution:

∆ADE ||| ∆ABC

𝐴𝐷

𝐴𝐵 =

𝐷𝐸

𝐵𝐶 =

𝐴𝐸

𝐴𝐶

12

𝐴𝐵 =

8

12 AB = 18

In ADE (Angular bisector theorem)

𝐷𝐴

𝐷𝐸 =

𝐴𝐶

𝐶𝐸

Substitute the values to get the value of AC

10. Let PQ & RS be tangents at the extremities of the

diameter PR of a circle of radius r. If PS & RQ

intersect at a point X on the circumference of the

circle, then 2r equals.

(a) .PQ RS (b) 𝑃𝑄+𝑅𝑆

2

(c) 2 (𝑃𝑄.𝑅𝑆)

(𝑃𝑄+𝑅𝑆) (d) 2 2

2

PQ RS

Solution:

Figure not drawn to scale:



TC | CAT Geometry


In ∆QPR,

tan 90 – 𝜃 = 𝑃𝑄

𝑃𝑅

cot 𝜃 = 𝑃𝑄

𝑃𝑅

tan 𝜃 = 𝑅𝑆

𝑃𝑅

cot 𝜃 x tan 𝜃 = 𝑃𝑄

𝑃𝑅 x

𝑅𝑆

𝑃𝑅

1 = 𝑃𝑄 𝑥 𝑅𝑆

𝑃𝑅2

PR2 = PQ x RS

PR = Diameter, which is 2r, so answer is option a)

11. A square is inscribed in a quarter of a circle in

such manner that two of its adjacent vertices lie

on the radii at an equal distance from the

centre, while the other two vertices lie on the

circular arc. If the square has side of length x,

then the radius of the circle is.

(a) 16 / 4x (b) 2.5x

(c) 2x (d) 3

2x

Solution: Figure not drawn to scale

OA = radius

OA = √𝑥2 + 2𝑥2 = 𝑥√2.5

Answer is Option b) 12. ABC is an isosceles triangle right angled at B. A

square is inscribed inside the triangle with three

vertices of the square on three sides of the

triangles as shown in the adjacent figure. It is

known that the ratio x to y equal to 2 to 1.

The ratio of the area of the square to the area of

the triangle is equal to:

(a) 2 : 5 (b) 1 : 10

(c) 1 : 3 (d) 2 : 3

Solution: We draw a perpendicular GF from the

square on the side AB.



TC | CAT Geometry


∆ EFG and ∆ EBD are congruent (three

angles equal and one side equal)

⇒ EF = y and FG = x

Now FG = AF (∠FAG = ∠AGF = 450)

⇒ AF = x

Therefore one side of the right triangle = 2x

+ y. The other side is also equal to 2x + y ⇒

Area of the triangle = (2𝑥+𝑦)2

2

Area of the square = x2 + y2

Ratio = 𝑥2+𝑦2

(2𝑥+𝑦)2 = 2 : 5, Option a)

13. A quadrilateral is obtained by joining the

midpoints of the adjacent sides of the rhombus

ABCD with angle A = 60 degrees. This process of

joining midpoints of the adjacent sides is

continued infinitely. If the sum of the areas of all

the above said quadrilateral including the

rhombus ABCD is 64 3 sq. cm, what is the sum

of the perimeters of all the quadrilaterals

including the rhombus ABCD? (in cm) .

(a) 16 (5 + √3) (b) 16 (5 - √3)

(c) 16 (3 + √5) (d) 16 (2 + √3)

(e) 16 (3 - √2)

Solution:

Figure 1 In the above image, the topmost figure is the

point of concern. The outermost diagram is that of a rhombus and if we join the midpoints of this rhombus, we get a rectangle (parallelogram), joining the mid points of this rectangle, you will get another rhombus and this process continues. If A1 is the area of the outer rhombus, then the value will be a2 Sin ϴ, where a is the side of the rhombus and ϴ is the angle between the two sides. Also, the rectangle formed inside the rhombus will have its area equal to half of that of the rhombus (can be proven), and the rhombus formed inside this rectangle will have its area half of that of the rectangle, which in turn means ¼ the area of the outer Rhombus. Area will be proportional to the length of the sides, generally the product of the sides, here the sides are of equal length, If there are two geometrical figures whose sides are in the ratio 2:1, then their areas will be in the ratio 4:1. Let the length of the side of the outer rhombus be 4a. Thus

A1 = (4a)2sin60 = 8√3 a2

Similarly, if A2 is the area of the inner rhombus, then

A2 = 8√3 (2) a2 = 4√3 a2 (half of the outer rhombus)

A3 = (2a)2 sin60 = 2√3 a2

This will be an infinite GP, thus total area = 8√3 a2 +

4√3 a2 + 2√3 a2+…………..so on = 64√3



TC | CAT Geometry


Solve for a, to get the value of a as 2.

Therefore the value of the sum of all the perimeters will be = 4(4a + 2a + a+ a/2….so on)

The above expression talks about only the rhombus, but then we need to consider the rectangles that are formed inside these rhombus by joining their midpoints.

So the rectangle formed inside the first rhombus will

have sides as √3a and 2a, refer the smaller diagram in figure 1. The subsequent rectangles will have their areas equal to half of their outer counterparts.

The sum of all their perimerters will be

2(2√3a + 2a) + 2 (√3a + a)…. so on

Therefore the total sum will be: Perimeter = 4(4a +

2a ….) + 2(2√3a + 2a) + 2 (√3a + a)….

= 4(4𝑎)

1−1

2

+ 4(√3+1)𝑎

1−1/2

=32a + 8√3 a + 8a

= 40 a + 8√3 a

= 8a(5 + √3)

= 16 (5 + √3)

Option a)

14. In the given figure, EB is parallel & equal in length

to DC, the length of ED is equal to the length of

DC, the area of triangle ADC = 8 units, the area of

triangle BDC is 3 units. And angle DAB is a right

angle. Then the area of triangle AEB is

A

B C

DE

O

(a) 3 units (b) 5 units (c) 2 units (d) 8 units (e) none of these

Solution:

∆ADC = ½ (AD)h = 8 ⟹ (AD)h = 16

∆BDC = ½ (BC)h = 3 ⟹ (BC)h = 6 ⟹ (ED)h = 6

∆ ABE = ½ (AE)h = ½ [AD – ED]h = ½ (16 – 6) = 5

Option b)

15. A square ABCD is constructed inside a triangle

PQR having sides 10, 17 & 21 as shown in the fig.

Find the perimeter of the square ABCD.

A

D C

B

R

P

Q21

1017

Solution:

Check for the triplets, 17, 15 and 8

Draw a perpendicular from P to S on QR. PSQ is a

right angled triangle, 17 is the hypotenuse, so the

other two sides will be 15 and 8, hence QR =

QS+SR, where QS= 15 and SR = 6 so that QR = 21

PS = 8 units

Now Triangle PAB and PQR are similar, apply the

concept of similarity, write the ratios and you will

get the value of one side of the square ABCD and

thereby the perimeter, answer should be 23.2

units

16. In a triangle ABC, D, E, F & G are points taken on

the sides of the triangle such that

90BED BGC DFC . Further, AB = AC,

BG = 6cm, DF = 4cm & BE = 2cm. What is the

length in cm of BD?



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B D C

F

G

E

A

(a) 4 (b) 2√2 (c) √3 (d) √5

Solution:

∆BGC ||| DFC

𝐵𝐺

𝐷𝐹 =

𝐵𝐶

𝐶𝐷 =

3

2 Let BC = 3x, CD = 2x, Hence BD will

be x

Join E & D Now

BDE ||| CFD

𝐵𝐸

𝐹𝐶=

𝐸𝐷

𝐹𝐷=

𝐵𝐷

𝐶𝐷=

𝑋

2𝑋=

1

2

ED = 2

In triangle BED, E is 900, and BE = ED =2, it is a right

angled isosceles triangle, hence BD = 2√2

Option b)

17. The length (in cm) of the side of a 100 sided

convex polygon are 1, 2, 3, …..100. A circle of

radius r cm is inscribed in this polygon. What is

the area of the polygon?

(a) 2525r (b) 2515r

(c) 2540r (d) 2520r

Solution:

By adding the area of all the triangles youll get area of the polygon and the height of each ∆= 𝑟 (radius of the circle) and base = length of side of the polygon.

∴ Area of polygon = ½ [(1)𝑟 + 2(𝑟)…………100(𝑟)]

= 𝑟(100)(101)

2𝑥2

= 2525 𝑟

18. ∆ ABC is equilateral triangle with area A.

1 1 1PQ R is drawn such that point 1 1 1, ,P Q R

intersect the side AB, BC, & CA in the ratio 2 :1 ,

respectively. Then another 2 2 2PQ R is drawn

such that points 2 2 2P Q R intersect the side

1 1 1 1 1 1, &PQ Q R R P in the ratio 2 :1,

respectively. If the process is repeated to infinite

times, find the sum of the areas of all triangle,

such drawn (i.e.

1 1 1 2 2 2ABC PQ R PQ R ) .

(a) 2A (b) 1.5 A

(c) 3A (d) 4

3 A

Solution:

Let the side of the triangle be ‘a’

AB = BC = CA = a



TC | CAT Geometry


Since the sides are divided in the ratio

2:1, we can write, AR1 = a/3 and R1C = 2a/3

∆ ABC = √3

4 a2

, let this area be A

∆ R1Q1C = 1

2(𝑎

3)(

2𝑎

3) sin600 =

𝑎2√3

9𝑥2

∆ P1Q1R1 = √3

4 a2 – 3 (

𝑎2√3

9𝑥2) = √3a2 (1/4

– 1/6)

= √3𝑎2

12 = A/3 =

One third the area of the bigger triangle.

Simlarly, this follows for other triangles

∴ ∆ ABC + ∆ P1Q1R1 + ∆ P2Q2R2 …… = A +

A/3 + A/9…so on

= 𝐴

1−1/3 =

3

2 A

19. Find the area of the shaded region in the figure,

if the side of the largest square is 4 cm & the

other squares are the largest squares that can be

drawn inside the respective quarter circle?

(a) 26.4 cm (b)

28 cm

(c) 24.4 cm (d)

212 cm

Solution:

If a circle is inscribed inside a square, ratio of area of

square to circle = 4:𝜋

On the other hand, if a square is inscribed inside a

circle, ratio of area of square to circle = 2:𝜋

In the figure below, area of inner square is half that

of the outer square, and this relation holds good for

the circles as well.

You now have a quarter square and circle inscribed

in it. Let us assume that we have a full square and a

circle inscribed in it.

Ao (complete squares) = 64

We know that 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑆𝑞𝑢𝑎𝑟𝑒

𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 =

4

𝜋 Area of

Circle = 𝜋

4 * Area of Square

Area of Circle = 16𝜋. Therefore the area of the remaining region is 64 - 16 𝜋. We are looking at ¼ of



TC | CAT Geometry


the total figure, hence the remaining region will be 1

4

(64 - 16 𝜋).

In fact the subsequent figures will also have the remaining region in the same order, i.e. they will be half of their preceding diagram

64 - 16 𝜋 1

4 (64 - 16 𝜋).

32 - 8 𝜋 1

4 (32 - 8 𝜋)

16 - 4 𝜋 ¼ (16 - 4 𝜋)

8– 2 𝜋 ¼ (8– 2 𝜋 )

The question is about sum of all the values on the

right side, which is 120−30𝜋

4 = 6.4

Answer is Option a)

20. A regular hexagon is circumscribed by a rectangle

such that all the six corners of the hexagon lie on

the rectangle. What is the ratio of the area of the

largest possible equilateral triangle that can be

cut from the rectangle to the area of the

hexagon?

(a) 3: 2 (b) 2:3

(c) 3 :1 (d) 2: 3

Solution:

The biggest triangle you can fit within the

rectangle is one that has its base identical to one

of the sides of the rectangle and the third vertex

lies on the opposite side.

There are enough proofs available for the same.

Here the question is equilateral triangle of

maximum area, so we need to find the longer

side of the rectangle so that, it will become the

base of the triangle.

You can see from the above figure where there is

a hexagon of side ‘a’ units, and if a rectangle is

circumscribed, the length of the largest side will

be 2a, hence the area of this triangle will be

√3

4(2a)2. And the area of the hexagon is 6 *

√3

4(a)2

Taking their ratio we get, 2:3, option b)

21. There is a square field with three cows tethered

at three different vertices of the fields. The ropes

used to thither the cows are all of the same

length. Each of the cows can reach the other two

cows & the ropes used have the minimum

possible length. If the area of the square field is2784m , what is the area of the field that cannot

be accessed by any of the three cows?

(a) 284m (b) 2168m

(c) 2161m (d) 2322m

Solution:

A, B and C are the positions where three cows are

tethered, A and C or A and B can meet each other at



TC | CAT Geometry


the mid point of one of the sides of the square field.

However for B and C to meet, it can happen only at

D, the area is given as 784, hence, each side is 28,

and CD = 14 √2 units, in fact this will be the length

of the rope to which all three cows are tethered. We

need to find out the area of the required region,

which is equal to Area of the triangle – [Area of two

sectors].

Area of each sector will be ½ * r2 * ϴ, here ϴ is

the angle subtended and it is in radians. Here ϴ will

be 45 degrees and it is written as 𝜋

4 radians.

Find out the area of the triangle and sectors,

substitute to get the answer as 84, option a)

22. A Ball of diameter 13cm is floating so that top of

the ball is 4cm above the smooth surface of pond.

What is the circumference (in cm) of the circle

formed by the contact of water surface with the

ball?

(a) 10 (b) 11

(c) 12 (d) 24

Solution:

The above figure summarizes the situation and

the answer is 12 , option c)

23. cot-1 [√1−sin 𝛼+√1+sin 𝛼

√1−𝑠𝑖𝑛 𝛼 −√1+𝑠𝑖𝑛 𝛼 ] =

(a) 2𝜋 – a (b) 𝜋 - 1

2𝑎 (c)

1

2𝑎 - 3𝜋

(d) None of these

Solution:

1 – sin 𝑎 = (cos2 𝑎

2 + sin2

𝑎

2)2

Similarly,

∴ 1 + sin 𝑎 = (cos 𝑎

2 - sin

𝑎

2)2

∴ cot-1 [√1−sin 𝑎+√1+sin 𝑎

√1−𝑠𝑖𝑛 𝑎 −√1+𝑠𝑖𝑛 𝑎 ]

= cot-1 (𝑐𝑜𝑠

𝑎

2 −sin

𝑎

2 +cos

𝑎

2 +𝑠𝑖𝑛

𝑎

2

𝑐𝑜𝑠𝑎

2 −𝑠𝑖𝑛

𝑎

2 −𝑐𝑜𝑠

𝑎

2 −sin

𝑎

2

)

= cot-1 (−𝑐𝑜𝑡𝑎

2)

= cot-1 [𝑐𝑜𝑡 (𝜋 −𝑎

2)]

= 𝜋 – 𝑎

2

Hence, Option b).

24. In triangle ABC, D and E are any points on AB

and AC such AD = AE. The bisector of angle C

meets DE at F. It is known that angle B = 600

What is the value of angle DFC?

Solution:



TC | CAT Geometry


(Figure not drawn accordingly)

∠B = 600 ⇒ ∠A + ∠C = 1200

In the figure 𝜃 is the external angle of ∆FEC

⇒ ∠EFC + 𝐶

2

𝜃 = 180−𝐴

2 = 90 -

𝐴

2 = ∠EFC +

𝐶

2

⇒ EFC = 90 – (𝐴

2+

𝐶

2) = 90 – 60 = 300

25. In triangle PQR, points X, Y and Z are on PQ, PR

and QR, respectively, such that PX = XQ, 𝑅𝑌

𝑌𝑃 =

𝑎

𝑏,

and 𝑄𝑍

𝑍𝑅 = 3. Also (area ∆PXY)2 = (area ∆QXZ) x

(area ∆RYZ)

The ratio of a: b is

(a) 𝟑+√𝟏𝟎𝟓

𝟔 (b)

𝟐+√𝟑𝟓

𝟔

(c) 𝟑+√𝟑𝟏

𝟑 (d)

√𝟏𝟎𝟓−𝟑

𝟔

Solution:

Area of a triangle = ½ * a * b * Sinϴ, Using that

concept, we can get

𝐴𝑟𝑒𝑎 ∆𝑃𝑋𝑇

𝐴𝑟𝑒𝑎 ∆𝑃𝑄𝑅 =

𝑏

2(𝑎+𝑏)

𝐴𝑟𝑒𝑎 ∆𝑅𝑍𝑌


𝑎

4(𝑎+𝑏)

𝐴𝑟𝑒𝑎 ∆𝑄𝑋𝑍


3

8

(𝑏

2(𝑎+𝑏))2 =

3

8 x

𝑎

4(𝑎+𝑏)

Solving the above to get 𝑎

𝑏. You will come across a

quadratic equation, on simplifying that you will get

the answer

26. In the following figure, PS bisects ∠QPR. The

area of ∆PQS = 40 sq. cm. and PR is 2.5 times of

PQ. Find the area of ∆PQR.

a) 35 sq. cm

b) 140 sq. cm

c) 70 sq. cm

d) 105 sq. cm

Solution:



TC | CAT Geometry


PS is the angle bisector of ∠OPR

⇒ 𝑃𝑄

𝑃𝑅 =

𝑄𝑆

𝑆𝑅 SR = 2.5QS

⇒ Area ∆PSR = 2.5 x 40 = 100

⇒ Area of ∆PQR = 100 + 40 = 140 sq. cm

27. In triangle ABC, AB = 10, AC = 7, and BC = 8. How

much should we slide it along AB so that the

area of the overlapping region (the shaded

triangle A’BD) is one – half the area of the

triangle ABC?

Solution:

Figure not drawn to scale

∆A ‘DB’ and ∆ACB’ are similar

⇒ 𝐴𝑟𝑒𝑎 ∆𝐴′𝐵′𝐷

𝐴𝑟𝑒𝑎 ∆𝐴𝐵′𝐶 =

(𝐴′𝐵′)2

(𝐴𝐵′)2 ⇒

𝐴′𝐵′

𝐴𝐵′ =

1

√2

⇒ 𝐴′𝐵′= 5√2 ⇒ AA’ = 10 - 5√2

28. Three circles are drawn touching each other,

their centers lying on a straight line. The line PT

is 16 units long and is tangent to the two smaller

circles, with points P and T lying on the larger

circle.

The area inside the largest circle but outside the

smaller two circles is equal to

a) 4𝜋

b) 8𝜋

c) 16𝜋

d) 32𝜋

Solution:

Let A, C, B be the centers of the smallest to the

largest circle respectively.

All O, A, B, C, O’ would be collinear.

Let r1, r2, r3 be the radii off circle 1, 2, 3

respectively.

So r, 2r1 + 2r2 = 2r3

r1 + r2 = r3

As PQ is tangent to both the circles in side and a

chord of circle 3 so D will be the midpoint of PQ.

Now, OD. DO’ = PD. DQ (Result that holds in

case of chords inside a circle)

2r1. 2r2 = 8 . 8

r1r2 = 16

Required area = 𝜋(r1+r2)2 – r12 – r2

2)

= 𝜋(2r1r2)

= 32 𝜋

29. Three squares of side lengths 3, 5 and 8 are kept

side by side. A corner of the smallest square is

joined to a corner of the biggest square, as

shown in the figure.



TC | CAT Geometry


What is the area of the shaded region?

Solution:

In ∆’s ABE & ADG,

the ∆’s are similar

∴ 𝐴𝐵

𝐵𝐸 =

𝐴𝐷

𝐷𝐺

3

BE =

16

8 ⇒ BE =

3

2 cms

Also ∆ABE - ∆ACF

Hence 𝐴𝐵

𝐵𝐸 =

𝐴𝐶

𝐶𝐹 ⇒

3

3/2 =

8

𝐶𝐹

CF = 4

∴ Area of EBFC = Area of ACF – Area of ABE

= 1

2.8.4 -

1

2 . 3.

3

2

= 16 - 9

4 = 13.75

30. Square ABCD is inscribed inside a circle. Another

square is inscribed between square ABCD and

the circle such that its two vertices are on the

circle and one side lies along AB, as shown in the

figure.

The ratio of the length of the sides of

the smaller square and bigger square is

a) 1/5

b) 2/7

c) 3/8

d) 4/9

Solution:

Let O be the centre of circle

Let the sides of bigger & smaller squares be x & y

cm.

Hence in ∆OAB

AB = y + 𝑥

2, OA =

𝑦

2, OB =

𝑥

√2

⇒ (𝑦

2)2 + (

𝑥

2 + y)2 = (

𝑥

√2)2

Solving we get 𝑥

𝑦 =

5

1



TC | CAT Geometry


31. A warship and a submarine (completely

submerged in water) are moving horizontally in a

straight line. The captain of the warship observes

that the submarine makes an angle of depression

of 300, and the distance between them from the

point of observation is 50 km. After 30 minutes,

the angle of depression becomes 600.

Find the distance between then after 30 min

from the initial point of reference.

Solution:

Angle of Elevation:

Consider an object that is at a higher level than

the point from which it is observed. Then the

angle that the line of sight makes with the

horizontal is called the angle of elevation.

Angle of Depression:

Consider an object that is at a lower level than

the point from which it is observed. Then the

angle that the line of sight makes with the

horizontal is called the angle of depression.

For this question, refer the below figure:

The vertical distance between the warship and

the submarine = 25 km

∴ When the angle of depression of the submarine

is 600, the distance between the warship and the

submarine is 25

𝑠𝑖𝑛600 = 50

√3 km

32. If both are moving in same direction and the

submarine is ahead of the warship in both the

situations, then the speed of the warship, if the

ratio of the speed of warship to that of the

submarine is 2:1, is:

Solution:

Let the speed of the submarine be x kmph.

Then the speed of the warship is 2x kmph.

Distance travelled by the warship in 30 minutes =

x km

Distance travelled by the submarine in 30

minutes = 0.5x km.

From the figure,

𝑥 = 25√3 + 0.5𝑥 - 25

√3

∴ 0.5√3𝑥 = 5

∴ x = 100

√3 kmph

∴ The speed of the warship = 200

√3 kmph



TC | CAT Geometry


33. Adil is standing on top of a railway bridge

watching an approaching train. The angle of

depression of the topmost point of the start of

the train as it comes into sight is 300. It changes

to 450 by the time it comes to a halt. The height

of the train is 10 feet and the bridge is 60 feet tall.

Find the horizontal distance between the point

that the train comes into sight and the point at

which it halts.

Neglect Adil’s height.

Solution:

From the figure, JP = 50 feet.

∠ JQP = 450

tan 450 = JP

PQ

∴ PQ = 50 feet

Also, ∠JKP = 300

tan 300 = JP

KP

∴ KP = 50√3

Required distance = KQ = KP – PQ

= 50√3 – 50 = 50 (√3 – 1) feet



Documents

TC | CAT Geometry Geometry 4 - Solutions the radius of the circle is. (a) x16 / 4S (b) x 2.5 (c) x 2 (d) 3 2 x Solution: Figure not drawn to scale OA = radius OA = √𝑥2+2𝑥2