**Complete Geometry Previous Questions (CAT PYQs 2017-22)**

Geometry is one of the most important topics in the CAT **Quants section**. If you’re weak in Geometry questions for CAT, make sure you learn all the basic concepts of solving the questions and also **all the important formulas**. Here, you can learn all the **important questions on the concept of CAT Geometry**. You can check out these CAT Geometry questions from the **CAT Previous year’s papers**. This post will look at important **Geometry questions** in the CAT quant section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this CAT **TSD** **Questions** **PDF **along with the detailed solutions (and video solutions) below, which is completely Free.

Geometry is a crucial component of the CAT (Common Admission Test) examination, and mastering it is key to achieving success. If you’re a CAT aspirant looking to sharpen your geometry skills and excel in the 2023 CAT exam, you’re in the right place. In this blog post, we have compiled a comprehensive set of Geometry questions from the **CAT papers of 2017 to 2022**. These questions are not just any Geometry problems; they are the real deal, giving you a taste of what to expect on the exam day. Let’s dive into the world of CAT Geometry, dissecting past year questions, and equipping you with the knowledge and strategies to tackle them confidently.

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**Question 1:Â **The area of the closed region bounded by the equation I x I + I y I = 2 in the two-dimensional plane is

a)Â $4\pi$ sq. units

b)Â 4Â sq. units

c)Â 8Â sq. units

d)Â $2\pi$Â sq. units

**1)Â AnswerÂ (C)**

**Solution:**

The following equation will form a square of side $ 2\sqrt{2} $.

The area of the square = $(2\sqrt{2})^2$ = 8 units.

**Question 2:Â **From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is

a)Â $225\sqrt{3}$

b)Â $\frac{500}{\sqrt{3}}$

c)Â $\frac{275}{\sqrt{3}}$

d)Â $\frac{250}{\sqrt{3}}$

**2)Â AnswerÂ (B)**

**Solution:**

The lengths are given as 40, 25 and 35.

The perimeter = 100

Semi-perimeter, s = 50

Area = $ \sqrt{50 * 10 * 25 * 15}$ = $250\sqrt{3}$

The triangle formed by the centroid and two vertices is removed.

Since the cenroid divides the median in the ratio 2 : 1

The remaining area will be two-thirds the area of the original triangle.

Remaining area = $\frac{2}{3} * 250\sqrt{3}$ = $\frac{500}{\sqrt{3}}$

**Question 3:Â **Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

a)Â $9\pi-18$

b)Â 18

c)Â $9\pi$

d)Â 9

**3)Â AnswerÂ (B)**

**Solution:**

The image of the figure is as shown.

AB = AC = 6cm. Thus, BC = $\sqrt{6^2 + 6^2}$ = 6âˆš2 cm

The required area = Area of semi-circle BQC – Area of quadrant BPC + Area of triangle ABC

__Area of semicircle BQC__

Diameter BC = 6âˆš2cm

Radius = 6âˆš2/2 = 3âˆš2 cm

Area = $\pi r^2 $/2 = $ \pi$ * $(3 \sqrt{2})^2 $/2 = 9$\pi$

__Area of quadrant BPC__

Area = $\pi r^2$/4 = $\pi*(6)^2$/4 = 9$\pi$

__Area of triangle ABC__

Area = 1/2Â * 6Â * 6 = 18

The required area = Area of semi-circle BQC – Area of quadrant BPC + Area of triangle ABC

=Â 9$\pi$ –Â 9$\pi$ + 18 = 18

**Question 4:Â **A solid metallic cube is melted to form five solid cubes whose volumes are in the ratio 1 : 1 : 8 : 27 : 27. The percentage by which the sum of the surface areas of these five cubes exceeds the surface area of the original cube is nearest to

a)Â 10

b)Â 50

c)Â 60

d)Â 20

**4)Â AnswerÂ (B)**

**Solution:**

Let the volumes of the five cubes be x, x, 8x, 27x and 27x.

Let the sides be a, a, 2a, 3a and 3a. ($x = a^3$)

Let the side of the original cube be A.

$A^3 = x + x + 8x + 27x + 27x$

$A = 4a$

Original surface area = $96a^2$

New surface area = $6(a^2 + a^2 + 4a^2 + 9a^2 + 9a^2)$ = $144a^2$

Percentage increase = $\frac{144 – 96}{96}*100 =$ 50%

**Question 5:Â **A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is $9\pi$ cubic centimeters. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is

**5)Â Answer:Â 6**

**Solution:**

The volume of a cylinder is $\pi * r^2 * 3 = 9 \pi$

r = $\sqrt{3}$ cm.

Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder.

Lets draw the figure.

Based on the Pythagoras theorem, the other leg will be 1 cm.

Thus, the height will be 3 + 1 + 2 = 6 cm

**Question 6:Â **Let ABC be a right-angled triangle with BC as the hypotenuse. Lengths of AB and AC are 15 km and 20 km, respectively. The minimum possible time, in minutes, required to reach the hypotenuse from A at a speed of 30 km per hour is

**6)Â Answer:Â 24**

**Solution:**

The length of the altitude from A to the hypotenuse will be the shortest distance.

This is a right triangle with sides 3 : 4 : 5.

Hence, the hypotenuse = $\sqrt{20^2+15^2}$ = 25 Km.

Length of the altitude = $\frac{15 * 20}{25}$ = 12 Km

(This is derived from equating area of triangle, $\frac{15\cdot20}{2}\ =\ \frac{25\cdot\text{altitude}}{2}$)

The time taken = $\frac{12}{30} * 60$ = 24 minutes

**Question 7:Â **The shortest distance of the point $(\frac{1}{2},1)$ from the curve y = I x -1I + I x + 1I is

a)Â 1

b)Â 0

c)Â $\sqrt{2}$

d)Â $\sqrt{\frac{3}{2}}$

**7)Â AnswerÂ (A)**

**Solution:**

The graph of the given function is as shown below.

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

**Question 8:Â **Let ABCDEF be a regular hexagon with each side of length 1 cm. The area (in sq cm) of a square with AC as one side is

a)Â $3\sqrt{2}$

b)Â $3$

c)Â $4$

d)Â $\sqrt{3}$

**8)Â AnswerÂ (B)**

**Solution:**

The length of the diagonals of a regular hexagon with side s are $\sqrt{3}s$.

Here length of AC =

$\sqrt{3}s$ =Â $\sqrt{3}$ cms

Hence area of the square = $\sqrt{3}^2$ = 3 sq cm

**Question 9:Â **The base of a vertical pillar with uniform cross section is a trapezium whose parallel sides are of lengths 10 cm and 20 cm while the other two sides are of equal length. The perpendicular distance between the parallel sides of the trapezium is 12 cm. If the height of the pillar is 20 cm, then the total area, in sq cm, of all six surfaces of the pillar is

a)Â 1300

b)Â 1340

c)Â 1480

d)Â 1520

**9)Â AnswerÂ (C)**

**Solution:**

See the diagram below

Length of side AD = $\sqrt{12^2 + 5^2 } = 13$

Area of the trapezium = 12 * (10 + 20)/2 = 180

Perimeter of the trapezium = 10 + 20 + 13 + 13 = 56

Area of the sides of the pillar = 56 * heightÂ = 56 * 20 = 1120.

Total are of the pillar = 1120 + area of base + area of the top = 1120 + 180 + 180 = 1480

**Question 10:Â **The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y =3x+c,then c is

a)Â -5

b)Â -6

c)Â -7

d)Â -8

**10)Â AnswerÂ (D)**

**Solution:**

The midpoint of one diagonal lies on the other diagonal.

Midpoint is ((2+6)/2, (5+3)/2) = (4,4)

Hence 4 = 3 * 4 + c => c = -8

**Question 11:Â **ABCD is a quadrilateral inscribed in a circle with centre O such that O lies inside the quadrilateral. If $\angle COD = 120$ degrees and $\angle BAC = 30$ degrees, then the value of $\angle BCD$ (in degrees) is

**11)Â Answer:Â 90**

**Solution:**

$\angle COD = 120$ => $\angle CAD = 120/2 = 60$ (The angle subtended by the chord DC at the major arc is half the angle subtended at the centre of the circle.)

$\angle BAC = 30$

$\angle BAD = \angle BAC + \angle CAD$ = 30 + 60 = 90.

$\angle BCD = 180 – \angle BAD$ = 180 – 90 = 90

**Question 12:Â **If three sides of a rectangular park have a total length 400 ft, then the area of the park is maximum when the length (in ft) of its longer side is

**12)Â Answer:Â 200**

**Solution:**

Let the length and breadth of the park be l,b, l > b

Case 1: 2l + b = 400

Area = lb. Area is maximum when 2l * b is maximum, which is maximum when 2l = b (using AM $ \geq $ GM inequality) => l = 100, b = 200. Which can’t happen since l > b

Case 2: l + 2b = 400

Area = lb. Area is maximum when l *2 b is

maximum, which is maximum when l = 2b (using AM $ \geq $ GM

inequality) => l = 200, b = 100.

Hence length of the longer side is 200 ft

**Question 13:Â **Let P be an interior point of a right-angled isosceles triangle ABC with hypotenuse AB. If the perpendicular distance of P from each of AB,BC,and CA is $4(\sqrt{2}-1)$ cm,then the area, in sq cm, of the triangle ABC is

**13)Â Answer:Â 16**

**Solution:**

Let the length of non-hypotenuse sides of the right angled triangle be $a$. Then the hypotenuse h = $\sqrt{2}a$

P is equidistant from all the side of the triangle. Hence P is the incenter and the perpendicular distance is the inradius.

In a right angled triangle, inradius = $\frac{a + b – h}{2}$

=> $\frac{a + a – \sqrt{2}a}{2} = 4(\sqrt{2}-1)$

=> $ \sqrt{2}a( \sqrt{2} – 1) = 8(\sqrt{2} -1)$

=> $ a = 4\sqrt{2}$

Area of the triangle = $\frac{1}{2}a^2$ = 16 sq cm

**Question 14:Â **In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

a)Â $\sqrt{13}$

b)Â $\sqrt{14}$

c)Â $\sqrt{11}$

d)Â $\sqrt{12}$

**14)Â AnswerÂ (A)**

**Solution:**

Given thatÂ two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.

In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.

We can see that M and N are the mid points of AB and CD respectively.

AM = 3 cm and CD = 2 cm. Let ‘OM’ be x cm.

In right angle triangle AMO,

$AO^2 = AM^2 + OM^2$

$\Rightarrow$Â $AO^2 = 3^2 + x^2$Â … (1)

In right angle triangle CNO,

$CO^2 = CN^2 + ON^2$

$\Rightarrow$ $CO^2 = 2^2 + (OM+MN)^2$

$\Rightarrow$Â $CO^2 = 2^2 + (x+1)^2$Â … (2)

We know that both AO and CO are the radius of the circle. HenceÂ $AO =Â CO$

Therefore, we can equate equation (1) and (2)

$3^2+x^2$=$2^2+(x+1)^2$

$\Rightarrow$ x = 2 cm

Therefore, the radius of the circle

$AO = \sqrt{AM^2 + OM^2}$

$\Rightarrow$Â $AO=\sqrt{3^2+2^2}=\sqrt{13}$. Hence, option A is the correct answer.

**Question 15:Â **Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

a)Â 24, 10

b)Â 25, 9

c)Â 25, 10

d)Â 24, 12

**15)Â AnswerÂ (A)**

**Solution:**

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

We know that AC is the diameter and $\angle$ ABC = 90Â°. AC = 2*13 = 26 cm

In right angle triangle ABC,

$AC^2 = AB^2 + BC^2$

$\Rightarrow$Â $AB^2+BC^2=26^2$

$\Rightarrow$Â $AB^2+BC^2=676$

Let us check with the options.

Option (A): $24^2+10^2 = 676$. Hence, this is a possible answer.

Option (B): $25^2+9^2 = 706Â \neq 676$.Â Â Hence, this is an incorrect pair.

Option (C): $25^2+10^2 = 725 \neq 676$.Â Hence, this is an incorrect pair.

Option (D): $24^2+12^2 = 720 \neq 676$.Â Hence, this is an incorrect pair.

Therefore, we can say that option A is the correct answer.

**Question 16:Â **Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

a)Â 3 : 8

b)Â 2 : 5

c)Â 4 : 9

d)Â 1 : 3

**16)Â AnswerÂ (D)**

**Solution:**

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x :Â 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that ‘x+1’ is the length of side of square ABCD.

Area of square ABCD = $(x+1)^2$ sq. units.

Therefore, area of square EFGH = $\dfrac{62.5}{100}*(x+1)^2$ = $\dfrac{5(x+1)^2}{8}$ … (1)

In right angle triangle EBF,

$EF^2 = EB^2 + BF^2$

$\Rightarrow$Â $EF = \sqrt{1^2+x^2}$

Therefore, the area of squareÂ EFGH = $EF^2$ =Â $x^2+1$ … (2)

By equating (1) and (2),

$x^2+1 =Â \dfrac{5(x+1)^2}{8}$

$\Rightarrow$ $8x^2+8 = 5x^2+10x+5$

$\Rightarrow$ $3x^2-10x+3 = 0$

$\Rightarrow$ $(x – 3)(3x – 1) = 0$

$\Rightarrow$ $x = 3$ or $1/3$

The ratio of length of EB to that of CG = 1 : x

EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer.

**Question 17:Â **Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,… will be

a)Â $188\sqrt{3}$

b)Â $248\sqrt{3}$

c)Â $164\sqrt{3}$

d)Â $192\sqrt{3}$

**17)Â AnswerÂ (D)**

**Solution:**

We can see that T$_{2}$ is formed by using the mid points of T$_{1}$. Hence, we can say that area of triangle of T$_{2}$ will be (1/4)th of the area of triangle T$_{1}$.

Area of triangle T$_{1}$ = $\dfrac{\sqrt{3}}{4}*(24)^2$ = $144\sqrt{3}$ sq. cm

Area of triangle T$_{2}$ = $\dfrac{144\sqrt{3}}{4}$ = $36\sqrt{3}$ sq. cm

Sum of the area of all triangles =Â Â T$_{1}$ +Â T$_{2}$ + T$_{3}$ + …

$\Rightarrow$Â T$_{1}$ +Â T$_{1}$/$4$ +Â T$_{1}$/$4^2$ + …

$\Rightarrow$ $\dfrac{T_{1}}{1 – 0.25}$

$\Rightarrow$ $\dfrac{4}{3}*T_{1}$

$\Rightarrow$ $\dfrac{4}{3}*144\sqrt{3}$

$\Rightarrow$ $192\sqrt{3}$

Hence, option D is the correct answer.

**Question 18:Â **A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With $\pi$ = 22/7, the volume, in cubic ft, of the remaining part of the cone is

**18)Â Answer:Â 198**

**Solution:**

We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft

In triangle OAB and OCD

$\dfrac{OA}{AB} =Â \dfrac{OC}{CD}$

$\Rightarrow$ AB = $\dfrac{3*4}{12}$ = 1 ft.

Therefore, the volume of remaining part = Volume of entire cone –Â Volume of smaller cone

$\Rightarrow$ $\dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$

$\Rightarrow$ $\dfrac{1}{3}*\pi*189$

$\Rightarrow$ $\dfrac{22}{7*3}*189$

$\Rightarrow$ $198$Â cubic ft

**Question 19:Â **In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

a)Â $32\sqrt{3}$

b)Â $18\sqrt{3}$

c)Â $24\sqrt{3}$

d)Â $12\sqrt{3}$

**19)Â AnswerÂ (A)**

**Solution:**

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

Given that area of parallelogram =Â 72 sq cm

Area of triangle ABC = $\dfrac{1}{2}$*area of parallelogram

(1/2)*AB*BC*sinABC =Â $\dfrac{1}{2}$*72

sinABC =Â $\dfrac{1}{2}$

$\angle$ ABC = 30Â°

Let us draw a perpendicular CQ from C to AB.

$\angle\ B\ =\ \angle\ D\ =\ 30^{\circ\ }$ ( opposite angles in a parallelogram)

$\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$Â ( Sum of angles in a triangle = 180)

In triangles APD and CQB, AP= CQ , AD=CB ( opposite sides of a parallelogram) ,Â $\angle\ QCB\ =\ \angle\ DAP\ =\ 60^{\circ\ }$

Hence APD and CQB are congruent triangles using SAS property.

Therefore, we can say that area of triangle APD =Â area of triangle CQB

In right angle triangle CQB,

QB = CBcos30Â° = $16*\dfrac{\sqrt{3}}{2}$ = $8\sqrt{3}$ cm

CQ = CBsin30Â° = $16*\dfrac{1}{2}$ = $8$Â cm

Therefore, area ofÂ triangle CQB = 1/2*CQ*QB = $1/2*8*8\sqrt{3}$ =Â $32\sqrt{3}$

Hence, we can say thatÂ area of triangle APD =Â $32\sqrt{3}$.

**Question 20:Â **In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

a)Â $ (\frac{\pi}{4\sqrt{3}})^{\frac{1}{2}}$

b)Â $(\frac{\pi}{6})^{\frac{1}{2}}$

c)Â $(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$

d)Â $(\frac{\pi}{4})^{\frac{1}{2}}$

**20)Â AnswerÂ (C)**

**Solution:**

It is given that radius of the circle = 1 cm

Chord AB subtends an angle of 60Â° on the centre of the given circle.Â R be the region bounded by the radii OA, OB and the arc AB.

Therefore, R = $\dfrac{60Â°}{360Â°}$*Area of the circle =Â $\dfrac{1}{6}$*$\pi*(1)^2$ = $\dfrac{\pi}{6}$ sq. cm

It is given that OC = OD andÂ area of triangle OCD is half that of R. Let OC = OD = x.

Area of triangle COD = $\dfrac{1}{2}*OC*OD*sin60Â°$

$\dfrac{\pi}{6*2}$ = $\dfrac{1}{2}*x*x*\dfrac{\sqrt{3}}{2}$

$\Rightarrow$ $x^2 = \dfrac{\pi}{3\sqrt{3}}$

$\Rightarrow$ $x$ = $(\frac{\pi}{3\sqrt{3}})^{\frac{1}{2}}$ cm. Hence, option C is the correct answer.

**Question 21:Â **A parallelogram ABCD has area 48 sqcm. If the length of CD is 8 cm and that of AD is s cm, then which one of the following is necessarily true?

a)Â $sâ‰ 6$

b)Â $s \geq 6$

c)Â $5 \leq s \leq 7$

d)Â $s \leq 6$

**21)Â AnswerÂ (B)**

**Solution:**

We can see that area of parallelogram ABCD = 2*Area of triangle ACD

48 = 2*Area of triangle ACD

Area of triangle ACDÂ = 24

$(1/2)*CD*DA*sinADC=24$

$AD*sinADC=6$

We know that $sin\theta$ $\leq$ 1, Hence, we can say that AD $\geq$ 6

$\Rightarrow$ s $\geq$ 6

**Question 22:Â **A chord of length 5 cm subtends an angle of 60Â° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120Â° at the centre of the same circle is

a)Â $5\sqrt{3}$

b)Â $2\pi$

c)Â $8$

d)Â $6\sqrt{2}$

**22)Â AnswerÂ (A)**

**Solution:**

We are given that AB = 5 cm and $\angle$ AOB = 60Â°

Let us draw OM such that OM $\perp$ AB.

In right angle triangle AMO,

$sin 30Â° = \dfrac{AM}{AO}$

$\Rightarrow$ AO = 2*AM = 2*2.5 = 5 cm. Therefore, we can say that the radius of the circle = 5 cm.

In right angle triangle PNO,

$sin 60Â° = \dfrac{PN}{PO}$

$\Rightarrow$ PN = $\dfrac{\sqrt{3}}{2}$*PO = $\dfrac{5\sqrt{3}}{2}$

Therefore, PQ = 2*PN = $5\sqrt{3}$ cm

**Question 23:Â **The area of a rectangle and the square of its perimeter are in the ratio 1 âˆ¶ 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio

a)Â 1:4

b)Â 2:9

c)Â 1:3

d)Â 3:8

**23)Â AnswerÂ (A)**

**Solution:**

Let ‘a’ and ‘b’ be the length of sides of the rectangle. (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

$\Rightarrow$ $\dfrac{a*b}{(2*(a+b))^2}=\dfrac{1}{25}$

$\Rightarrow$ $25ab=4(a+b)^2$

$\Rightarrow$ $4a^2-17ab+4b^2=0$

$\Rightarrow$ $(4a-b)(a-4b)=0$

$\Rightarrow$ $a = 4b$ or $\dfrac{b}{4}$

We initially assumed that a > b, thereforeÂ a $\neq$ $\dfrac{b}{4}$.

Hence, a = 4b

$\Rightarrow$ b : a = 1 : 4

**Question 24:Â **On a triangle ABC, a circle with diameter BC is drawn, intersecting AB and AC at points P and Q, respectively. If the lengths of AB, AC, and CP are 30 cm, 25 cm, and 20 cm respectively, then the length of BQ, in cm, is

**24)Â Answer:Â 24**

**Solution:**

Let us draw the diagram according to the available information.

We can see that triangle BPC and BQC are inscribed inside a semicircle. Hence, we can say that

$\angle$ BPC = $\angle$Â BQC = 90Â°

Therefore, we can say that BQÂ $\perp$ AC and CP $\perp$ AB.

In triangle ABC,

Area of triangle = (1/2)*Base*Height =Â (1/2)*AB*CP =Â (1/2)*AC*BQ

$\Rightarrow$ BQ = $\dfrac{AB*CP}{AC}$ =Â $\dfrac{30*20}{25}$ = 24 cm.

**Question 25:Â **A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is

a)Â $8$ units

b)Â $4$ units

c)Â $2\sqrt{2}$ units

d)Â $4\sqrt{2}$ units

**25)Â AnswerÂ (B)**

**Solution:**

We know that area of the triangle = 32 sq. units, BC = 8 units

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle.

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units.

**Question 26:Â **From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72Ï€ sq cm is removed. The perimeter of the leftover portion, in cm, is

a)Â 80 + 16Ï€

b)Â 86 + 8Ï€

c)Â 88 + 12Ï€

d)Â 82 + 24Ï€

**26)Â AnswerÂ (C)**

**Solution:**

Area of the semicircle with AB as a diameter = $\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$

$\Rightarrow$ $\dfrac{1}{2}*\pi*(\dfrac{AB^2}{4})$ = $72*\pi$

$\Rightarrow$ $AB = 24 cm$

Given that area of the rectangle ABCD = 768 sq.cm

$\Rightarrow$ AB*BC = 768

$\Rightarrow$ BC = 32 cm

We can see that the perimeter of the remaining shape = AD + DC + BC + Arc(AB)

$\Rightarrow$ 32+24+32+$\pi*24/2$

$\Rightarrow$ $88+12\pi$

**Question 27:Â **Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

a)Â $\frac{1}{\surd2}$

b)Â $\frac{\pi}{3}$

c)Â $\surd2$

d)Â 1

**27)Â AnswerÂ (D)**

**Solution:**

Let ‘h’ be the height of the triangle ABC, semiperimeter(S) $= \frac{4+4+r+4+4+r}{2} = 8+r$,

$a=4+r, b=4+r, c=8$

Area of triangle ABCÂ $=\ \ \sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$

$= \sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$ =Â $\ \frac{\ 1}{2}\times\ \left(4+4\right)\times\ height$

Height (h) =Â $\sqrt{\ \left(8+r\right)r}$

Now, $ h + r = 4 \longrightarrowÂ Â \sqrt{\ \left(8+r\right)r} + r = 4$ (Considering the height of the triangle)

$\sqrt{\ \left(8+r\right)r}$=4-r

16r=16

r=1

Alternatively,

$\text{AE}^@+\text{EC}^2=\text{AC}^2 \longrightarrowÂ 4^2+\left(4-r\right)^2 =Â \left(4+r\right)^2Â \longrightarrow\ \longrightarrow\ \longrightarrow\ r=1$

**Question 28:Â **The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

a)Â 12

b)Â $10\surd2$

c)Â $8\surd3$

d)Â $5\surd5$

**28)Â AnswerÂ (B)**

**Solution:**

It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square

AB is the height of the equilateral triangle i.e 10$\sqrt{\ 3}$

AOB is a right angle, so applying the Pythagorean formula, we get

$OA^2+ OB^2= AB^2$

$h^2$ + 100= 300

h=10$\sqrt{\ 2}$

**Question 29:Â **Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then the maximum possible length of AP, in cm, is

a)Â 10

b)Â 5

c)Â $8\surd2$

d)Â $6\surd2$

**29)Â AnswerÂ (A)**

**Solution:**

Let p be the length of AP.

It is given thatÂ $\angle\ BAC\ =90\ $ andÂ $\angle\ APC\ =90\ $

LetÂ $\angle\ ABC\ =\theta\ $, thenÂ $\angle\ BAP\ =90-\theta\ $ andÂ $\angle\ BCA\ =90-\theta\ $

SoÂ $\angle\ PAC\ =\theta\ $

Triangles BPA and APC are similar

$p^2=x\left(20-x\right)$

We have to maximize the value of p, which will be maximum when x=20-x

x=10

**Question 30:Â **A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is

a)Â $1026(1 + \pi)$

b)Â $8464\pi$

c)Â $928\pi$

d)Â $1044(4 + \pi)$

**30)Â AnswerÂ (A)**

**Solution:**

It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)

=27

The number of iron cylinders =Â $\ \frac{\ 405}{27}$ = 15

The number of aluminium cylinders =Â $\ \frac{\ 783}{27}$ = 29

The number of copper cylinders =Â $\ \frac{351\ }{27}$ = 13

15*$\pi\ r^2h=$ 405

15*$\pi\ 9*h=$ 405

$\pi\ h=3$

Now we have to calculate the total surface area of all the cylinders

Total number of cylinders = 15+29+13 = 57

Total surface area of the cylinder = 57*($2\pi\ rh+2\pi\ r^2$)

=57(2*3*3 + 2*9*$\pi\ $)

=$1026(1 + \pi)$

**Question 31:Â **In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

a)Â 78

b)Â 80

c)Â 72

d)Â 68

**31)Â AnswerÂ (C)**

**Solution:**

It is given that AD and BE are medians which are perpendicular to each other.

The lengths of AD and BE are 12cm and 9cm respectively.

It is known that the centroid G divides the median in the ratio of 2:1

Area of $\triangle$ ABC = 2* Area of the triangle ABD

Area of $\triangle\ $ABD = Area ofÂ $\triangle\ $ AGB + Area ofÂ $\triangle\ $ BGD

SinceÂ $\angle\ AGB\ =\ \angle\ BGD\ =\ 90$ (Given)

Area of $\triangle\ $ AGB =Â $\ \frac{\ 1}{2}\times\ 8\times\ 6$ = 24

Area of $\triangle\ $ BGD =Â $\ \frac{\ 1}{2}\times\ 6\times\ 4$ = 12

Area of $\triangle\ $ABD = 24+12=36

Area ofÂ $\triangle\ ABC\ =\ 2\times\ 36=72$

**Question 32:Â **Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each interior angle of B is $\frac{3}{2}$ times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is

**32)Â Answer:Â 150**

**Solution:**

Each interior angle in an n-sided polygon =Â $\frac{\left(n-2\right)180\ }{n}$

It is given thatÂ each interior angle of B is $\frac{3}{2}$ times each interior angle of A and b = 2a

$\frac{\left(b-2\right)180\ }{b}$ =Â $\ \frac{\ 3}{2}\times\ $Â $\frac{\left(a-2\right)180\ }{a}$

$2\times\ \left(b-2\right)\times\ a\ =\ 3\times\ \left(a-2\right)\times\ b$

2(ab-2a) = 3(ab-2b)

ab-6b+4a=0

a*2a-12a+4a=0

$2a^2-8a=0$

a(2a-8) = 0

a cannot be zero so 2a=8

a=4, b = 4*2=8

a+b = 12

Each interior angle of a regular polygon with 12 sides =Â $\ \frac{\ \left(12-2\right)\times\ 180}{12}$

=150

**Question 33:Â **AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

a)Â 9.3

b)Â 7.8

c)Â 9.1

d)Â 8.5

**33)Â AnswerÂ (C)**

**Solution:**

Since AB is a diameter,Â AQB and APB will right angles.

In right triangle APB, AP =Â $\sqrt{10^2-6^2}=8$

Now, 2AQ=APÂ => AQ= 8/2=4

In right triangle AQB,Â AP = $\sqrt{10^2-4^2}=9.165$ =9.1 (Approx)

**Question 34:Â **In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

a)Â 2.5

b)Â 1.5

c)Â 3.5

d)Â 0.5

**34)Â AnswerÂ (D)**

**Solution:**

In figure AE*BE=CE*DEÂ Â (**The Â ** **intersecting chords theorem**)

=> 7*15= x(20.5-x)Â Â Â Â Â Â Â Â Â (Assuming AE=x)

=> 210=x(41-2x)

=>Â $2x^2$-41x+210=0

=> x=10 or x=10.5Â Â => AE=10 or AE=10.5Â Â Â Hence BE = 20.5-10=10.5Â or BE = 20.5-10.5=10

Required difference= 10.5-10=0.5

**Question 35:Â **Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

a)Â 2 : 3

b)Â 4 : 5

c)Â 5 : 6

d)Â 3 : 4

**35)Â AnswerÂ (A)**

**Solution:**

The given figure can be divided into 9 regions or equilateral triangles of equal areas as shown below,

Now the hexagon consists of 6 regions and the triangle consists of 9 regions.

Hence the ratio of areas = 6/9 =2:3

**Question 36:Â **If the rectangular faces of a brick have their diagonals in the ratio $3 : 2 \surd3 : \surd{15}$, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

a)Â $\sqrt{3} : 2$

b)Â $1 : \sqrt{3}$

c)Â $2 : \sqrt{5}$

d)Â $\sqrt{2} : \sqrt{3}$

**36)Â AnswerÂ (B)**

**Solution:**

Assuming the dimensions of the brick are a, b and c and the diagonals areÂ 3, 2 $\surd3$ and $\surd{15}$

Hence,Â $a^{2\ }+\ b^2$ =Â $3^2$Â Â ……(1)

$b^{2\ }+\ c^2$ = $(2\sqrt{3})^2$Â ……(2)

$c^{2\ }+\ a^2$ = $(\sqrt{15})^2$Â ……(3)

Adding the three equations, 2($a^2+b^2+c^2$) = 9+12+15=36

=>$a^2+b^2+c^2$ = 18……(4)

Subtracting (1) from (4), we get $c^2$ = 9Â Â =>c=3

Subtracting (2) from (4), we get $a^2$ = 6Â Â =>a=$\sqrt{6}$

Subtracting (3) from (4), we get $b^2$ = 3Â Â =>b=$\sqrt{3}$

The ratio of the length of the shortest edge of the brick to that of its longest edge is =Â $\ \frac{\ \sqrt{3}}{3}$ =Â $1 : \sqrt{3}$

**Question 37:Â **Let T be the triangle formed by the straight line 3x + 5y – 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is

**37)Â Answer:Â 9**

**Solution:**

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$\ \frac{\ 1}{2}\ $* the length of the hypotenuse = $\ \frac{\ 1}{2}$($\sqrt{\ 15^2+9^2}$) =Â $\ \frac{\ 1}{2}\ $*$\sqrt{\ 306}$ =Â $\ \frac{\ 1}{\ 2}\times\ $17.49 = 8.74

Hence, the integer close to L = 9

**Question 38:Â **A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

a)Â 243

b)Â 232

c)Â 256

d)Â 264

**38)Â AnswerÂ (A)**

**Solution:**

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$

Volume of upper cone =Â $\frac{\pi}{3}.r^2.9$

Difference=Â $\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$ =>Â $\frac{\pi}{3}\cdot r^2=1$

Volume of larger cone =Â $\frac{\pi}{3}\cdot9r^2\cdot27=243$

**Question 39:Â **On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is

a)Â $3\sqrt{\pi}(5+\frac{12}{\pi})$

b)Â $4\sqrt{\pi}(3+\frac{9}{\pi})$

c)Â $3\sqrt{\pi}(\frac{5}{2}+\frac{6}{\pi})$

d)Â $5\sqrt{\pi}(3+\frac{9}{\pi})$

**39)Â AnswerÂ (A)**

**Solution:**

Let ABCD be the rectangle with length 2l and breadth 2b respectively.

Area of the circle i.e. area of painted region =Â $\pi\ b^2$.

Given, 4lb-$\pi\ b^2$=(2/3)$\pi\ b^2$.

=> 4lb=(5/3)$\pi\ b^2$.

=>l=$\frac{5\pi}{12}b$.

Given, 4lb=135 => 4*$\frac{5\pi}{12}b^2$=135 => b=Â $\frac{9}{\sqrt{\ \pi\ }}$

=> l=$\frac{15}{4}\sqrt{\ \pi\ }$

Perimeter of rectangle =4(l+b)=4($\frac{15}{4}\sqrt{\ \pi\ }$+$\frac{9}{\sqrt{\ \pi\ }}$ )=$3\sqrt{\pi}(5+\frac{12}{\pi})$.

Hence option A is correct.

**Question 40:Â **A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

a)Â $\frac{6\pi}{25}$

b)Â $\frac{5\pi}{18}$

c)Â $\frac{3\pi}{25}$

d)Â $\frac{2\pi}{15}$

**40)Â AnswerÂ (A)**

**Solution:**

Let the length of radius be ‘r’.

From the above diagram,

$x^2+r^2=6^2\ $….(i)

$\left(10-x\right)^2+r^2=8^2\ $—-(ii)

Subtracting (i) from (ii), we get:

x=3.6 =>Â $r^2=36-\left(3.6\right)^2$ ==>Â $r^2=36-\left(3.6\right)^2\ =23.04$.

Area of circle =Â $\pi\ r^2=23.04\pi\ $

Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

.’. Ratio of areas = 23.04$\pi\ $/96=$\frac{6\pi}{25}$

**Question 41:Â **The points (2,1) and (-3,-4) are opposite vertices of a parallelogram.If the other two vertices lie on the line $x+9y+c=0$, then c is

a)Â 12

b)Â 13

c)Â 15

d)Â 14

**41)Â AnswerÂ (D)**

**Solution:**

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint ofÂ (2,1) and (-3,-4) lie onÂ $x+9y+c=0$

midpoint of (2,1) and (-3,-4) = ($\frac{2-3}{2},\frac{1-4}{2}$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

**Question 42:Â **In a trapezium $ABCD$, $AB$ is parallel to $DC$, $BC$ is perpendicular to $DC$ and $\angle BAD=45^{0}$. If $DC$ = 5cm, $BC$ = 4 cm,the area of the trapezium in sq cm is

**42)Â Answer:Â 28**

**Solution:**

Given, BC = DE = 4

CD = BE = 5

In triangle ADE, EAD=45^{0}$

$\tan\ 45\ =\ \frac{DE}{AE}$ => AE = 4

Area of trapezium = Area of rectangle BCDE + Area of triangle AED

= 20 + 8 = 28

**Question 43:Â **The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is

a)Â $\frac{14\pi}{3}$

b)Â $\frac{123\pi}{7}$

c)Â $\frac{12\pi}{5}$

d)Â $\frac{205\pi}{9}$

**43)Â AnswerÂ (D)**

**Solution:**

Equation of circleÂ $x^2+y^2+2gx+2fy+c=0$

It passes throughÂ (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $c=0$

=> On substituting the value (4,0) in the above equation, we obtain:Â $16+0+8g+0 = 0$ ; $g=-2$

=>Â On substituting the value (3,9) in the above equation, we obtain: $9+81-12+18f = 0$ ;Â $f= -13/3$

Radius of the circle **rÂ **=Â $\sqrt{\ g^2+f^2-c}$ =>Â $r^2=\frac{205}{9}$

Therefore, Area =Â $\pi\ r^2=\frac{205\pi\ }{9}$

**Question 44:Â **Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C thatÂ passes through points A and B where A is located 4 meters north of O and B is locatedÂ 3 meters east of O. Then, the length of PQ, in meters, is nearest to

a)Â 8.8

b)Â 7.8

c)Â 6.6

d)Â 7.2

**44)Â AnswerÂ (A)**

**Solution:**

We can form the following figure based on the given information:

Since OA = 4 m and OB=3 m; AB = 5 m. OR bisects the chord into PC and QC.

Since AB = 5 m, we have $a+b = 5Â Â Â Â …(i)$Â Also,Â $4^2\ -k^2=a^2…\left(ii\right)$ andÂ $3^2\ -k^2=b^2…\left(iii\right)$

Subtracting (iii) from (ii), we get:Â $a^2\ -b^2=7…\left(iv\right)$

Substituting (i) inÂ (iv), we get $a – b = 1.4Â Â Â Â …(v)$;Â $\left[\left(a+b\right)\left(a\ -b\right)=7;\ \therefore\ \left(a-b\right)=\frac{7}{5}\right]$

Solving (i) and (v), we obtain the value of $a=3.2$ and $b=1.8$

Hence,Â $k^2\ =\ 5.76$

Moving on to the larger triangleÂ $\triangle\ POC$, we haveÂ $5^2-k^2=\left(x+a\right)^2$;

Substituting the previous values, we get:Â $(25-5.76)=\left(x+3.2\right)^2$

$\sqrt{19.24}=\left(x+3.2\right)$Â or $x = 1.19 m$

Similarly, solving for y usingÂ $\triangle\ QOC$, we get $y=2.59 m$

Therefore, $PQ = 5+2.59+1.19 = 8.78Â \approx\ 8.8 m$

Hence, Option A is the correct answer.

**Question 45:Â **The sum of the perimeters of an equilateral triangle and a rectangle is 90cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfying the relationship $R=T^{2}$. If the sides of the rectangle are in the ratio 1:3, then the length, in cm, of the longer side of the rectangle, is

a)Â 27

b)Â 21

c)Â 24

d)Â 18

**45)Â AnswerÂ (A)**

**Solution:**

Let the sides of the rectangleÂ be “a” and “3a” m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be “m” cm. Hence the perimeter of the equilateral triangle is “3m” cm. Now we know that 8a+3m=90……(1)

Moreover area of the equilateral triangle isÂ $\frac{\sqrt{\ 3}}{4}m^2$ and area of the rectangle isÂ $3a^2$

According to the relation givenÂ $\left(\frac{\sqrt{\ 3}}{4}m^2\right)^{^2}=\ 3a^2$

$\frac{3}{16}m^4=\ 3a^2\ or\ a^2=\frac{m^4}{16}$

$a=\frac{m^2}{4}$

Substituting this in (1) we getÂ $2m^2+3m-90\ =0$ solving this we get m=6 (ignoring the negative value since side can’t be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

**Question 46:Â **Let C1 and C2 be concentric circles such that the diameter of C1 is 2cm longer than that of C2. If a chord of C1 has length 6cm and is a tangent to C2, then the diameter, in cm, of C1 is

**46)Â Answer:Â 10**

**Solution:**

Now we know that the perpendicular from the centre to a chord bisects the chord.Â Hence at the point of intersection of tangent, the chord will be divided into two parts of 3 cm each. As you can clearly see in the diagram, a right angled triangle is formed there.

HenceÂ Â $\left(r+1\right)^2=r^2+9$ orÂ $r^{2\ }+1+2r\ =\ r^2+9\ or\ 2r=8\ or\ r=4cm$

Hence the radius of the larger circle is 5cm and diameter is 10cm.

**Question 47:Â **From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

a)Â $\frac{\sqrt{3}s^{2}}{2}$

b)Â $\frac{2s^{2}}{\sqrt{3}}$

c)Â $\frac{s^{2}}{2\sqrt{3}}$

d)Â $\frac{s^{2}}{\sqrt{3}}$

**47)Â AnswerÂ (D)**

**Solution:**

Based on the question:Â AD, CE and BF are the three altitudes of the triangle. It has been stated that {GD+GE+GF = s}

Now since the triangle is equilateral, let the length of each side be “a”. So area of triangle will be

$\frac{1}{2}\times\ GD\times\ a+\ \frac{1}{2}\times\ GE\times\ a\ +\frac{1}{2}\times\ GF\times\ a=\frac{\sqrt{\ 3}}{4}a^2$

NowÂ $GD+GE+GF=\frac{\sqrt{\ 3}a}{2}$ orÂ $s=\frac{\sqrt{\ 3}a}{2}\ or\ a=\frac{2s}{\sqrt{\ 3}}$

Given the area of the equilateralÂ triangle =Â $\ \frac{\sqrt{3}}{4}a^2\ $ ; substituting the value of ‘a’ from above, we get the area {in terms ‘s’}=Â $\frac{s^2}{\sqrt{3}}$

**Question 48:Â **If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

a)Â $4\sqrt{6}$

b)Â $6\sqrt{6}$

c)Â $\sqrt{6}$

d)Â $2\sqrt{6}$

**48)Â AnswerÂ (D)**

**Solution:**

Area of a regular hexagon =Â $\frac{3\sqrt{3}}{2}x^2$

Area of an equilateral triangle =Â $\frac{\sqrt{3}}{4}\left(a\right)^2$ ; where a = side of the triangle

Since the area of the two figures are equal, we can equate them as folllows:Â $\frac{3\sqrt{3}}{2}x^2=\frac{\sqrt{3}}{4}\left(12\right)^2$

On simplifying:Â $x^2=24\ $

$\therefore\ x=2\sqrt{6}$

**Question 49:Â **Suppose the length of each side of a regular hexagon ABCDEF is 2 cm.It T is the mid point of CD,then the length of AT, in cm, is

a)Â $\sqrt{13}$

b)Â $\sqrt{14}$

c)Â $\sqrt{12}$

d)Â $\sqrt{15}$

**49)Â AnswerÂ (A)**

**Solution:**

Since a regular hexagon can be considered to be made up of 6 equilateral triangles, a line joining the farthest vertices of a hexagon can be considered to be made up using the sides of two opposite equilateral triangle forming the hexagon. Hence, its length should be twice the side of the hexagon, in this case, 4 cm.

Now, AD divided the hexagon into two symmetrical halves. Hence, AD bisects angle D, and hence, angle ADC isÂ $60^{\circ\ }$.

We can find out the value of AT using cosine formula:

$AT^2=\ 4^2+1^2-2\times\ 1\times\ 4\cos60$

$AT^2=\ 17-4=13$

$AT=\sqrt{\ 13}$

**Question 50:Â **A circle of diameter 8 inches isÂ inscribed in a triangle ABC where $\angle ABC = 90^\circ$. If BC = 10 inches then the area of the triangle in square inches is

**50)Â Answer:Â 120**

**Solution:**

We know that Inradius=$\frac{\left(Perpendicular+Base-Hypotenuse\right)}{2}$

$4=\frac{\left(p+10-h\right)}{2}$

h-p = 2 or h= p+2.

Now,Â $p^2+100\ =\ h^2$

$p^2+100\ =\ \left(p+2\right)^2$

$p^2+100\ =\ p^2+4p+4$

4p = 96

p=24.

Hence, Area =Â $\frac{1}{2}\times\ 10\times\ 24\ =\ 120$.

**Question 51:Â **If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

a)Â $\sqrt{37}+\sqrt{13}$

b)Â $\sqrt{13}+\sqrt{12}$

c)Â $\frac{\sqrt{37}+\sqrt{13}}{2}$

d)Â $\frac{\sqrt{13}+\sqrt{12}}{2}$

**51)Â AnswerÂ (A)**

**Solution:**

All the sides of the rhombus are equal.

The area of a rhombus isÂ $12\ cm^2$

Considering d1 to be the length of the longer diagonal, d2 to be the length of the shorter diagonal.

The area of a rhombus isÂ $\left(\frac{1}{2}\right)\left(d1\right)\cdot\left(d2\right)\ =\ 12$

d1*d2 = 24.

The length of the side of a rhombus is given byÂ $\frac{\sqrt{\ d1^2+d2^2}}{2}$. This is because the two diagonals and a side from a right-angled triangle with sides d1/2, d2/2 and the side length.

$\frac{\sqrt{\ d1^2+d2^2}}{2}=\ 5$

HenceÂ $\sqrt{\ d1^2+d2^2}\ =\ 10$

$d1^2+d2^2\ =\ 100$

Using d1*d2 = 24, 2*d1*d2 = 48.

$d1^2+d2^2\ +2\cdot d1\cdot d2=\ 100+48\ =\ 148$

$d1^2+d2^2\ -2\cdot d1\cdot d2=\ 100-48\ =\ 52$

$d1+d2\ =\ \sqrt{\ 148}$Â (1)

d1-d2 =Â $\sqrt{52}$Â Â (2)

(1) + (2)= 2*(d1) = 2*($\sqrt{\ 37}+\sqrt{\ 13}$)

d1 =Â $\sqrt{\ 37}+\sqrt{\ 13}$

or

In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.

The area of a Rhombus is :Â $\left(\frac{1}{2}\right)\cdot\left(2a\right)\cdot\left(2b\right)\ =\ 12$

ab =6.

The length of each side is :Â $\sqrt{\ a^2+b^2}$ = 5,Â $a^2+b^{2\ }=\ 25,\ $

$\left(a+b\right)^2=\ 37,\ \left(a+b\right)\ =\ \sqrt{\ 37}$

($\left(a-b\right)^2=\ 13,\ a-b\ =\ \sqrt{\ 13}$

$2a\ =\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$,Â $2b\ =\ \left(\sqrt{\ 37}-\sqrt{\ 13}\right)$.

2a is longer diagonal which is equal toÂ $\ \left(\sqrt{\ 37}+\sqrt{\ 13}\right)$

**Question 52:Â **The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is

a)Â 30

b)Â 40

c)Â 25

d)Â 20

**52)Â AnswerÂ (A)**

**Solution:**

We are given that :

Let DP =x

So AB =x

Now DP=CP

So CD = 2x

Now let height of trapezium be h

we can say A(Parallelogram ABPD ) = xh

And A (BPC) =$\frac{1}{2}xh$

Now by condition$xh-\frac{1}{2}xh=10$

$\frac{xh}{2}=10$

so xh =20

Now therefore area of trapezium ABCD = $\frac{1}{2}\left(x+2x\right)h\ =\ \frac{3}{2}xh\ =\ 30$

**Question 53:Â **Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

**53)Â Answer:Â 30**

**Solution:**

We have :

Now area of ADE =Â $\frac{1}{2}\times\ AD\times\ AE\times\ \sin A$

=$\frac{1}{2}\times\ 2x\times\ 2y\times\ \sin A=8$

we get xy sinA =4

Now Area of triangle ABC = $\frac{1}{2}AB\times\ AC\times\ \sin A$

we get $\frac{1}{2}\times3x\ \times5y\ \sin A=\frac{15}{2}xy\ \sin A=\ \frac{15}{2}\times\ 4$

we get Area of ABC = 30

**Question 54:Â **The cost of fencing a rectangular plot is â‚¹ 200 per ft along one side, and â‚¹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

a)Â 120000

b)Â 90000

c)Â 100000

d)Â 160000

**54)Â AnswerÂ (A)**

**Solution:**

Let us draw the rectangle.

Now, definitely, three sides should be fenced at Rs 100/ft, and one side should be fenced at Rs 200/ft.

In this question, we are going to assume that the L is greater than B.

Hence, the one side painted at Rs 200/ft should be B to minimise costs.

Hence, the total cost = 200B + 100B + 100L + 100L = 300B + 200L

Now, L x B = 60000

B = 60000/L

Hence, total cost = 300B + 200L = 18000000/L + 200L

To minimise this cost, we can use AM>=GM,

$\frac{\frac{18000000}{L}+200L}{2}\ge\sqrt{\ \frac{18000000}{L}\times\ 200L}$

$\frac{18000000}{L}+200L\ge2\sqrt{\ 18000000\times\ 200}$

$\frac{18000000}{L}+200L\ge2\times\ 60000$

Hence, minimum cost = Rs 120000.

**Question 55:Â **A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of â‚¹125 per m, is

**55)Â Answer:Â 3500**

**Solution:**

We can say 40m is the perimeter of the park

so side of rhombus = 10

NowÂ $\frac{1}{2}\times\ d_1\times\ d_2\ =\ 96$

so we getÂ $\ d_1\times\ d_2\ =\ 192$Â Â (1)

And as we know diagonals of a rhombus are perpendicular bisectors of each other :

soÂ $\ \frac{d_1^2}{4}+\ \frac{d_2^2}{4}=\ 100$

so we getÂ $\ d_1^2+\ d_2^2=\ 400$Â Â Â (2)

Solving (1) and (2)

We get $d_1=12$ and$d_2=16$

Now the cost, in INR, of laying electric wires along its two diagonals, at the rate of â‚¹125 per m, is= (12+16)(125) =3500

**Question 56:Â **If a triangle ABC, $\angle BCA=50^{0}$. D and E are points on $AB$ and $AC$, respectively, such that $AD=DE$. If F is a point on $BC$ such that $BD=DF$, then $\angle FDE$, in degrees, is equal to

a)Â 72

b)Â 80

c)Â 100

d)Â 96

**56)Â AnswerÂ (B)**

**Solution:**

We need to find out p.

Angle ADE = 180 – 2x

Angle BDF = 180 –Â 2y

Now, 180 – 2y + p + 180 – 2x = 180 [Straight line = 180 deg]

p = 2x + 2y – 180

Also, x + y + 50 = 180 [Sum of the angles of triangle = 180]

x + y = 130

p = 260 – 180 = 80 degrees.

**Question 57:Â **Let ABCD be a parallelogram. The lengths of the side AD and the diogonal AC are 10cm and 20cm, respectively. If the angle $\angle ADC$ is equal to $30^{0}$ then the area of the parallelogram, in sq.cm is

a)Â $\frac{25(\sqrt{5}+\sqrt{15})}{2}$

b)Â $25(\sqrt{3}+\sqrt{15})$

c)Â $\frac{25(\sqrt{3}+\sqrt{15})}{2}$

d)Â ${25(\sqrt{5}+\sqrt{15})}$

**57)Â AnswerÂ (B)**

**Solution:**

Applying cosine rule in triangle ACD,

$100+X^2-2\times\ 10\times\ X\cos30=400$

$X^2-10X\sqrt{\ 3}-300=0$

Solving, we get X =Â $\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)$

Hence, area = 10Xsin 30 = $\frac{\left(\frac{10\sqrt{\ 3}+10\sqrt{\ 15}}{2}\right)10}{2}$

= $25(\sqrt{3}+\sqrt{15})$

**Question 58:Â **A trapezium $ABCD$ has side $AD$ parallel to $BC, \angle BAD = 90^\circ, BC = 3$ cm and $AD= 8$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is

**58)Â Answer:Â 66**

**Solution:**

CD =Â $\sqrt{\ y^2+25}$

$11+y+\sqrt{y^2+25}=36$

$\sqrt{y^2+25}=25-y$

$y^2+25=25^2+y^2-50y$

2y = 24

y = 12

Area of trapezium =Â $3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$

**Question 59:Â **All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is

a)Â $\frac{P^2}{16} – R^2$

b)Â $\frac{P^2}{8} – 2R^2$

c)Â $\frac{P^2}{2} – 2PR$

d)Â $\frac{P^2}{8} – \frac{R^2}{2}$

**59)Â AnswerÂ (B)**

**Solution:**

$A=lb$

$l^2+b^2=4r^2$

$P\ =2\left(l+b\right)$

$\frac{P}{2}=l+b$

Squaring on both the sides, we get

$\frac{P^2}{4}=l^2+b^2+2lb$

$\frac{P^2}{4}=4r^2+2lb$

$\frac{P^2}{8}-2r^2=lb$

The answer is option B.

**Question 60:Â **Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (âˆ’2, 8), respectively. Then, the coordinates of the vertex D are

a)Â (0, 11)

b)Â (4, 5)

c)Â (âˆ’3, 4)

d)Â (âˆ’4, 5)

**60)Â AnswerÂ (D)**

**Solution:**

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC =Â $\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$

Let the coordinates of vertex D be (x,y)

$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$

x = -4 and y = 5

The answer is option D.

**Question 61:Â **In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $\angle BAC = 45^{\circ}$ and $\angle ABC=\theta\ $, then $\frac{AD}{BE}$ equals

a)Â $\sqrt{2} \cos \theta$

b)Â $\frac{(\sin \theta + \cos \theta)}{\sqrt{2}}$

c)Â 1

d)Â $\sqrt{2} \sin \theta$

**61)Â AnswerÂ (D)**

**Solution:**

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x

In right-angled triangle ABD, it is given $\angle ABC=\theta\ $

$\sin\theta=\frac{AD}{AB}\ $

$\sin\theta=\frac{AD}{x\sqrt{\ 2}}\ $

$\sqrt{\ 2}\sin\theta=\frac{AD}{BE}\ $

The answer is option D.

**Question 62:Â **Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

**62)Â Answer:Â 10**

**Solution:**

Let the number of sides of polygons A and B be n and 2n, respectively.

$\ \dfrac{\ \ \dfrac{\ \left(n-2\right)180}{n}}{\ \dfrac{\ \left(2n-2\right)180}{2n}}=\dfrac{3}{4}$

$\ \dfrac{\ \ \ n-2}{\ \ n-1}=\dfrac{3}{4}$

$\ \ \ \ 4n-8=3n-3$

$n=5$

The number of sides of polygon B is 2*5, i.e. 10.

**Question 63:Â **The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

a)Â $\sqrt{8}$

b)Â $\sqrt{6}$

c)Â $\sqrt{7}$

d)Â $\sqrt{5}$

**63)Â AnswerÂ (C)**

**Solution:**

Area of triangle ABD is twice the area of triangle ACD

$\angle\ ADB=\theta\ $

$\frac{1}{2}\left(AD\right)\left(BD\right)\sin\theta\ =2\left(\frac{1}{2}\left(AD\left(CD\right)\sin\left(180-\theta\ \right)\right)\right)$

$BD\ =2CD$

Therefore, BD = 2 and CD = 1

$\angle\ ABC=\angle\ ACB=60^{\circ\ }$

Applying cosine rule in triangle ADC, we get

$\cos\angle\ ACD=\ \frac{\ AC^2+CD^2-AD^2}{2\left(AC\right)\left(CD\right)}$

$\frac{1}{2}=\ \frac{\ 9+1-AD^2}{6}$

$AD^2=7$

$AD=\sqrt{\ 7}$

The answer is option C.

**Question 64:Â **Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is

**64)Â Answer:Â 9**

**Solution:**

Area of ABD : Area of BDC = 1:1

Therefore, area of ABD = 54

Area of ADE : Area of EDB = 1:1

Therefore, area of ADE = 27

O is the centroid and it divides the medians in the ratio of 2:1

Area of BEO : Area of EOD = 2:1

Area of EOD = 9

**Question 65:Â **The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

a)Â 3

b)Â 4

c)Â 6

d)Â 5

**65)Â AnswerÂ (D)**

**Solution:**

Sum of the three sides of a quadrilateral is greater than the fourth side.

Therefore, let the fourth side be

1+2+4>d or d<7

1+2+d>4 or d>1

Possible values of d are 2, 3, 4, 5 and 6.

**Question 66:Â **In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through Band C. Then the area, in sq. cm, of the overlapping region between the two circles is

a)Â $16\pi$

b)Â $16(\pi – 1)$

c)Â $32(\pi – 1)$

d)Â $32\pi$

**66)Â AnswerÂ (C)**

**Solution:**

BC is the diameter of circle C2 so we can say thatÂ $\angle BAC=90^{\circ\ }$ as angle in the semi circle isÂ $90^{\circ\ }$

Therefore overlapping area =Â $\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1

AB= AC = 8 cm and asÂ $\angle BAC=90^{\circ\ }$, so we can conclude that BC=Â $8\sqrt{2}$ cm

Radius of C2 = Half of length ofÂ BC =Â $4\sqrt{2}$ cm

Area of C2 =Â $\pi\left(4\sqrt{2}\right)^2=32\pi$Â $cm^2$

A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm

Area of the minor sector made be BC in C1 =Â $\frac{1}{4}$(Area of circle C1) – Area of triangle ABC =Â $\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$Â $cm^2$

Therefore,

Overlapping area between the two circles= $\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1

=Â $\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$Â $cm^2$