Free Geometry Set II - 01 Practice Test - CAT

Question 1

A square has two of its vertices on a circle and the other two are on the tangent to the circle. Find the area of the square if the radius of the circle is 5 units.

A. 64 sq units

B. 36 sq units

C. 50 sq units

D. 100 sq units

SOLUTION

Solution : A

Area of the square: (x + 5) $^{2}$

So the answer has to be a perfect square: 36 or 64

If area is 36 then x = 1; so EF = 6, OE = 1, OA = 5 and EA = 4

Angle OFB = ABF = BAE = 90, so angle OEA = 90

OE, EA and OA do not form Pythagoras triplet.

If area is 64 then x = 3, EF = 8, EA = 4, OA = 5

Shortcut 1

Reverse Gear approach-With a radius of 5, we will get a square of side 5%undefined2. area will be 50, but we know the side has to be slightly greater than 5%undefined2. hence, choose an option which is slightly greater than 50; i.e. 64

Shortcut 2

Graphical division:- The square gets divided into 8 parts. 4 parts are each right angled triangles of sides 3,4,5 and hence with area 6. 4 of them in total have an area of 24. The other 4 are triangles of the form 4, 5, $\sqrt{41}$ and whose areas are 10 each. Total area = (6+10)*4 = 64.

Question 2

What is the maximum number of times x circles of the same size can intersect each other?

A. 2x

B. (x - 1)

C. x(x-1)

D. x(x+1)/2

SOLUTION

Solution : C

Deducing a Pattern:

Taking two circles. The number of points of intersection is 2 at most.

If u consider 3 circles, the number of points of intersection = 6 ( the third circles will have 4 points of intersection with the other 2 circles and those two circles will have 2 points together).

Shortcut:- Assumption & Reverse Gear:

Taking two circles. The number of points of intersection is 2 at most. From this itself, option a, b and d can be eliminated giving answer c.

Question 3

The interior angles of a polygon are in AP. If the smallest angle is 120 $^{\circ}$ and the common difference is 5 $^{\circ}$ , then find the possible number of sides of that polygon?

A. 16

B. 9

C. Both a and b

D. None of these

SOLUTION

Solution : B

Sum of the interior angles of a polygon = (n-2) 180

$\frac{n}{2}$ * (2a+ (n-1)d)= (n-2)180

$\frac{n}{2}$ * [2*(120) + (n-1)5]=(n-2)180

n[48+(n-1)] = (n-2)72

n $^{2}$ -25n + 144=0

n=9 and n=16

when n= 16, the greatest angle will be equal to a+15d = 120 + 15 x 5 = 195 and no interior angle of a

polygon can be equal to or greater than 180. hence answer =b

Question 4

Concentric circles are drawn with radii 1, 2, 3 … 100. The interior of the smallest circle is colored black and the annular regions are colored alternately red and black, so that no two adjacent regions are the same color. The total area of the red regions divided by the area of the largest circle is

\frac{1}{2}

\frac{51}{100}

\frac{101}{200}

\frac{50}{101}

SOLUTION

Solution : C

Areas of circles follow the sequence 1,4,9,16,…………100 $^{2}$

Areas of Black & Red annuals – 1,3,5,7,9……………199 – General term 2n – 1

Red annuals – 3, 7, 11, …………….. 199 Sum = $(\frac{50}{2})$ *(3 + 199)

Largest circle’s area = 100 $^{2}$

Ratio = $\frac{101}{200}$

Note : Ignore pi as it is a ratio of areas

Shortcut: Reverse gear approach –

We know that the denominator is 100 $^{2}$ and num is an integer, hence the denominator cannot be 101 as it is in no-way a factor of 10000. Hence option d) has been eliminated.

We need to find $\frac{R}{(R + B)}$ . R + B = 10000. Each of the red annual is greater in area by 2 units than the subsequent black annual and there are 50 such cases. Hence R = 5050 and B = 4950. $\frac{R}{(R + B)}$ . = $\frac{5050}{10000}$ = $\frac{101}{200}$

Question 5

There is a circle of radius 10 units that circumscribes an equilateral triangle. A quadrilateral is drawn by joining mid-points of two adjacent sides of the triangle. Find the area of the quadrilateral

\frac{225 \sqrt{3}}{2}

\frac{225 \sqrt{3}}{4}

\frac{225 \sqrt{3}}{8}

\frac{225 \sqrt{3}}{12}

SOLUTION

Solution : B

In an equilateral triangle circum radius = $\frac{2}{3}$ h,

10 = $\frac{2}{3}$ * $\frac{\sqrt{3}}{2} S$

Side = 10$\sqrt 3$ units.

Area of Triangle = $\frac{\sqrt{3}}{4} (10 \sqrt{3})^{2}$

By graphical division, we need $\frac{3}{4}$ $^{t h}$ of the area of the triangle = $\frac{3}{4} \times \frac{\sqrt{3}}{4} \times (10 \sqrt{3})^{2}$ = $\frac{225 \sqrt{3}}{4}$

Question 6

Find the side of a piece of cloth in the shape of an equilateral triangle, whose area costs as much to paint at Rs 10/ square metre as it would cost to lay a border around the three sides at Rs 25 per metre?

A. 13.33 m

B. 17.32 m

C. 14.14 m

D. 8.66 m

SOLUTION

Solution : B

Option (b)

Perimeter = 3a

Area = $\frac{\sqrt{3}}{4 a^{2}}$

10x $\frac{\sqrt{3}}{4 a^{2}}$ = 25x3a

a= $\frac{30}{\sqrt{3}}$ = 17.32 m

Question 7

If sin $α$ = P and tan $α$ = Q, find the value of $\frac{1}{P^{2}} - \frac{1}{Q^{2}}$ ?

A. 0

B. 1

C. 2

D. -1

SOLUTION

Solution : B

Option (b)

Assume a 45-45-90 triangle

sin $α$ = P = $\frac{1}{\sqrt{2}}$ . (Sin= opposite side/ hypotenuse)

tan $α$ = Q = 1 (Tan = opposite side/ adjacent side)

$\frac{1}{P^{2}} - \frac{1}{Q^{2}}$ = 2-1 =1

Shortcut

$c o s e c^{2} α$ – $c o t^{2} α$ = 1 is a trigonometric identity

If you are not aware of this identity, substitute $α$ = 45.

Question 8

In $Δ$ ABC, D,E and F are taken on AB, AC and BC respectively, so that EF=CF and DF=BF. If $∠$ A=40 $^{\circ}$ , then find $∠$ DFE?

A. 100

^{\circ}

B. 120

^{\circ}

C. 90

^{\circ}

D. 140

^{\circ}

SOLUTION

Solution : A

Option (a)

Take DBF = a $^{\circ}$ and ECF = b $^{\circ}$ .
40 + a $^{\circ}$ + b $^{\circ}$ = 180 $\Rightarrow$ a $^{\circ}$ + b $^{\circ}$ = 140
DFB + EFC + DFE = 180 $\Rightarrow$ 180 – 2a $^{\circ}$ + 180 – 2b $^{\circ}$ + DFE = 180 $\Rightarrow$ 180 $^{\circ}$ – 2(a + b) = - DFE
DFE = 280 – 180 = 100 $^{\circ}$

Question 9

What is the circumradius of a triangle whose sides are 7, 24 and 25 respectively?

A. 18

B. 12.5

C. 12

D. 14

SOLUTION

Solution : B

Option b
It’s important that we note that it is a Pythagoras triplet

7 $^{2}$ = 25 $^{2}$ +24 $^{2}$

Now,

In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle.

As the hypotenuse is 25, the circum radius is 12.5

Question 10

If O is the centre of a circle and the radius of the circle is 4 units, find the value of length of AC.

A. 4(

\sqrt{3}

+ 1)

B. 9

\sqrt{2}

C. 10.5

D. none of these

SOLUTION

Solution : D

Option (d)
AC has to less than 8 as the diameter will be of 8 units. So the correct option is “D” as all are greater than 8

Question 11

A trapezium DEFG is circumscribed about a circle that has centre C and radius 2 cm. If DE = 3 cm and the measure of $∠$ DEF= $∠$ EFG=90 $^{\circ}$ , then find the area of trapezium DEFG

A. 18 cm

^{2}

B. 16 cm

^{2}

C. 15 cm

^{2}

D. 20 cm

^{2}

SOLUTION

Solution : A

Draw perpendicular from D to GF meeting at X.

Also, let GN = GP = k

Since, DE = XF = 3 cm

NX = 3 – 2 = 1 = DM = DP

Therefore, GX = k – 1

$Δ$ DXG is right angled triangle.

Hence, (k + 1) $^{2}$ – (k – 1) $^{2}$ = 42

Hence, k = 4 and so GF = 6 cm

Area of trapezium DEFG = $\frac{(3 + 6)}{2}$ * 4 = 18 cm $^{2}$

Question 12

P,Q,R,S,T are points on a circle with centre O so that $∠$ PTQ= 2 x $∠$ RTS and PQ %undefined RS. Find $∠$ SOQ, if P,O,Q are in a straight line

A. 45

^{\circ}

B. 90

^{\circ}

C. 60

^{\circ}

D. 30

^{\circ}

SOLUTION

Solution : A

$∠$ PTQ= 90 (angle in a semi circle)

, hence $∠$ RTS = 45.

$∠$ ROS = 90 ( Inscribed angle= 2 *Angle at circumference)

OR=OS=OR (radius)

$∠$ RSO = $∠$ SOQ= 45 $^{\circ}$ (interior alternate angles)

Option (a)

Question 13

In an equilateral triangle ΔABC, D divides BC in the ratio 2 : 1 while E divides AC in the ratio 1 : 2. Find BE : ED ?

A. 4:3

B. 2:

\sqrt{3}

\sqrt{7}

\sqrt{3}

D. Cannot be determined

SOLUTION

Solution : C

Let the side of the equilateral triangle ΔABC be 3 units. Clearly, BD = 2, CD = 1 and CE = 2.

CD/CE = $\frac{1}{2}$ So $∠$ DEC=30 $^{\circ}$

Hence, ΔCDE is a right angled triangle.

Similarly, ΔBDE is also a right angled triangle.

BE = $\sqrt{7}$ and ED = $\sqrt{3}$ hence BE:ED = $\sqrt{7}$ : $\sqrt{3}$

Question 14

In a triangle ABC, M is the mid-point of BC. If $∠$ AMB = 45 $^{\circ}$ , and $∠$ ACM = 30 $^{\circ}$ , then $∠$ BAM is ( $∠$ B is not obtuse)

A. 30

^{\circ}

B. 45

^{\circ}

C. < 30

^{\circ}

\geq

^{\circ}

SOLUTION

Solution : D

Consider any triangle ABC, AM is a line connecting A to mid point of BC. Drop a perpendicular from A to BC touching BC at D. The angle ADM is a right angled triangle with 45 – 45 – 90 angles. Hence DAM = 45 and BAM > 45. BAM= 45 $^{\circ}$ If $∠$ ABC =90 $^{\circ}$ . hence, answer is option (d)

Question 15

What is the minimum area of quadrilateral ABCD, if the area of a pair of diagonally opposite triangles (Formed by joining the diagonals of the quadrilateral) is 12 and 27 respectively?

A. 70

B. 73

C. 75

D. 77

SOLUTION

Solution : C

Option c

In a quadrilateral, product of area of diagonally opposite triangles is same.

So, the product of other set of opposite triangles will be 12*27

Now, we have to find minimum a and b such that a*b=12*27 and a+b is minimum.

So a = b = 18 (as 12*27 = 324)

Hence the area of the quadrilateral = 18 + 18 + 12 + 27 = 75

Free Geometry Set II - 01 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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