# Free Basic Geometry - 02 Practice Test - CAT

Areas of three adjacent sides of a cuboid are a,b,c. Volume of the cuboid is N. The value of N2 equals to?

A. abc
B. (ab+ac+bc)
C. (a3+b3+c3)
D. None of these

#### SOLUTION

Solution : A

Assume sides to be x,y,z. Then xy=a, yz=b, xz=c. Multiplying: (xyz)2 = abc; =N2. (Vol=xyz)

Hence Ans: (a)

A string of certain length n makes one perfect turn around a cube of side A, starting from corner X and ending at Y (as in the fig). Find the length of the string. A. 2A
B. 17A
C. A
D. 17(A + 1)

#### SOLUTION

Solution : B

The string is wrapped equally around the cube 4 times. So the part ab = A/4. The hypotenuse of the triangle (A2)+A216The length of the string is four times this hypotenuse.

4×(17A)4=17A

Hence Ans: b

Three coins are placed in such a way that one coin touches the other two (in the same plane). If the radius of each coin is r, what is the area of the triangle that circumscribes this arrangement of coins.

A. 3(2r+2r3)216
B. 3(r+2r3)24
C. 3(2r+2r3)24
D. 3(2r+r3)216

#### SOLUTION

Solution : B

Soln: The angle shown is 30 degrees. Thus AB=r3 (30-60-90 triangle).

Thus side of a triangle: 2r + 2r3.

Area of the equilateral triangle: 3(r+2r3)24Hence Ans: b

A sphere of radius 13 cm is cut by a plane whose distance from the centre of the sphere is 5cm. What is the circumference of the section so obtained?

A. 10π
B. 12π
C. 24π
D. 26π

#### SOLUTION

Solution : C From the fig: X = 13252 = 12x Thus, circumference= 24 π

Hence Ans: c

Two spheres of radii 6 cm and 1 cm are inscribed inside a right circular cone. The bigger sphere touches the smaller sphere and also the base of the cone. What is the height of the cone?

A. 14 cm
B. 12 cm
C. 856 cm
D. 725 cm

#### SOLUTION

Solution : D

Soln: the triangles shown are similar. Hence x+(1+6)x=16;x=75; Height of cone:13 + (75)=725. Hence, Ans: d)

A square hole of cross sectional area 4 cm2 is drilled across a cube with its length parallel to a side of the cube. If an edge of the cube measures 5 units, what is the total surface area of the structure so formed?

A. 158 cm2
B. 190 cm2
C. 166 cm2
D. 182 cm2

#### SOLUTION

Solution : D

Ans: d) 182 cm2

New SA=(original SA)+(SA of square channel formed through the cube)-(squares on 2 faces)

= (25x6)+(2x5x4)-(4+4) ;=182 sq units

What’s the area of the shaded region? Given that the side of the square = 1 unit. A. π2
B. 12
C. (π4) - (12)
D. (π2) - 1

#### SOLUTION

Solution : D

Ans: d)

Soln: Consider the double shaded area. Area: (Area of quadrant) – (Area of triangle).

(π4)-(12 x 1 x 1). Total shaded area is twice of this

= 2 (π4) - (12 x 1 x 1); = π2 -1,

A goat is tied to two poles P and Q with 15 m ropes. P and Q are 20 m apart. What is the area the goat is likely to graze. A. B. C. D. #### SOLUTION

Solution : C

Ans: c.

An equilateral triangle plate is to be cut into n number of identical small equilateral triangle plates. Which of the following can be a value of n.

A. 212
B. 216
C. 256
D. 312

#### SOLUTION

Solution : C

Ans: c…. 256

Solution: The total no of plates has to be a perfect square, you can try proving this by using the unitary approach (i.e. by using smaller numbers)

A fan has 3 equally spaced blades. The central dial has an area of 3π cm2. Length of each blade is (20-3)cm. If a triangle is drawn from the tips of the blades, find its area.

A. 900 sq cm
B. 300 3 sq cm
C. 900 sq cm
D. 4003 sq units

#### SOLUTION

Solution : B

Ans: b) 3002 sq cm

Soln: Radius of the dial at the centre: 3ππ total length of the blade from the centre: (20-3)+3; = 20.

Half of each side: 30 cos 30. = 103. Total length of each side: 203.

Thus , area: (34)(203)2. =3002 sq cm

If (9p2+4), (2p+7) and (8p2-1) are three sides of a triangle. Which of the following CANNOT be a value for p?

A. 2
B. 2
C. 22
D. None of these

#### SOLUTION

Solution : C

Ans: c 22 .

Soln: Reverse Gear Approach: Use answer options

The largest side should be smaller than sum of other 2 sides.
The sides are 76, 42+7, 63; Here the sum of last two sides is 75.6 which is lesser than 76

AB is a line segment with P as midpoint. Three semicircles are drawn with AP,PB and AB as diameters. Another circle is drawn so as to touch all three semicircles. What’s the radius of this smaller circle?

A. 14AB
B. 16AB
C. 17AB
D. 18AB

#### SOLUTION

Solution : B

Ans: b) 16AB.

Soln: Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.

Since PRB is right angled, (r+x)2=(2rx)2+r2 .

Solving we get x=r3=AB6

If a, b, c are sides of a triangle such that a2+b2+c2=ab+bc+ac. Then the triangles are:

A. equilateral
B. isosceles
C. right angled
D. obtuse angled

#### SOLUTION

Solution : A

Ans: a) equilateral.

Substitute values and check. 2,2,2 for equilateral; 1,1,2 for isosceles, 3,4,5 for right angled. The equation holds only for equilateral triangles

Find the radius of the smallest circle if the equilateral triangle shown is of side 3 units. A. 3
B. 3+12
C. 312
D. None of these

#### SOLUTION

Solution : D

Solution:- Calculate the altitude of the equilateral triangle. Then according to the conditions given,
Diameter of the circle circumscribing the equilateral triangle= Altitude of the equilateral triangle + Diameter of the smaller cirlce.
Solve the above equation for the required answer.

Alternate Soln: The triangle’s dimension is used to calculate the radii of the outermost and the second circle. Radius of the bigger circle: R=(1.5)cos30; =3 units.

r = 1.5 tan 30. =  323 r= 32.

34

Hence ans: d (none of these)

Which of the following is not true?
a) Given that the sides of a scalene triangle are integers, then one of the sides can =1
b) The 3 medians of a triangle divide it into 6 similar triangles
c) An isosceles triangle can never be obtuse

A. Only A
B. only A and C
C. Only C
D. A, B and C

#### SOLUTION

Solution : B

option (b)

a is false as if one side is 1, another side is a, then the third side < a+1. this is not possible as it needs to be an integer value and since it’s a scalene triangle, the third side cannot be 1+a

b is true

c is false as there is a possibility of 120, 30 , 30