Basic Geometry(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

BASIC GEOMETRY MCQs

Total Questions : 93 | Page 9 of 10 pages

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.

$\frac{\sqrt{3} (2 r + 2 r \sqrt{3})^{2}}{16}$
$\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}$
$\frac{\sqrt{3} (2 r + 2 r \sqrt{3})^{2}}{4}$
$\frac{\sqrt{3} (2 r + r \sqrt{3})^{2}}{16}$

Discuss Question

Answer: Option B. ->

\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}

:
B

Soln:

The angle shown is 30 degrees. Thus AB=r $\sqrt{3}$ (30-60-90 triangle).

Thus side of a triangle: 2r + 2r $\sqrt{3}$ .

Area of the equilateral triangle: $\frac{\sqrt{3} (r + 2 r \sqrt{3})^{2}}{4}$ Hence Ans: b

Question 82.

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?

14 cm
12 cm
$\frac{85}{6}$ cm
$\frac{72}{5}$ cm

Discuss Question

Answer: Option D. ->

\frac{72}{5}

cm
:
D

Soln: the triangles shown are similar. Hence $\frac{x + (1 + 6)}{x} = \frac{1}{6}$ ;x= $\frac{7}{5}$ ; Height of cone:13 + $(\frac{7}{5}) = \frac{72}{5}$ .

Hence, Ans: d)

Question 83.

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?

10 $π$
12 $π$
24 $π$
26 $π$

Discuss Question

Answer: Option C. -> 24

π

:
C

From the fig: X = $\sqrt{13^{2} - 5^{2}}$ = 12x Thus, circumference= 24 $π$

Hence Ans: c

Question 84.

A square hole of cross sectional area 4 cm $^{2}$ 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?

158 cm $^{2}$
190 cm $^{2}$
166 cm $^{2}$
182 cm $^{2}$

Discuss Question

Answer: Option D. -> 182 cm

^{2}

:
D

Ans: d) 182 cm $^{2}$

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

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

Question 85.

What’s the area of the shaded region? Given that the side of the square = 1 unit.

$\frac{π}{2}$
$\frac{1}{2}$
( $\frac{π}{4}$ ) - ( $\frac{1}{2}$ )
( $\frac{π}{2}$ ) - 1

Discuss Question

Answer: Option D. -> (

\frac{π}{2}

) - 1
:
D

Ans: d)

Soln: Consider the double shaded area.

Area: (Area of quadrant) – (Area of triangle).

( $\frac{π}{4}$ )-( $\frac{1}{2}$ x 1 x 1). Total shaded area is twice of this

= 2 ( $\frac{π}{4}$ ) - ( $\frac{1}{2}$ x 1 x 1); = $\frac{π}{2}$ -1,

Question 86.

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.

212
216
256
312

Discuss Question

Answer: Option C. -> 256
:
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)

Question 87.

A fan has 3 equally spaced blades. The central dial has an area of 3 $π$ cm $^{2}$ . Length of each blade is (20- $\sqrt{3}$ )cm. If a triangle is drawn from the tips of the blades, find its area.

900 sq cm
300 $\sqrt{3}$ sq cm
900 sq cm
400 $\sqrt{3}$ sq units

Discuss Question

Answer: Option B. -> 300

\sqrt{3}

sq cm
:
B

Ans: b) 300 $\sqrt{2}$ sq cm

Soln: Radius of the dial at the centre: $\sqrt{\frac{3 π}{π}}$

total length of the blade from the centre: (20- $\sqrt{3}$ )+ $\sqrt{3}$ ; = 20.

Half of each side: 30 cos 30. = 10 $\sqrt{3}$ . Total length of each side: 20 $\sqrt{3}$ .

Thus , area: $(\frac{\sqrt{3}}{4}) (20 \sqrt{3})^{2}$ . =300 $\sqrt{2}$ sq cm

Question 88.

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.

Discuss Question

Answer: Option C. ->

:
C

Ans: c.

Question 89.

If a, b, c are sides of a triangle such that $a^{2} + b^{2} + c^{2}$ =ab+bc+ac. Then the triangles are:

equilateral
isosceles
right angled
obtuse angled

Discuss Question

Answer: Option A. -> equilateral
:
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

Question 90.

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

2
$\sqrt{2}$
2 $\sqrt{2}$
None of these

Discuss Question

Answer: Option C. -> 2

\sqrt{2}

:
C

Ans: c 2 $\sqrt{2}$ .

Soln: Reverse Gear Approach: Use answer options

The largest side should be smaller than sum of other 2 sides.
The sides are 76, 4 $\sqrt{2}$ +7, 63; Here the sum of last two sides is 75.6 which is lesser than 76