Mensuration Questions and answers

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Aptitude

11). The ratio of two unequal sides of a rectangle is 1:2. If its perimeter is 24, then length of a dlagonal is

A). \( \Large \frac{2}{\sqrt{5}} cm \)

B). \( \Large \frac{4}{\sqrt{5}} cm \)

C). \( \Large 2\sqrt{5} cm \)

D). \( \Large 4\sqrt{5} cm \)

View Answer

Correct Answer: \( \Large 4\sqrt{5} cm \)

\( \Large 2 \left(length+breadth\right) =24 \)

=> \( \Large length\ + breadth\ =\ 12 \)

Therefore, Breadth = \( \Large \frac{1}{3} \times 12 =\ 4\ cm \)

and length = \( \Large \frac{2}{3} \times 12 =\ 8\ cm \)

Therefore, Diagonal of the rectangle

= \( \Large \sqrt{4^{2}+8^{2}}=\sqrt{16+64} \)

= \( \Large \sqrt{80} = 4\sqrt{5} cm \)

12). ABCD is a parallelogram of area S. E and F are the middle points of the sides AD and BC respectively. If G is any point on the line EF, then area of \( \Large \triangle AGB \) is equal to

A). \( \Large \frac{S}{2} \)

B). \( \Large \frac{S}{3} \)

C). \( \Large \frac{S}{4} \)

D). \( \Large \frac{3S}{4} \)

View Answer

Correct Answer: \( \Large \frac{S}{4} \)

Area of ABCD = S

Therefore, Area of ABFE = \( \Large \frac{1}{2}S \)

Since both \( \Large \triangle AGB\ and\ \triangle GFB \) stand on the same base and between same parallels, therefore

Area of AGB = \( \Large \frac{1}{2}\ area\ of\ AEFB \)

= \( \Large \frac{1}{2} \times \left[ \frac{1}{2}S \right] = \frac{S}{4} \)

13). A rectangular lawn is 80 metres long and 60 metres wide. The time taken by a man to walk along its diagonal at the speed of 18 km. per hour is

A). 25 sec

B). 20 sec

C). 18 sec

D). 10 sec

View Answer

Correct Answer: 20 sec

\( \Large d = \sqrt{ \left(80\right)^{2}+ \left(60\right)^{2} } = \sqrt{10000} =\ 100\ m \)

Therefore, \( \Large v = 18 km/hr = 18 \times \frac{5}{18}m/s =\ 5\ sec. \)

Therefore, \( \Large t = \frac{d}{v} = \frac{100}{5} =\ 20\ sec. \)

14). If \( \Large \triangle ABC \) is an equilateral triangle of area \( \Large 36\sqrt{3} cm^{2} \), then area of the inscribed circle is

A). \( \Large 48 \pi cm^{2} \)

B). \( \Large 36 \pi cm^{2} \)

C). \( \Large 24 \pi cm^{2} \)

D). \( \Large 12 \pi cm^{2} \)

View Answer

Correct Answer: \( \Large 12 \pi cm^{2} \)

Let a be the side of the equilateral triangle.

Therefore, Its area = \( \Large \frac{a^{2}\sqrt{3}}{4} \)

\( \Large \therefore \frac{a^{2}\sqrt{3}}{4} = 36\sqrt{3} \)

=> \( \Large a^{2} = 36 \times 4 \)

=> \( \Large a = 6 \times 2 = 12 cm\)

Let r be the radius of the inscribed circle.

Therefore, \( \Large r = \frac{\triangle}{s}=\frac{36\sqrt{3}}{\frac{1}{2} \left(12+12+12\right) } = 2\sqrt{3} cm \)

Area of inscribed circle = \( \Large \pi r^{2}= \pi \times \left(2\sqrt{3}\right)^{2}=12 \pi cm^{2} \)

15). A ladder 2.0m long is placed in a street so as to reach a window 1.6 m high and on turning the ladder to the other side of a street, it reaches a point 1.2 m high. The width of the street is

A). 2.8 m

B). 2.4 m

C). 2.0 m

D). 1.6 m

View Answer

16). Area of the parallelogram given below is

A). 40 sq. cm

B). 20 sq. cm

C). 32 sq. cm

D). 64 sq. cm

View Answer

17). If three solid bodies; a sphere, right circular cylinder and a right circular cone are of equal radius and equal surface area, then their heights are in the ratio

A). \( \Large 2 : 1 : 2\sqrt{2} \)

B). \( \Large \sqrt{2} : 1 : 2 \)

C). \( \Large 2 : 1 : 3\sqrt{2} \)

D). \( \Large 6\sqrt{2} : 3\sqrt{3} : 4 \)

View Answer

Correct Answer: \( \Large 2 : 1 : 2\sqrt{2} \)

Let r be radius of each of sphere, cylinder and cone, respectively

with heights 2r, \( \Large h_{1} \), \( \Large h_{2} \).

By hypothesis,

\( \Large 4 \pi r^{2}=2 \pi rh_{1}+2 \pi r^{2} \)

= \( \Large \pi r\sqrt{r^{2}+h^{2}_{2}}+ \pi r^{2} \)

=> \( \Large 4r = 2 \left(h_{1}+r\right)=\left[ \sqrt{r^{2}+h^{2}_{2}}+r \right] \)

Therefore, \( \Large h_{1}=r\ and\ 4r = \sqrt{r^{2}+h^{2}_{2}}+r \)

=> \( \Large 3r = \sqrt{r^{2}+h^{2}_{2}} \)

Squaring \( \Large 9r^{2}=r^{2}+h^{2}_{2} \)

=> \( \Large h^{2}_{2}=8r^{2} \)

=> \( \Large h_{2}=2\sqrt{2}r \)

Hence, ratio of their heights

= \( \Large 2r : r 2\sqrt{2}r \)

=> \( \Large 2\ :\ 1\ :\ 2\sqrt{2} \)

18). A square and an equilateral triangle are inscribed in a circle. If a and b denote lengths of their sides, then

A). \( \Large a^{2}=\frac{b^{2}}{2} \)

B). \( \Large \frac{a^{2}}{2}=b^{2} \)

C). \( \Large 3b^{2}=2a^{2} \)

D). \( \Large 3a^{2}=2b^{2} \)

View Answer

19). The slant height of a right circular cone is 10 cm and its height is 8 cm. It is cut by a plane parallel to its base passing through the midpoint of the height. Ratio of the volume of the cone to that of the frustum of the cone cut is

A). 2 : 1

B). 8 : 7

C). 4 : 3

D). 3 : 2

View Answer

Correct Answer: 8 : 7

Given: OM = 8, OP = 4 and OB = 10

\( \Large BM^{2}= OB^{2}-OM^{2}=100-64=36 \)

Therefore, BM = 6.

Then PC = 3

Therefore, Volume of cone OAB

= \( \Large \frac{1}{3} \times \pi \times 6^{2} \times 8 \)

= \( \Large 96\ \pi \)

Volume of cone OCD

=\( \Large \frac{1}{3} \times \pi \times 3^{2} \times 4 = 12\ \pi \)

Volume of frustum ABCD = \( \Large 96 \pi -12 \pi = 84 \pi \)

Required ratio = \( \Large \frac{96 \pi }{84 \pi } = \frac{8}{7} \)

20). A solid sphere of radius r is sliced by the planes passing through its centre and perpendicular to each other. The total surface area of each of the pieces so formed is

A). \( \Large \frac{2}{3} \pi r^{2} \)

B). \( \Large 2 \pi r^{2} \)

C). \( \Large \frac{4}{3} \pi r^{2} \)

D). \( \Large \pi r^{2} \)

View Answer

Correct Answer: \( \Large 2 \pi r^{2} \)

Surface of one piece consists of sum of one fourth of the surface of the sphere and two semicircular areas.

Surface area of each piece

\( \Large \frac{1}{4}.4 \pi r^{2}+2.\frac{1}{2} \pi r^{2} \)

= \( \Large 2 \pi r^{2} \)

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