Practice MCQ Questions and Answer on Area

121.

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

• (A) 16 cm
• (B) 18 cm
• (C) 24 cm
• (D) Data inadequate

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 $$\eqalign{ & \frac{{2\left( {l + b} \right)}}{b} = \frac{5}{1} \cr & \Rightarrow 2l + 2b = 5b \cr & \Rightarrow 3b = 2l \cr & b = \frac{2}{3}l \cr & {\text{Then,}} \cr & {\text{Area = }}216\,c{m^2} \cr & \Rightarrow l \times b = 216 \cr & \Rightarrow l \times \frac{2}{3}l = 216 \cr & \Rightarrow {l^2} = 324 \cr & l = 18\,cm \cr} $$

122.

A hall, whose length is 16 m and the breadth is twice its height, takes 168 m of paper with 2 m as its width to cover its four walls. The area of the floor is :

• (A) 96 m2
• (B) 190 m2
• (C) 192 m2
• (D) 216 m2

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 Let the height of the room be x metres Then, breadth of the room = (2x) metres Area of 4 walls :$$\eqalign{ & = \left[ {2\left( {16 + 2x} \right) \times x} \right]{m^2} \cr & = \left( {32x + 4{x^2}} \right){m^2} \cr} $$ $$\eqalign{ & \therefore \left( {32x + 4{x^2}} \right) = 168 \times 2 \cr & \Rightarrow {x^2} + 8x - 84 = 0 \cr & \Rightarrow {x^2} + 14x - 6x - 84 = 0 \cr & \Rightarrow x\left( {x + 14} \right) - 6\left( {x + 14} \right) = 0 \cr & \Rightarrow \left( {x + 14} \right)\left( {x - 6} \right) = 0 \cr & \Rightarrow x = 6 \cr} $$ Area of the floor : = (16 × 12) m2 = 192 m2

123.

The area of two similar triangles are 12 cm² and 48 cm². If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is :

• (A) 0.525 cm
• (B) 4.2 cm
• (C) 4.41 cm
• (D) 8.4 cm

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 Note : The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Let the length of the required altitude be x cm Then, $$\eqalign{ & \frac{{12}}{{48}} = \frac{{{{\left( {2.1} \right)}^2}}}{{{x^2}}} \cr & \Rightarrow {x^2} = \left( {4.41 \times 4} \right) \cr & \Rightarrow x = 2.1 \times 2 \cr & \Rightarrow x = 4.2\,cm \cr} $$

124.

The length of a room is double its breadth. The cost of colouring the ceiling at Rs. 25 per sq. meter is Rs. 5000 and the cost of painting the four walls at Rs. 240 per sq. metre is Rs. 64800. Find the height of the room :

• (A) 3.5 m
• (B) 4 m
• (C) 4.5 m
• (D) 5 m

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 Let the breadth and height of the room be b metres and h metres respectively. Then, length of the room = (2b) metres Area of the ceiling : $$\eqalign{ & = \left( {2b \times b} \right)m \cr & = \left( {2{b^2}} \right){m^2} \cr} $$ $$\eqalign{ & 2{b^2} = \frac{{5000}}{{25}} \cr & 2{b^2} = 200 \cr & {b^2} = 100 \cr & b = 10 \cr} $$ So, length = 20 m, breadth = 10 m Area of 4 walls : $$\eqalign{ & = \left[ {2\left( {20 + 10} \right) \times h} \right]{m^2} \cr & = \left( {60h} \right){m^2} \cr} $$ $$\eqalign{ & \therefore 60h = \frac{{64800}}{{240}} \cr & \Rightarrow 60h = 270 \cr & \Rightarrow h = \frac{{270}}{{60}} \cr & \Rightarrow h = 4.5\,m \cr} $$

125.

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, the n the area of the park (in square metre) is :

• (A) 15360 square metre
• (B) 153600 square metre
• (C) 30720 square metre
• (D) 307200 square metre

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 $${\text{Perimeter}} = $$   $${\text{Distance covered in 8 minutes}}$$$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{12000}}{{60}} \times 8} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1600{\text{ }}\,m \cr} $$ Let the length = 3x metres and breadth = 2x metres Then, 2 (3x + 2x) = 1600 or, x = 160 Length = 480 m and breadth = 320 m ∴ Area = (480 × 320) m2            = 153600 square metre

126.

A carpenter is designing a table. The table will be in the form of a rectangle whose length is 4 feet more than its width. How long should the table be if the carpenter wants the area of the table to be 45 sq. ft ?

• (A) 6 ft
• (B) 9 ft
• (C) 11 ft
• (D) 13 ft

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 Let the width of the table be x feet. Then, length of the table = (x + 4) ft $$\eqalign{ & \therefore x\left( {x + 4} \right) = 45 \cr & \Rightarrow {x^2} + 4x - 45 = 0 \cr & \Rightarrow {x^2} + 9x - 5x - 45 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 5\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr} $$ Hence, length of the table = (5 + 4) = 9 feet

127.

In a triangle ABC, a line XY is drawn parallel to BC meeting AB in X and AC in Y. The area of the triangle AXY is half of the area of the triangle ABC. XY divides AB in the ratio of :

• (A) $$1:\sqrt{2}$$
• (B) $$\sqrt{2}:\left(\sqrt{2}-1\right)$$
• (C) $$1:\left(\sqrt{2}-1\right)$$
• (D) $$\sqrt{2}:\sqrt{3}$$

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 Note : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Since, XY || BC, we have : $$\eqalign{ & \angle AXY = \angle ABC{\text{ and }} \cr & \angle AYX = \angle ACB \cr & {\text{Aslo, }}\angle A = \angle A\,\,{\text{(common)}} \cr & {\text{So, }}\vartriangle AXY \sim \,\vartriangle ABC \cr & {\text{Let area (}}\vartriangle ABC{\text{)}} = x\,\text{sq. units} \cr & {\text{Then,}} \cr & {\text{Area (}}\vartriangle AXY{\text{) = }}\frac{x}{2}\text{sq. units} \cr & \frac{{{{\left( {AB} \right)}^2}}}{{{{\left( {AX} \right)}^2}}} = \frac{x}{{\left( {\frac{x}{2}} \right)}} \cr & \Rightarrow \frac{{AB}}{{AX}} = \sqrt 2 \cr & \Rightarrow \frac{{AX + BX}}{{AX}} = \sqrt 2 \cr & \Rightarrow 1 + \frac{{BX}}{{AX}} = \sqrt 2 \cr & \Rightarrow \frac{{BX}}{{AX}} = \left( {\sqrt 2 - 1} \right) \cr & \Rightarrow \frac{{AX}}{{BX}} = \frac{1}{{\left( {\sqrt 2 - 1} \right)}}{\text{ Or 1}}:\left( {\sqrt 2 - 1} \right) \cr} $$

128.

The dimensions of a room are 12.5 metres by 9 metres by 7 metres. There are 2 doors and 4 windows in the room; each door measures 2.5 metres by 1.2 metres and each window 1.5 metres by 1 metre. Find the cost of painting the walls at Rs. 3.50 per square metre.

• (A) Rs. 150.50
• (B) Rs. 1011.50
• (C) Rs. 1101.50
• (D) Cannot be determined

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 Area of 4 walls : $$\eqalign{ & = 2\left( {l + b} \right) \times h \cr & = \left[ {2\left( {12.5 + 9} \right) \times 7} \right]{m^2} \cr & = 301\,{m^2} \cr} $$ Area of 2 doors and 4 windows : $$\eqalign{ & = \left[ {2\left( {2.5 \times 1.2} \right) + 4\left( {1.5 \times 1} \right)} \right]{m^2} \cr & = 12\,{m^2} \cr} $$ ∴ Area to be painted : $$\eqalign{ & = \left( {301 - 12} \right){m^2} \cr & = 289\,{m^2} \cr} $$ Cost of painting : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {289 \times 3.50} \right) \cr & = {\text{Rs}}{\text{. }}1011.50 \cr} $$

129.

A vertical rod of height 33 metres is bent to form a semi-circular shape so that the top touches the ground. The distance between the top head and the base on the ground is :

• (A) 10.5 m
• (B) 12 m
• (C) 21 m
• (D) 33 m

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 $$\eqalign{ & \pi R = 33 \cr & \Rightarrow R = \left( {\frac{{33 \times 7}}{{22}}} \right) \cr & \Rightarrow R = \left( {\frac{{21}}{2}} \right)m \cr} $$ ∴ Required distance = 2R = 21 m

130.

A small disc of radius r is cut out from a disc of radius R. The weight of the disc which now has a hole in it, is reduced to $$\frac{24}{25}$$ of the original weight. If R = xr, what is the value of x ?

• (A) 4
• (B) 4.5
• (C) 24
• (D) 25

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Answer & Solution Answer: Option E Solution: Since weight of the disc is proportional to its area, we have : $$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$