Practice MCQ Questions and Answer on Area

131.

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

• (A) 2.91 m
• (B) 3 m
• (C) 5.82 m
• (D) None of these

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 Area of the park = (60 x 40) m2 = 2400 m2 Area of the lawn = 2109 m2 ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2 Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⇒ x2 - 100x + 291 = 0 ⇒ (x - 97)(x - 3) = 0 ⇒ x = 3 m

132.

The circumference of a circle is 100 cm. The side of a square inscribed in the circle is :

• (A) $$50\sqrt{2}c{m}^2$$
• (B) $$\frac{100}{\pi }c{m}^2$$
• (C) $$\frac{50\sqrt{2}}{\pi }c{m}^2$$
• (D) $$\frac{100\sqrt{2}}{\pi }c{m}^2$$

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 $$\eqalign{ & 2\pi R = 100 \cr & R = \frac{{100}}{{2\pi }} = \frac{{50}}{\pi } \cr & R = \frac{1}{2} \times {\text{diagonal}} \cr & \Rightarrow {\text{Diagonal}} = 2R \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 50}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{\pi } \cr} $$ $$\eqalign{ & \therefore {\text{Area of the square}} = \frac{1}{2} \times {\left( {{\text{diagonal}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times {\left( {\frac{{100}}{\pi }} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 2 }} \times \frac{{100}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 2 }}{\pi }cm \cr} $$

133.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $$12\sqrt{2}$$ cm, then the area of the triangle is :

• (A) $$24\sqrt{2}$$ cm2
• (B) $$24\sqrt{3}$$ cm2
• (C) $$48\sqrt{3}$$ cm2
• (D) $$64\sqrt{3}$$ cm2

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 Let the side of the square be a cm Then, its diagonal = $$\sqrt 2 $$ a cm Now, $$\sqrt 2 $$ a = $$12\sqrt 2 $$ ⇒ a = 12 cm Perimeter of the square = 4a = 48 cm Perimeter of the equilateral triangle = 48 cm Each side of the triangle = 16 cm Area of the triangle : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times 16 \times 16} \right)c{m^2} \cr & = \left( {64\sqrt 3 } \right)c{m^2} \cr} $$

134.

The sides of a triangle are consecutive integers. The perimeter of the triangle is 120 cm. Find the length of the greatest side :

• (A) 39 cm
• (B) 40 cm
• (C) 41 cm
• (D) 42 cm

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 Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively. Then, $$\eqalign{ & x + \left( {x + 1} \right) + \left( {x + 2} \right) = 120 \cr & \Rightarrow 3x + 3 = 120 \cr & \Rightarrow 3x = 117 \cr & \Rightarrow x = 39 \cr} $$ ∴ Length of greatest side : = (39 + 2) cm = 41 cm

135.

The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

• (A) 42$$\sqrt{2}$$ sq. cm
• (B) 42 sq. cm
• (C) 84 sq. cm
• (D) 48 sq. cm

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 Let each equal side of isosceles triangle be x cm Perimeter of an isosceles triangle = 36 cm $$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$ BD = DC = 7cm From ΔABD By using Pythagoras theorem : $$\eqalign{ & AD = \sqrt {A{B^2} - B{D^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{11}^2} - {7^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {121 - 49} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {72} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3 \times 2\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\sqrt 2 \,cm \cr} $$ ∴ Area of ΔABC $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$

136.

A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is :

• (A) 15 m × 6.67 m
• (B) 20 m × 5 m
• (C) 30 m × 3.33 m
• (D) 40 m × 2.5 m

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 We have : $$\eqalign{ & 2b + l = 30 \cr & \Rightarrow l = 30 - 2b \cr} $$ $$\eqalign{ & {\text{Area}} = {\text{100 }}{m^2} \cr & \Rightarrow l \times b = 100 \cr & \Rightarrow b\left( {30 - 2b} \right) = 100 \cr & \Rightarrow {b^2} - 15b + 50 = 0 \cr & \Rightarrow \left( {b - 10} \right)\left( {b - 5} \right) = 0 \cr & \Rightarrow b = 10{\text{ or }}b = 5 \cr} $$ When, b = 10, $$l$$ = 10 and when b = 5, $$l$$ = 20 Since the garden is rectangular, so its dimension is 20 m × 5 m

137.

A circular pond has area equal to 616 m². A circular stage is made at the centre of the pond whose radius is equal to half the radius of the pond. What is the area where water is present ?

• (A) 454 sq. m
• (B) 462 sq. m
• (C) 532 sq. m
• (D) 564 sq. m

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 Let the radius of the pond be R metres Then, $$\eqalign{ & \pi {R^2} = 616 \cr & \Rightarrow {R^2} = \frac{{616 \times 7}}{{22}} \cr & \Rightarrow {R^2} = 196 \cr & \Rightarrow R = 14\,m \cr} $$ Radius of the stage : $$\eqalign{ & = \left( {\frac{{14}}{2}} \right)m \cr & = 7\,m \cr} $$ ∴ Area where water is present : $$\eqalign{ & = \pi \left( {{{14}^2} - {7^2}} \right){m^2} \cr & = \left( {\frac{{22}}{7} \times 21 \times 7} \right){m^2} \cr & = 462\,{m^2} \cr} $$

138.

A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq.ft per day, they approximately what time will be taken by the cow to graze the whole field ?

• (A) 2 days
• (B) 6 days
• (C) 18 days
• (D) 24 days

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 Area of the field grazed : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 14 \times 14} \right)sq.ft \cr & = 616\,sq.ft \cr} $$ Number of days taken to graze the field : $$\eqalign{ & = \frac{{616}}{{100}}\text{days} \cr & = 6\,\text{days}(\text{approx}) \cr} $$

139.

The length, breadth and height of the room are in the ratio 3 : 2 : 1. The breadth and height of the room are halved and length of the room is doubled. The area of the four walls of the room will :

• (A) decrease by 13.64%
• (B) decrease by 15%
• (C) decrease by 18.75%
• (D) decrease by 30%

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 Let the length, breadth and height of the room be 3x, 2x and x respectively. Area of 4 walls : $$\eqalign{ & = 2\left( {l + b} \right) \times h \cr & = 2\left( {3x + 2x} \right) \times x \cr & = 10{x^2} \cr} $$ New length = 6x New breadth = x New height = $$\frac{x}{2}$$ New area of four walls : $$\eqalign{ & = \left[ {2\left( {6x + x} \right)\frac{x}{2}} \right] \cr & = 7{x^2} \cr} $$ Decrease in area : $$\eqalign{ & = \left( {10{x^2} - 7{x^2}} \right) \cr & = 3{x^2} \cr} $$ ∴ Decrease% : $$\eqalign{ & = \left( {\frac{{3{x^2}}}{{10{x^2}}} \times 100} \right)\% \cr & = 30\% \cr} $$

140.

A piece of wire when bent to from a circle will have a radius of 84 cm. If the wire is bent to form a square, the length of a side of the square is :

• (A) 216 cm
• (B) 133 cm
• (C) 132 cm
• (D) 168 cm

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 Given length of the piece of wire = 84 cm Length of the piece of wire = Circumference of circle $$\eqalign{ & = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 84 \cr & = 528\,cm \cr} $$ Length of each side of square = a ∴ Perimeter of square = 4a = 528 cm ∴ Side of square = $$\frac{{528}}{4}$$ = 132 cm