Practice MCQ Questions and Answer on Area

1.

The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area ?

• (A) $$37\frac{1}{2}\mathrm{\%}$$
• (B) $$60\mathrm{\%}$$
• (C) $$75\mathrm{\%}$$
• (D) $$120\mathrm{\%}$$

Show Answer

 Let original length = x and original breadth = y Then, original area = xy New length : $$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$ Let the new breadth = z Then, $$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$ ∴ Decrease in breadth : $$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$

2.

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

Show Answer

 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 $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$

3.

The perimeter of a circular field and a square field are equal. If the area of the square field is 12100 m², the area of the circular field will be :

• (A) 15200 m2
• (B) 15300 m2
• (C) 15400 m2
• (D) 15500 m2

Show Answer

 Side of the square field = $$\sqrt {12100} $$   m = 110 m Perimeter of the circle field := Perimeter of the square field = (4 × 110) m = 440 m $$\eqalign{ & 2\pi R = 440 \cr & \Rightarrow R = \frac{{440 \times 7}}{{2 \times 22}} = 70\,m \cr} $$ ∴ Area of the circular field : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 70 \times 70} \right){m^2} \cr & = 15400\,{m^2} \cr} $$

4.

The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by :

• (A) 35%
• (B) 42%
• (C) 62%
• (D) 82%

Show Answer

 Let length = $$l$$ metres and breadth = b metres Then, original area = (lb) m2 New length : $$\eqalign{ & = \left( {140\% {\text{ of }}l} \right)m \cr & = \left( {\frac{{140}}{{100}} \times l} \right)m \cr & = \frac{{7l}}{5}m \cr} $$ New breadth : $$\eqalign{ & = \left( {130\% {\text{ of }}b} \right)m \cr & = \left( {\frac{{130}}{{100}} \times b} \right)m \cr & = \frac{{13l}}{{10}}m \cr} $$ New area : $$\eqalign{ & = \left( {\frac{{7l}}{5} \times \frac{{13b}}{{10}}} \right){m^2} \cr & = \left( {\frac{{91lb}}{{50}}} \right){m^2} \cr} $$ Increase : $$\eqalign{ & = \left( {\frac{{91lb}}{{50}} - lb} \right) \cr & = \frac{{41}}{{50}}lb \cr} $$ ∴ Increase % :$$\eqalign{ & = \left( {\frac{{41}}{{50}} \times lb \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 82\% \cr} $$

5.

The ratio of the areas of the in-circle and the circum-circle of a square is :

• (A) 1 : 2
• (B) $$\sqrt{2}:1$$
• (C) $$1:\sqrt{2}$$
• (D) 2 : 1

Show Answer

 Let r1 and r2 be the radii of the in-circle and circum-circle of a square respectively and let each side of the square be a. Then, $$\eqalign{ & {r_1} = \frac{a}{2} \cr & {r_2} = \frac{1}{2} \times {\text{diagonal of the sequence}} \cr & {r_2} = \frac{1}{2} \times \sqrt 2 a \cr & {r_2} = \frac{{\sqrt 2 a}}{2}cm \cr} $$ ∴ Required ratio : $$\eqalign{ & = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr & = 1:2 \cr} $$

6.

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

Show Answer

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

7.

The area of a right-angled triangle is 20 sq.cm and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is :

• (A) $$\frac{41}{\sqrt{34}}\text{ cm}$$
• (B) $$\sqrt{\frac{41}{40}}\text{ cm}$$
• (C) $$\frac{29}{\sqrt{20}}\text{ cm}$$
• (D) $$\frac{20}{\sqrt{29}}\text{ cm}$$

Show Answer

 Let the length of the other side containing the right angle be x cm Then, $$\eqalign{ & \frac{1}{2} \times 4 \times x = 20 \cr & \Rightarrow x = 10{\text{ cm}} \cr} $$ Hypotenuse : $$\eqalign{ & = \sqrt {{{10}^2} + {4^2}} {\text{ cm}} \cr & = \sqrt {116} {\text{ cm}} \cr & = 2\sqrt {29} \,{\text{cm}} \cr} $$ Let the altitude on the hypotenuse be h cm Then,$$\eqalign{ & \frac{1}{2} \times 2\sqrt {29} \times h = 20 \cr & \Rightarrow h = \frac{{20}}{{\sqrt {29} }}{\text{ cm}} \cr} $$

8.

If the diagonal and the area of a rectangle are 25 m² and 168 m², what is the length of the rectangle ?

• (A) 12 m
• (B) 17 m
• (C) 24 m
• (D) 31 m

Show Answer

 Let the length of the rectangle be x metres. Then, breath of the rectangle = $$\left( {\frac{{168}}{x}} \right)m$$$$\eqalign{ & \therefore \sqrt {{x^2} + {{\left( {\frac{{168}}{x}} \right)}^2}} = 25 \cr & \Rightarrow \sqrt {{x^2} + \frac{{28224}}{{{x^2}}}} = 25 \cr & \Rightarrow {x^2} + \frac{{28224}}{{{x^2}}} = 625 \cr & \Rightarrow {x^4} - 625{x^2} + 28224 = 0 \cr & \Rightarrow {x^4} - 576{x^2} - 49{x^2} + 28224 = 0 \cr & \Rightarrow {x^2}\left( {{x^2} - 576} \right) - 49\left( {{x^2} - 576} \right) = 0 \cr & \Rightarrow \left( {{x^2} - 576} \right)\left( {{x^2} - 49} \right) = 0 \cr & \Rightarrow {x^2} = 576\,\,or\,\,{x^2} = 49 \cr & \Rightarrow x = 24\,\,\,or\,\,\,x = 7 \cr} $$ Hence, length = 24 m and breadth = 7 m

9.

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

Show Answer

 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

10.

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

Show Answer

 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