Practice MCQ Questions and Answer on Area

1.

A rectangle becomes a square when its length is reduced by 10 units and its breadth is increased by 5 units. but by this process the area of the rectangle is reduced by 210 sq.units. The area of the rectangle (A) in square units is :

• (A) 2950 > A 2900
• (B) 2900 > A > 2875
• (C) 2925 A > 2875
• (D) 2925 > A > 2900

Show Answer

 Let the length and breadth of the rectangle be $$l$$ and b respectively $$\eqalign{ & l - 10 = b + 5 \cr & \Rightarrow l - b = 15.....(i) \cr & {\text{And,}} \cr & \Rightarrow lb - \left( {l - 10} \right)\left( {b + 5} \right) = 210 \cr & \Rightarrow lb - \left( {lb + 5l - 10b - 50} \right) = 210 \cr & \Rightarrow - 5l + 10b = 160 \cr & \Rightarrow - l + 2b = 32.....(ii) \cr} $$ Adding (i) and (ii), we get : b = 47 Putting b = 47 in equation (i), we get $$l$$ = 62 Hence, area of the rectangle : = $$lb$$ = (62 × 47) sq.units = 2914 sq.units Clearly, 2925 > A > 2900

2.

The perimeter of an isosceles triangle is equal to 14 cm and the lateral side is to the base in the ratio 5 : 4. The area of the triangle is :

• (A) 21 cm2
• (B) $$0.5\sqrt{21}$$ cm2
• (C) $$1.5\sqrt{21}$$ cm2
• (D) $$2\sqrt{21}$$ cm2

Show Answer

 Let the sides of the triangle be 5x, 5x and 4x cm respectively Then, 5x + 5x = 4x + 14 ⇒ 14 x = 14 ⇒ x = 1 So, a = 5 cm, b = 5 cm, c = 4 cm $$\eqalign{ & s = \frac{{{\text{a + b + c}}}}{2} \cr & \,\,\,\,\,\,\,\, = \left( {\frac{{14}}{2}} \right)cm \cr & \,\,\,\,\,\,\,\, = 7cm \cr} $$ (s - a) = 2 cm, (s - b) = 2 cm, (s - c) = 3 cm ∴ Area of the triangle : $$\eqalign{ & = \sqrt {7 \times 2 \times 2 \times 3} \cr & = 2\sqrt {21} \,c{m^2} \cr} $$

3.

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 : $$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$

4.

The circumference of a circle is equal to the side of a square whose area measures 407044 sq.cm. What is the area of the circle ?

• (A) 22583.2 sq.cm
• (B) 32378.5 sq.cm
• (C) 39483.4 sq.cm
• (D) 41263.5 sq.cm

Show Answer

 Area of the square = 407044 cm2 Side of the square = $$\sqrt {407044} $$  cm = 638 cm Circumference of circle = 638 cm Let the radius of the circle be R cm Then, $$\eqalign{ & 2\pi R = 638 \cr & \Rightarrow R = \frac{{638 \times 7}}{{2 \times 22}} \cr & \Rightarrow R = 101.5\,cm \cr} $$ ∴ Area of the circle : $$\eqalign{ & = \left( {\frac{{22}}{7} \times 101.5 \times 101.5} \right)c{m^2} \cr & = 32378.5\,\,c{m^2} \cr} $$

5.

The radius of a circle is 20% more than the height of a right-angled triangle. The base of the triangle is 36 cm. If the area of triangle and circle be equal, what will be area of circle ?

• (A) 72 cm2
• (B) 128 cm2
• (C) 144 cm2
• (D) 216 cm2

Show Answer

 Let the height of the triangle be x cm Then, radius of the circle = (120% of x) cm = $$\left( {\frac{{6x}}{5}} \right)$$ cm $$\eqalign{ & \therefore \frac{1}{2} \times 36 \times x = \frac{{22}}{7} \times \frac{{6x}}{5} \times \frac{{6x}}{5} \cr & \Rightarrow x = \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)cm \cr} $$ So, radius of the circle : $$\eqalign{ & = \left[ {\frac{6}{5} \times \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)} \right]cm \cr & = \left( {\frac{{105}}{{22}}} \right)cm \cr} $$ ∴ Area of the circle :$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}}} \right)c{m^2} \cr & = \left( {\frac{{1575}}{{22}}} \right)c{m^2} \cr & = 71.6\,c{m^2} \approx 72\,c{m^2} \cr} $$

6.

If every side of a triangle is doubled, the area of the new triangle is K times the area of the old one. K is equal to :

• (A) $$\sqrt{2}$$
• (B) 2
• (C) 3
• (D) 4

Show Answer

 $$\eqalign{ & {{\text{A}}_{\text{1}}} = \frac{{\sqrt 3 }}{2}{a^2}{\text{ and }} \cr & {{\text{A}}_{\text{2}}} = \frac{{\sqrt 3 }}{2}{\left( {2a} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \times \frac{{\sqrt 3 }}{2}{a^2} \cr & \,\,\,\,\,\,\,\,\,\, = 4{{\text{A}}_{\text{1}}} \cr & \therefore {\text{K}} = 4 \cr} $$

7.

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

8.

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

• (A) 40%
• (B) 42%
• (C) 44%
• (D) 46%

Show Answer

 Let original length = x metres and original breadth = y metres $$\eqalign{ & {\text{Original}}\,{\text{are}} = \left( {xy} \right){m^2} \cr & {\text{New}}\,{\text{length}} = \left( {\frac{{120}}{{100}}x} \right)m = \left( {\frac{6}{5}x} \right)m \cr & {\text{New}}\,{\text{breadth}} = \left( {\frac{{120}}{{100}}y} \right)m = \left( {\frac{6}{5}y} \right)m \cr & {\text{New}}\,{\text{area}} = \left( {\frac{6}{5}x \times \frac{6}{5}y} \right){m^2} = \left( {\frac{{36}}{{25}}xy} \right){m^2} \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{the}}\,{\text{original}}\,{\text{area = }}xy \cr & {\text{and}}\,{\text{new}}\,{\text{area}}\,\frac{{36}}{{25}}xy\,{\text{is}} \cr & = \left( {\frac{{36}}{{25}}} \right)xy - xy \cr & = xy\left( {\frac{{36}}{{25}} - 1} \right) \cr & = xy\left( {\frac{{11}}{{25}}} \right)\,or\,\left( {\frac{{11}}{{25}}} \right)xy \cr & \therefore {\text{Increase}}\,\% \cr & = \left( {\frac{{11}}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 44\% \cr} $$

9.

A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top and bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing ?

• (A) 60%
• (B) 65%
• (C) 70%
• (D) 80%

Show Answer

 Area of the sheet = (30 × 15) cm2 = 450 cm2 Area used for typing = [(30 - 5) × (15 - 2.5)] cm2 = 312.5 cm2 ∴ Required percentage : $$\eqalign{ & = \left( {\frac{{312.5}}{{450}} \times 100} \right)\% \cr & = 69.4\% \approx 70\% \cr} $$

10.

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 : $$\eqalign{ & = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr & = 1:2 \cr} $$