Area - Aptitude MCQ Questions and Solutions with Explanations

Let AD = x and DC = y

Then,
$$AE = ED = \frac{x}{2}{\text{ and}}$$     $$DE = FC = \frac{y}{2}$$
$${\text{Area }}\left( {\vartriangle EDF} \right) = \left( {\frac{1}{2} \times \frac{x}{2} \times \frac{x}{2}} \right)$$     $$ = \frac{{xy}}{8}$$
$${\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = $$     $${\text{Area }}\left( {\vartriangle ADC} \right) - $$     $${\text{Area}}\left( {\vartriangle EDF} \right)$$
$$\eqalign{ & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{xy}}{2} - \frac{{xy}}{8} \cr & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{3xy}}{8} \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{xy}}{8}:\frac{{3xy}}{8} \cr & = 1:3 \cr} $$

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 :
$$\eqalign{ & = \pi \left( {{{14}^2} - {7^2}} \right){m^2} \cr & = \left( {\frac{{22}}{7} \times 21 \times 7} \right){m^2} \cr & = 462\,{m^2} \cr} $$

Let each wheel x revolutions per second
Then,
$$\left[ {\left( {2\pi \times \frac{7}{2} \times x} \right) + \left( {2\pi \times 7 \times x} \right)} \right]$$     $$ \times 10$$ $$ = 1980$$
$$ \Rightarrow \left( {\frac{{22}}{7} \times 7 \times x} \right) + $$     $$\left( {2 \times \frac{{22}}{7} \times 7 \times x} \right)$$   $$ = 198$$
$$\eqalign{ & \Rightarrow 66x = 198 \cr & \Rightarrow x = 3 \cr} $$
Distance moved by smaller wheel in 3 revolutions :
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{2} \times 3} \right)cm \cr & = 66\,cm \cr} $$
∴ Speed of smaller wheel :
$$\eqalign{ & = \frac{{66}}{3}cm/\sec \cr & = 22\,cm/\sec \cr} $$

Let the original radius of the circle be R cm
Then,
$$\eqalign{ & \pi \left[ {{{\left( {R + 1} \right)}^2} - {R^2}} \right] = 22 \cr & \Rightarrow \left( {2R + 1} \right) = \frac{{22 \times 7}}{{22}} \cr & \Rightarrow 2R + 1 = 7 \cr & \Rightarrow 2R = 6 \cr & \Rightarrow R = 3\,cm \cr} $$

Let the radius of the circle be r and the side of the square be a
Then,
Diagonal of the square = 2r
⇒ $$\sqrt 2 $$ a = 2r
⇒ a = $$\frac{2}{{\sqrt 2 }}$$ r
⇒ a = $$\sqrt 2 $$ r
Area of one shaded portion :
$$\eqalign{ & = \frac{1}{4}\left[ {\pi {r^2} - {{\left( {\sqrt 2 r} \right)}^2}} \right] \cr & = \frac{1}{4}\left( {\pi - 2} \right){r^2} \cr & \therefore \frac{1}{4}\left( {\pi - 2} \right){r^2} = \frac{4}{7} \cr & \Rightarrow \left( {\frac{{22}}{7} - 2} \right){r^2} = \frac{{16}}{7} \cr & \Rightarrow {r^2} = \frac{{16}}{7} \times \frac{7}{8} \times 2 \cr & \Rightarrow r = \sqrt 2 \cr} $$

Let the original circumference be x units.
Then,
New circumference = 120% of x = $$\left( {\frac{{6x}}{5}} \right)$$
Let original radius = r and new radius = R
$$\eqalign{ & 2\pi r = x \cr & \Rightarrow r = \frac{{7x}}{{2 \times 22}} \cr & \Rightarrow r = \frac{{7x}}{{44}} \cr} $$
And,
$$\eqalign{ & 2\pi R = \frac{{6x}}{5} \cr & \Rightarrow R = \frac{{6x}}{5} \times \frac{7}{{2 \times 22}} \cr & \Rightarrow R = \frac{{21x}}{{110}} \cr} $$
Original are :
$$\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times \frac{{7x}}{{44}} \times \frac{{7x}}{{44}}} \right) \cr & = \frac{{7{x^2}}}{{88}} \cr} $$
New area :
$$\eqalign{ & = \pi {R^2} \cr & = \left( {\frac{{22}}{7} \times \frac{{21x}}{{110}} \times \frac{{21x}}{{110}}} \right) \cr & = \frac{{63{x^2}}}{{550}} \cr} $$
Increase area :
$$\eqalign{ & = \left( {\frac{{63{x^2}}}{{550}} - \frac{{7{x^2}}}{{88}}} \right) \cr & = \frac{{77{x^2}}}{{2200}} \cr} $$
∴ Increase % :
$$\eqalign{ & = \left( {\frac{{77{x^2}}}{{2200}} \times \frac{{88}}{{7{x^2}}} \times 100} \right)\% \cr & = 44\% \cr} $$

Given :
$$\eqalign{ & \Rightarrow \frac{{{\text{Circumference of circle}}}}{{{\text{Diameter of circle }}}} = \frac{{22}}{7} \cr & \Rightarrow \frac{{{\text{Circumference of circle}}}}{{{\text{Twice of radius}}}} = \frac{{22}}{7} \cr & \Rightarrow \frac{{1\frac{4}{7}}}{{2r}} = \frac{{22}}{7} \cr & \Rightarrow \frac{{\frac{{11}}{7}}}{{2r}} = \frac{{22}}{7} \cr & \Rightarrow \frac{{11}}{{14r}} = \frac{{22}}{7} \cr & \Rightarrow 14r \times 22 = 11 \times 7 \cr & \Rightarrow r = \frac{{11 \times 7}}{{14 \times 22}} \cr & \Rightarrow r = \frac{1}{4}\,m \cr} $$

Side of square = a units
Side of hexagon = b units
According to the question,
$$\eqalign{ & 4a = 6b \cr & \Rightarrow \frac{a}{b} = \frac{6}{4} = \frac{3}{2} \cr} $$
$$\eqalign{ & \therefore \frac{{{\text{Area of hexagon}}}}{{{\text{Area of square}}}} \cr & = \frac{{6 \times \frac{{\sqrt 3 }}{4} \times {b^2}}}{{{a^2}}} \cr & = \frac{{6 \times \sqrt 3 \times 2 \times 2}}{{4 \times 3 \times 3}} \cr & = \frac{{2\sqrt 3 }}{3} \cr} $$
Hence required ratio = $$2\sqrt 3 :3$$

Let the length and breadth of the field be (5x) metres and (3x) metres respectively
Then,
2(5x + 3x) = 480
⇒ 8x = 240
⇒ x = 30
So, length = 150 m, breadth = 90 m
∴ Area of the field :
= (150 × 90) sq. m
= 13500 sq. m

Let the length and breadth of the plot be l and b metres respectively
Then, lb = 480
And,
$$\eqalign{ & \left( {l + 5} \right)\left( {b + 5} \right) = 725 \cr & \Rightarrow lb + 5\left( {l + b} \right) + 25 = 725 \cr & \Rightarrow 5\left( {l + b} \right) + 505 = 725 \cr & \Rightarrow \left( {l + b} \right) = \frac{{220}}{5} \cr & \Rightarrow \left( {l + b} \right) = 44\,\,\,\,\left[ {\because lb = 480} \right] \cr} $$
∴ Length of the fence :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = \left( {2 \times 44} \right)m \cr & = 88\,m \cr} $$