Area - Aptitude MCQ Questions and Solutions with Explanations

Side of the square :
$$\eqalign{ & = \left( {\frac{{48}}{4}} \right)cm \cr & = 12\,cm \cr} $$
Area of the square :
$$\eqalign{ & = \left( {12 \times 12} \right)c{m^2} \cr & = 144\,c{m^2} \cr} $$

Required area :
= [(2 × 3) + (3 × 3) + (2 × 3)] cm²
= (6 + 9 + 6) cm²
= 21 cm²

Let the length and width of the rectangle be $$l$$ cm and b cm respectively.
Then,
$$\eqalign{ & \left( {l - 4} \right)\left( {b + 3} \right) = lb \cr & \Rightarrow lb + 3l - 4b - 12 = lb \cr & \Rightarrow 3l - 4b = 12.....(i) \cr & {\text{And, }} \cr & l - 4 = b + 3 \cr & \Rightarrow l - b = 7.....(ii) \cr} $$
Multiplying (ii) by 4 and subtracting (i) from it, we get :
$$l$$ = 16
Putting $$l$$ = 16 in (ii), we get : b = 9
∴ Perimeter of the original rectangle :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = [2\left( {16 + 9} \right)]cm \cr & = 50\,cm \cr} $$

$$\eqalign{ & \frac{1}{2} \times {\text{Base}} \times {\text{Height}} = p \cr & \Rightarrow \frac{1}{2} \times x \times {\text{Height}} = p \cr & \Rightarrow {\text{Height}} = \frac{{2p}}{x} \cr} $$

$$\eqalign{ & 9y = \frac{{\sqrt 3 }}{4} \times 6 \times 6 \cr & \Rightarrow y = \left( {\frac{{\sqrt 3 }}{4} \times 6 \times 6 \times \frac{1}{9}} \right) \cr & \Rightarrow y = \sqrt 3 \cr} $$

Let initial base = b cm and initial height = h cm
Then,
Initial area :
$$ = \left( {\frac{{1bh}}{2}} \right)c{m^2}$$
New base :
$$\eqalign{ & = \left( {140\% {\text{ of }}b} \right)cm \cr & = \left( {\frac{{140b}}{{100}}} \right)cm \cr & = \left( {\frac{{7b}}{5}} \right)cm \cr} $$
New height :
$$\eqalign{ & = \left( {60\% {\text{ of }}h} \right)cm \cr & = \left( {\frac{{60h}}{{100}}} \right)cm \cr & = \left( {\frac{{3h}}{5}} \right)cm \cr} $$
New area :
$$\eqalign{ & = \left( {\frac{1}{2} \times \frac{{7b}}{5} \times \frac{{3h}}{5}} \right)c{m^2} \cr & = \left( {\frac{{21bh}}{{50}}} \right)c{m^2} \cr} $$
Area decreased :
$$\eqalign{ & = \left( {\frac{{1bh}}{2} - \frac{{21bh}}{{50}}} \right)c{m^2} \cr & = \left( {\frac{{4bh}}{{50}}} \right)c{m^2} \cr} $$
Percentage decrease :
$$\eqalign{ & = \left( {\frac{{4bh}}{{50}} \times \frac{2}{{bh}} \times 100} \right)\% \cr & = 16\% \cr} $$

Radius :
$$\eqalign{ & = \frac{{3.5}}{2}cm \cr & = \frac{7}{4}cm \cr} $$
∴ Circumference :
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times \frac{7}{4}} \right)cm \cr & = 11\,cm \cr} $$

Let the radius of the circle be R cm
Then,
$$\eqalign{ & \pi {R^2} = 39424 \cr & \Rightarrow {R^2} = 39424 \times \frac{7}{{22}} = 12544 \cr & \Rightarrow R = 112\,cm \cr} $$
Perimeter of the square = 112 cm
Side of the square :
$$\eqalign{ & = \left( {\frac{{112}}{4}} \right)cm \cr & = 28\,cm \cr} $$
∴ Area of the square :
$$\eqalign{ & = \left( {28 \times 28} \right)c{m^2} \cr & = 784\,c{m^2} \cr} $$

Required number of rotations :
$$\eqalign{ & = \frac{{{\text{Circumference of bigger ring}}}}{{{\text{Circumference of smaller ring}}}} \cr & = \frac{{2 \times \pi \times 10}}{{2 \times \pi \times 2}} \cr & = 5 \cr} $$

Given :
$$\eqalign{ & \pi R + 2R = 36 \cr & \Leftrightarrow \left( {\pi + 2} \right)R = 36 \cr & \Leftrightarrow R = \frac{{36}}{{\left( {\frac{{22}}{7} + 2} \right)}}cm \cr & \Leftrightarrow R = 7\,cm \cr} $$
Required area :
$$\eqalign{ & = \frac{{\pi {R^2}}}{2} \cr & = \left( {\frac{{22}}{7} \times \frac{{7 \times 7}}{2}} \right)c{m^2} \cr & = 77\,c{m^2} \cr} $$