Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \Rightarrow 2\pi {R_1} = 132 \cr & \Rightarrow {R_1} = \frac{{132 \times 7}}{{2 \times 22}} \cr & \Rightarrow {R_1} = 21\,m \cr & \Rightarrow 2\pi {R_2} = 176 \cr & \Rightarrow {R_2} = \frac{{176 \times 7}}{{2 \times 22}} \cr & \Rightarrow {R_2} = 28\,m \cr & \therefore {\text{Required difference :}} \cr & = \pi \left( {R_2^2 - R_2^2} \right) \cr & = \pi \left( {{R_2} + {R_1}} \right)\left( {{R_2} - {R_1}} \right) \cr & = \left( {\frac{{22}}{7} \times 49 \times 7} \right){m^2} \cr & = 1078\,{m^2} \cr} $$

Let the radii of the two circle be r and 3r respectively.
Then, required ratio :
$$ = \frac{{\pi {r^2}}}{{\pi {{\left( {3r} \right)}^2}}} = \frac{{\pi {r^2}}}{{9\pi {r^2}}} = 1:9$$

Let the radius of the ground be R metres.
Then,
$$\eqalign{ & \Rightarrow 2\pi R = 88 \cr & \Rightarrow R = \left( {\frac{{88 \times 7}}{{2 \times 22}}} \right) \cr & \Rightarrow R = 14\,m \cr} $$
Area of land strip :
$$\eqalign{ & = \pi \left[ {{{\left( {14} \right)}^2} - {{\left( {11} \right)}^2}} \right]{m^2} \cr & = \left( {\frac{{22}}{7} \times 25 \times 3} \right){m^2} \cr & = \left( {\frac{{1650}}{7}} \right){m^2} \cr} $$
∴ Cost of levelling :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {\frac{{1650}}{7} \times 7} \right) \cr & = {\text{Rs}}{\text{. 1650}} \cr} $$

$$2\pi r = 100$$
So, length of the arc :
$$\eqalign{ & = \frac{{2\pi r\theta }}{{360}} \cr & = \left( {\frac{{100 \times 20}}{{360}}} \right){\text{units}} \cr & = \left( {\frac{{50}}{9}} \right){\text{ units}} \cr & = 5.55{\text{ units}} \cr} $$

$$\eqalign{ & {\text{Radius}} = \frac{{{\text{Area}}}}{{{\text{Semi - perimeter}}}} \cr & {\text{Radius}} = \left( {{\text{Area}} \times \frac{2}{{{\text{Area}}}}} \right) \cr & {\text{Radius}} = 2 \cr} $$

Let length of each side of square = 2π
According to the question,
$$\eqalign{ & \frac{{\pi {r^2}}}{2} + \frac{{\pi {r^2}}}{2} + 42 = {\text{Area of square}} \cr & \Rightarrow \pi {r^2} + 42 = 4{r^2} \cr & \Rightarrow 4{r^2} - \pi {r^2} = 42 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 42 \cr & \Rightarrow {r^2}\left( {\frac{{28 - 22}}{7}} \right) = 42 \cr & \Rightarrow \frac{{6{r^2}}}{7} = 42 \cr & \Rightarrow {r^2} = \frac{{42 \times 7}}{6} \cr & \Rightarrow {r^2} = 7 \times 7 \cr & \Rightarrow r = 7 \cr & \therefore 2r = 14\,cm \cr} $$

Given base of triangle and its height is 15 cm and 12 cm respectively
Area of first triangle :
$$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 15 \times 12 \cr & = 90{\text{ sq}}{\text{. cm}} \cr} $$
According to the question,
Let height of triangle be h cm
Area of new triangle = 180 sq. cm
Base = 20 sq. cm
$$\eqalign{ & \Rightarrow 180 = \frac{1}{2} \times 20 \times {\text{Height}} \cr & \Rightarrow {\text{h}} = \frac{{2 \times 180}}{{20}} \cr & \Rightarrow {\text{h}} = 18{\text{ cm}} \cr} $$

We have :
(l - b) = 23 and 2(l + b) = 206 or (l + b) = 103
Solving the two equations, we get :
l = 63 and b = 40
∴ Area = (l × b) = (63 × 40)m² = 2520 m²

Let breadth = x cm
Then, length = 3x cm
$$\eqalign{ & \Rightarrow {x^2} + {\left( {3x} \right)^2} = {\left( {8\sqrt {10} } \right)^2} \cr & \Rightarrow 10{x^2} = 640 \cr & \Rightarrow {x^2} = 64 \cr & \Rightarrow x = 8 \cr} $$
So, length = 24 cm and breadth = 8 cm
∴ Perimeter = [2(24 + 8)] cm = 64 cm

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