Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \frac{1}{2} \times {\text{Base}} \times {\text{Height}} = 60 \times 60 \cr & \Rightarrow \frac{1}{2} \times {\text{Base}} \times 90 = 3600 \cr & \Rightarrow {\text{Base}} = \left( {\frac{{3600 \times 2}}{{90}}} \right) \cr & \Rightarrow {\text{Base}} = 80\,m \cr} $$

Original area = $$\frac{1}{2}ab\sin \theta $$
New area :
$$\eqalign{ & = \frac{1}{2} \times \left( {2a} \right) \times \left( {2b} \right)\sin \theta \cr & = 4\left( {\frac{1}{2}ab\sin \theta } \right) \cr & = 4 \times {\text{ original area}} \cr} $$

Area of the field grazed :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 14 \times 14} \right)sq.ft \cr & = 616\,sq.ft \cr} $$
Number of days taken to graze the field :
$$\eqalign{ & = \frac{{616}}{{100}}\text{days} \cr & = 6\,\text{days}(\text{approx}) \cr} $$

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 :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 101.5 \times 101.5} \right)c{m^2} \cr & = 32378.5\,\,c{m^2} \cr} $$

Time taken by front wheel to complete one revolution :
$$\eqalign{ & = \left( {\frac{{3.5}}{{15}}} \right)\sec \cr & = \frac{7}{{30}}\operatorname{sec} \cr} $$
Time taken by rear wheel to complete one revolution :
$$\eqalign{ & = \left( {\frac{3}{{15}}} \right)\sec \cr & = \frac{1}{5}\sec \cr} $$
∴ Required time :
$$\eqalign{ & = \left[ {{\text{L}}{\text{.C}}{\text{.M}}{\text{. of }}\frac{7}{{30}}{\text{and}}\frac{1}{5}} \right]\sec \cr & = \left( {\frac{{{\text{L}}{\text{.C}}{\text{.M}}{\text{. of 7 and 1}}}}{{{\text{H}}{\text{.C}}{\text{.F}}{\text{. of 30 and 5}}}}} \right)\sec \cr & = \frac{7}{5}\sec \cr & = 1.4\,\sec \cr} $$

Area of the shaded region = Area of the semi-circle with radius a units $$-$$ Area of the triangle with base 2a units and height a units
$$\eqalign{ & = \left( {\frac{{\pi {a^2}}}{2} - \frac{1}{2} \times 2a \times a} \right)sq.units \cr & = \left( {\frac{{\pi {a^2}}}{2} - {a^2}} \right)sq.units \cr & = {a^2}\left( {\frac{\pi }{2} - 1} \right)sq.units \cr} $$

Radius of incircle $$ = \frac{a}{{2\sqrt 3 }}$$
Area of incircle $$ = \left( {\frac{{\pi \times {a^2}}}{{12}}} \right)c{m^2}$$
$$\eqalign{ & \therefore \frac{{\pi {a^2}}}{{12}} = 154 \cr & \Rightarrow {a^2} = \frac{{154 \times 12 \times 7}}{{22}} \cr & \Rightarrow a = 14\sqrt 3 \cr} $$
∴ Perimeter of the triangle :
$$\eqalign{ & = \left( {3 \times 14\sqrt 3 } \right)cm \cr & = \left( {42 \times 1.732} \right)cm \cr & = 72.7\,cm\,(\text{approx}) \cr} $$

Let the diameter of the round fort be D
Distance through the middle passage = D
Roundabout distance = $$\frac{{\pi D}}{2}$$
Time taken to cover distance D = 14 min
Time taken to cover distance :
$$\eqalign{ & = \frac{{\pi D}}{2} \cr & = \frac{{14}}{D} \times \frac{{\pi D}}{2} \cr & = 7\pi \cr & = \left( {7 \times \frac{{22}}{7}} \right)\min \cr & = 22\,\min \cr} $$
∴ Required time difference = (22 - 14) min = 8 min

Total cost to construct a boundary wall around a rectangular plot = Rs. 46000
Rate of construction per metre = Rs. 200
Perimeter of rectangular plot = $$\frac{46000}{200}$$   = 230 m
Let the length and breadth of rectangular plot be 13x metre and 10x metre respectively
$$\eqalign{ & \therefore 2\left( {13x + 10x} \right) = 230 \cr & \Rightarrow 2 \times 23x = 230 \cr & \Rightarrow x = \frac{{230}}{{2 \times 23}} \cr & \Rightarrow x = 5 \cr} $$
∴ Length = 13 × 5 = 65 cm
Breadth = 10 × 5 = 50 m
∴ Area of plot = 65 × 50 = 3250 sq. m²

Required perimeter :
= (AB + BC + CP + PQ + QR + RA)
= AB + BC + (CP + QR) + (PQ + RA)
= AB + BC + AB + BC
= [2 (8 + 4)] cm
= 24 cm