Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr & \Rightarrow {l^2} + {b^2} = 41 \cr & Also,\,\,lb = 20 \cr & {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2lb \cr & {\left( {l + b} \right)^2} = 41 + 40 \cr & {\left( {l + b} \right)^2} = 81 \cr & \left( {l + b} \right) = 9 \cr & \therefore {\text{ Perimeter}} = 2\left( {l + b} \right) = 18\,cm \cr} $$

Let the length of the field be x metres
Then, breadth of the field = $$\frac{3x}{4}$$ metres
$$\eqalign{ & x \times \frac{{3x}}{4} = 300 \cr & \Rightarrow {x^2} = 300 \times \frac{4}{3} \cr & \Rightarrow {x^2} = 400 \cr & \Rightarrow x = 20 \cr} $$
So, length = 20 m, breadth = 15 m
∴ Area of the garden :
$$\eqalign{ & = \left[ {\left\{ {\left( {20 + 3} \right) \times \left( {15 + 3} \right)} \right\} - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left[ {\left( {23 \times 18} \right) - \left( {20 \times 15} \right)} \right]{m^2} \cr & = \left( {414 - 300} \right){m^2} \cr & = 114\,{m^2} \cr} $$

length of rectangle = 4ft = (4 ×12) inch = 48 inch
length of rectangle = 6ft = (6 ×12) inch = 72 inch
∴ Number of squares :
$$\eqalign{ & = \frac{{48 \times 72}}{{\frac{1}{2} \times \frac{1}{2}}} \cr & = 13824 \cr} $$

Let the length of diagonal of the bigger square be x and that of the smaller square be y.
Then,
$$A = \frac{1}{2}{x^2}\,\,or\,\,x = \sqrt {2A} $$
And,
$$\frac{A}{2} = \frac{1}{2}{y^2}\,\,or\,\,y = \sqrt A $$
$$\eqalign{ & \therefore {\text{ Required ratio :}} \cr & = \frac{y}{x} = \frac{{\sqrt A }}{{\sqrt {2A} }} = 1:\sqrt 2 {\text{ }} \cr} $$

Let the breadth and height of the room be b metres and h metres respectively.
Then, length of the room = (2b) metres
Area of the ceiling :
$$\eqalign{ & = \left( {2b \times b} \right)m \cr & = \left( {2{b^2}} \right){m^2} \cr} $$
$$\eqalign{ & 2{b^2} = \frac{{5000}}{{25}} \cr & 2{b^2} = 200 \cr & {b^2} = 100 \cr & b = 10 \cr} $$
So, length = 20 m, breadth = 10 m
Area of 4 walls :
$$\eqalign{ & = \left[ {2\left( {20 + 10} \right) \times h} \right]{m^2} \cr & = \left( {60h} \right){m^2} \cr} $$
$$\eqalign{ & \therefore 60h = \frac{{64800}}{{240}} \cr & \Rightarrow 60h = 270 \cr & \Rightarrow h = \frac{{270}}{{60}} \cr & \Rightarrow h = 4.5\,m \cr} $$

Ratio of sides = $$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$$   = 6 : 4 : 3
Perimeter = 52 cm
So, sides are :
$$\eqalign{ & \left( {52 \times \frac{6}{{13}}} \right)cm \cr & \left( {52 \times \frac{4}{{13}}} \right)cm\,\& \cr & \left( {52 \times \frac{3}{{13}}} \right)cm \cr} $$
So, a = 24 cm, b = 16 cm, and c = 12 cm
∴ Length of smallest side = 12 cm

Let the side of the triangle be a
Then,
$$\eqalign{ & {a^2} = {\left( {\frac{a}{2}} \right)^2} + {x^2} \cr & \Leftrightarrow \frac{{3{a^2}}}{4} = {x^2} \cr & \Leftrightarrow {a^2} = \frac{{4{x^2}}}{3} \cr} $$
∴ Area :
$$\eqalign{ & = \frac{{\sqrt 3 }}{4}{a^2} \cr & = \frac{{\sqrt 3 }}{4} \times \frac{{4{x^2}}}{3} \cr & = \frac{{{x^2}}}{{\sqrt 3 }} \cr & = \frac{{\sqrt 3 {x^2}}}{3} \cr} $$

A square and a rhombus on the same base are equal in area.

Let the radius of the new park be R m
Then,
$$\eqalign{ & \pi {R^2} = \pi \times {8^2} + \pi \times {6^2} \cr & \Rightarrow \pi {R^2} = 100\pi \cr & \Rightarrow {R^2} = 100 \cr & \Rightarrow R = 10 \cr} $$

Distance covered in 4 sec :
$$\eqalign{ & = \left( {\frac{{30}}{7} \times 7} \right)m \cr & = 30\,m \cr} $$
Distance covered in 1 sec = $$\frac{{30}}{4}$$
Distance covered in 1 revolution :
$$\eqalign{ & = \left( {\frac{{30}}{4}} \right)m \cr & = \frac{{15}}{2}m \cr} $$
∴ Required speed :
$$\eqalign{ & = \left( {\frac{{15}}{2}} \right)m/s \cr & = \left( {\frac{{15}}{2} \times \frac{{18}}{5}} \right)km/hr \cr & = 27\,km/hr \cr} $$