Area - Aptitude MCQ Questions and Solutions with Explanations

Let the original length = x and original breadth = y
Decrease in area :
$$\eqalign{ & = xy - \left( {\frac{{80}}{{100}}x \times \frac{{90}}{{100}}y} \right) \cr & = \left( {xy - \frac{{18}}{{25}}xy} \right) \cr & = \frac{7}{{25}}xy \cr} $$
∴ Decrease% :
$$\eqalign{ & = \left( {\frac{7}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 28\% \cr} $$

Area of square ABCD = 36 m²
AB = $$\sqrt {36} $$ m = 6 m
AE = $$\frac{1}{2}$$ × AB = 3 m
Area of rectangle AEFG = 36 m²
∴ AE × EF = 36
⇒ EF = $$\frac{36}{3}$$ = 12 m
Perimeter of rectangle AEFG :
= 2 (AE + EF)
= [2 (3 + 12)] m
= 30 m

Area of the square :
$$\eqalign{ & = \left[ {\frac{1}{2} \times {{\left( {5.2} \right)}^2}} \right]{\text{ sq}}{\text{.cm}} \cr & {\text{ = }}\left( {\frac{1}{2} \times 27.04} \right){\text{ sq}}{\text{.cm}} \cr & = 13.52{\text{ sq}}{\text{.cm}} \cr} $$

Let the length of each side of the square be x cm
Then, length of rectangle = (x + 5) cm and its breadth = (x - 3) cm
$$\eqalign{ & \therefore \left( {x + 5} \right)\left( {x - 3} \right) = {x^2} \cr & \Rightarrow {x^2} + 2x - 15 = {x^2} \cr & \Rightarrow x = \frac{{15}}{2} \cr} $$
∴ Length :
$$\eqalign{ & = \left( {\frac{{15}}{2} + 5} \right)cm \cr & = \frac{{25}}{2}cm \cr} $$
Breadth :
$$\eqalign{ & = \left( {\frac{{15}}{2} - 3} \right)cm \cr & = \frac{9}{2}cm \cr} $$
Hence, perimeter :
$$\eqalign{ & = 2\left( {l + b} \right) \cr & = 2\left( {\frac{{25}}{2} + \frac{9}{2}} \right)cm \cr & = 34\,cm \cr} $$

$$\eqalign{ & AD = \sqrt {{4^2} + {6^2}} cm \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {52} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {13} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times 3.6} \right)cm \cr & \,\,\,\,\,\,\,\,\,\,\, = 7.2\,cm \cr} $$

Area of the whole figure :
$$ = {\text{Area (}}\vartriangle {\text{AED) + }}$$   $${\text{ Area (}}\square {\text{ ABCD) + }}$$   $${\text{Area (}}\vartriangle {\text{BFC)}}$$
$$ = \left[ {\left( {\frac{1}{2} \times 4 \times 6} \right) + \left( {12 \times 7.2} \right) + \left( {\frac{1}{2} \times 4 \times 6} \right)} \right]c{m^2}$$
$$\eqalign{ & = \left( {24 + 86.4} \right)c{m^2} \cr & = 110.4\,c{m^2} \cr} $$

Area of the triangle :
$$\eqalign{ & = \left( {\frac{1}{2} \times 12 \times 5} \right)c{m^2} \cr & = 30\,c{m^2} \cr} $$
Hypotenuse :
$$\eqalign{ & = \sqrt {{{12}^2} + {5^2}} \,cm \cr & = \sqrt {169} \,cm \cr & = 13\,cm \cr} $$
Let the perpendicular distance of the hypotenuse from the opposite vertex be x cm
Then,
$$\eqalign{ & \Rightarrow \frac{1}{2} \times 13 \times x = 30 \cr & \Rightarrow x = \frac{{60}}{{13}} \cr & \Rightarrow x = 4\frac{8}{{13}}\,cm \cr} $$

$$\eqalign{ & 2\pi R = 2\left( {l + b} \right) \cr & \Rightarrow 2\pi R = 2(26 + 18)cm \cr & \Rightarrow R = \left( {\frac{{88}}{{2 \times 22}} \times 7} \right)cm \cr & \Rightarrow R = 14\,cm \cr} $$
∴ Area of the circle :
$$\eqalign{ & = \pi {R^2} \cr & = \left( {\frac{{22}}{7} \times 14 \times 14} \right)c{m^2} \cr & = 616\,c{m^2} \cr} $$

$$\eqalign{ & \Rightarrow \frac{{\pi R_1^2}}{{\pi R_2^2}} = \frac{4}{1} \cr & \Rightarrow \frac{{R_1^2}}{{R_2^2}} = \frac{4}{1} \cr & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{2}{1}{\text{ Or }}2:1 \cr} $$

Let the circumference of front wheel be x metres

Then, Circumference of rear wheel = (x - 1) metres
$$\eqalign{ & \therefore \frac{{600}}{x} - \frac{{600}}{{\left( {x + 1} \right)}} = 30 \cr & \Rightarrow \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{{20}} \cr & \Rightarrow x\left( {x - 1} \right) = 20 \cr & \Rightarrow \left( {{x^2} + x - 20} \right) = 0 \cr & \Rightarrow \left( {x + 5} \right)\left( {x - 4} \right) = 0 \cr & \Rightarrow x = 4\,m \cr} $$

Let the radius of each of the circle and the semi-circle be r units
Diagonal of the first square = 2r units
Let the side of the second be a units

Then,
$$\eqalign{ & {r^2} = {a^2} + {\left( {\frac{a}{2}} \right)^2} \cr & \Rightarrow {r^2} = \frac{{5{a^2}}}{4} \cr & \Rightarrow {a^2} = \frac{{4{r^2}}}{5} \cr} $$
∴ Ratio of the areas of the two squares :
$$\eqalign{ & = \frac{{\frac{1}{2} \times {{\left( {2r} \right)}^2}}}{{{a^2}}} \cr & = \frac{{2{r^2}}}{{\left( {\frac{{4{r^2}}}{5}} \right)}} = \frac{5}{2} \cr & = \frac{5}{2} \cr & = 5:2 \cr} $$