Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Required % }} = \left[ {\frac{{\pi \times {{\left( 5 \right)}^2}}}{{2\pi \times 5}} \times 100} \right]\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 250\% \cr} $$

Length of wire :
$$\eqalign{ & = 2\pi \times R \cr & = \left( {2 \times \frac{{22}}{7} \times 56} \right)cm \cr & = 352\,cm \cr} $$
Side of the square :
$$\eqalign{ & = \frac{{352}}{4}\,cm \cr & = 88\,cm \cr} $$
Area of the square :
$$\eqalign{ & = \left( {88 \times 88} \right)c{m^2} \cr & = 7744\,c{m^2} \cr} $$

Let the rear wheel make x revolutions
Then, the front wheel makes (x + 5) revolutions
(x + 5) × 40 = 48x
⇒ 8x = 200
⇒ x = 25
Distance travelled by the cart :
= (48 × 25) ft
= 1200 ft

Let the length of the side of the square be x cm
Then, radius of each semi-circle = $$\left( {\frac{x}{2}} \right)$$ cm
Total area :
$$\eqalign{ & = \left[ {{x^2} + \frac{\pi }{2}{{\left( {\frac{x}{2}} \right)}^2} + \frac{\pi }{2}{{\left( {\frac{x}{2}} \right)}^2}} \right]c{m^2} \cr & = \left( {{x^2} + \frac{\pi }{4} \times {x^2}} \right)c{m^2} \cr} $$

$$\eqalign{ & \therefore {x^2} + \frac{\pi }{4} \times {x^2} = 350 \cr & \Rightarrow {x^2} + \frac{{22{x^2}}}{{28}} = 350 \cr & \Rightarrow {x^2} + \frac{{11{x^2}}}{{14}} = 350 \cr & \Rightarrow \frac{{25{x^2}}}{{14}} = 350 \cr & \Rightarrow {x^2} = \left( {\frac{{350 \times 14}}{{25}}} \right) \cr & \Rightarrow {x^2} = 196 \cr & \Rightarrow x = 14\,cm \cr} $$

Radius of circum-circle :
$$\eqalign{ & = \frac{a}{{\sqrt 3 }} \cr & = \frac{{12}}{{\sqrt 3 }}\,cm \cr & = 4\sqrt 3 \,cm \cr} $$

Required area = (Area of an equilateral Δ of side 7 cm) - (3 × Area of sector with θ = 60° and r = 3.5 cm)
$$ = \left[ {\left( {\frac{{\sqrt 3 }}{4} \times 7 \times 7} \right) - \left( {3 \times \frac{{22}}{7} \times 3.5 \times 3.5 \times \frac{{60}}{{360}}} \right)} \right]c{m^2}$$
$$\eqalign{ & = \left( {\frac{{49\sqrt 3 }}{4} - 11 \times 0.5 \times 3.5} \right)c{m^2} \cr & = (21.217 - 19.25)c{m^2} \cr & = 1.967\,c{m^2} \cr} $$

Height of triangle = perimeter of square
Diagonal of square = $$8\sqrt 2 $$ m
∴ Length of each side of square :
$$ = \frac{{8\sqrt 2 }}{{\sqrt 2 }} = 8\,m$$
∴ Perimeter of square = 4 × 8 = 32 m = Height
Area of other square = 729
Side of square = $$\sqrt {729} $$  = 27 m = base of triangle
∴ Area of triangle :
$$\eqalign{ & = \frac{1}{2} \times {\text{Base}} \times {\text{Height}} \cr & = \frac{1}{2} \times 27 \times 32 \cr & = 432{\text{ sq}}{\text{.metre}} \cr} $$

Let the length and breadth of the rectangle be 3x and 2x respectively
Then,
Perimeter = 2 (3x + 2x) = 10x
And,
Area = (3x × 2x) = 6x²
$$\eqalign{ & \therefore \frac{{10x}}{{6{x^2}}} = \frac{5}{9} \cr & \Rightarrow 30x = 90 \cr & \Rightarrow x = 3 \cr} $$
So, breadth = (2 × 3) m = 6 m

Let the original length and breadth of the rectangle be $$l$$ and b respectively
New length :
$$ = 150\% {\text{ of }}l = \frac{{3l}}{2}$$
New breadth :
$$ = 75\% {\text{ of }}b = \frac{{3b}}{4}$$
Original area = $$lb$$
New area :
$$\eqalign{ & = \left( {\frac{{3l}}{2} \times \frac{{3b}}{4}} \right) \cr & = \frac{{9lb}}{8} \cr} $$
Increase in area :
$$\eqalign{ & = \left( {\frac{{9lb}}{8} - lb} \right) \cr & = \frac{{lb}}{8} \cr} $$
∴ Increase % :
$$\eqalign{ & = \left( {\frac{{lb}}{8} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 12.5\% \cr} $$

Area of square = $$\sqrt {1024} $$  cm = 32 cm
Length of rectangle = (2 × 32) cm = 64 cm
Breadth of rectangle = (32 - 12) cm = 20 cm
∴ Required ratio = 64 : 20 = 16 : 5