Area - Aptitude MCQ Questions and Solutions with Explanations

Area of 4 walls :
$$\eqalign{ & = 2\left( {l + b} \right) \times h \cr & = \left[ {2\left( {12.5 + 9} \right) \times 7} \right]{m^2} \cr & = 301\,{m^2} \cr} $$
Area of 2 doors and 4 windows :
$$\eqalign{ & = \left[ {2\left( {2.5 \times 1.2} \right) + 4\left( {1.5 \times 1} \right)} \right]{m^2} \cr & = 12\,{m^2} \cr} $$
∴ Area to be painted :
$$\eqalign{ & = \left( {301 - 12} \right){m^2} \cr & = 289\,{m^2} \cr} $$
Cost of painting :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {289 \times 3.50} \right) \cr & = {\text{Rs}}{\text{. }}1011.50 \cr} $$

Let the sides of the triangles be x cm, (x + 1) cm and (x + 2) cm respectively.
Then,
$$\eqalign{ & x + \left( {x + 1} \right) + \left( {x + 2} \right) = 120 \cr & \Rightarrow 3x + 3 = 120 \cr & \Rightarrow 3x = 117 \cr & \Rightarrow x = 39 \cr} $$
∴ Length of greatest side :
= (39 + 2) cm
= 41 cm

Let the length of side of the square be a units
Then,
$$BE = EC = DF = FC = \frac{a}{2}$$
$$\eqalign{ & AE = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BE} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{a^2} + {{\left( {\frac{a}{2}} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{a^2} + \frac{{{a^2}}}{4}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{5{a^2}}}{4}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt 5 a}}{2} \cr} $$
Similarly, $$AF = \frac{{\sqrt 5 a}}{2}$$
$$\eqalign{ & EF = \sqrt {{{\left( {CE} \right)}^2} + {{\left( {CF} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{a}{2}} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{2{a^2}}}{4}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{a}{{\sqrt 2 }} \cr & EX = \frac{1}{2}EF = \frac{a}{{2\sqrt 2 }} \cr & AX = \sqrt {{{\left( {AE} \right)}^2} - {{\left( {EX} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {\frac{{\sqrt 5 a}}{2}} \right)}^2} - {{\left( {\frac{a}{{2\sqrt 2 }}} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{5{a^2}}}{4} - \frac{{{a^2}}}{8}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\frac{{9{a^2}}}{8}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{3a}}{{2\sqrt 2 }} \cr} $$
$$\eqalign{ & \therefore \,{\text{Area of }}\left( {\vartriangle AEF} \right): \cr & = \frac{1}{2} \times EF \times AX \cr & = \frac{1}{2} \times \frac{a}{{\sqrt 2 }} \times \frac{{3a}}{{2\sqrt 2 }} \cr & = \frac{{3{a^2}}}{8} \cr} $$
$$\eqalign{ & {\text{Required ratio :}} \cr & = \frac{{3{a^2}}}{8}:{a^2} \cr & = 3:8 \cr} $$

Let the length and breadth of the rectangle be (9x) cm and (7x) cm respectively.
Then,
$$\eqalign{ & 9x \times 7x = 252 \cr & \Rightarrow 63{x^2} = 252 \cr & \Rightarrow {x^2} = 4 \cr & \Rightarrow x = 2 \cr} $$
So, length = 18 cm, breadth = 14 cm
∴ Perimeter :
= 2(18 + 14) cm
= 64 cm

We have :
$$lb$$ = 60 and $$\sqrt {{l^2} + {b^2}} + l = 5b$$
Now,
$$\eqalign{ & {l^2} + {b^2} = {\left( {5b - l} \right)^2} \cr & \Rightarrow 24{b^2} - 10lb = 0 \cr & \Rightarrow 24{b^2} - 600 = 0 \cr & \Rightarrow {b^2} = 25 \cr & \Rightarrow b = 5 \cr & \therefore l = \left( {\frac{{60}}{b}} \right) \cr & \,\,\,\,\,\,\,\, = \left( {\frac{{60}}{5}} \right)m \cr & \,\,\,\,\,\,\,\, = 12\,m \cr} $$
So, length of the carpet = 12 m

Length of the fence :
= 2(60 + 40) m
= 200 m
Cost of fencing :
= Rs. (200 × 50)
= Rs. 10000
Area of the road :
= [(64 × 44) - (60 × 40)] m²
= (2816 - 2400) m²
= 416 m²
Let the cost of tiling the road be Rs. x per sq.m
∴ 416x + 10000 = 51600
⇒ 416x = 41600
⇒ x = Rs. 100

Let the side of each square formed by fencing parallel to breadth be x metres and that of each square formed by fencing parallel to length be y metres.

Then,
$$\eqalign{ & 3{x^2} + 3{y^2} = 4320 \cr & \Rightarrow {x^2} + {y^2} = 1440.....(i) \cr} $$
And,
$$\eqalign{ & x\left( {3x + y} \right) = 4320 \cr & \Rightarrow 3{x^2} + xy = 3\left( {{x^2} + {y^2}} \right) \cr & \Rightarrow xy = 3{y^2} \cr & \Rightarrow x = 3y.....(ii) \cr} $$
From (i) and (ii), we have :
$$\eqalign{ & {\left( {3y} \right)^2} + {y^2} = 1440 \cr & \Rightarrow 10{y^2} = 1440 \cr & \Rightarrow {y^2} = 144 \cr & \Rightarrow y = 12 \cr} $$
$${\text{So, }}x = 36$$
Length of rectangular plot :
$$\eqalign{ & = 3x + y \cr & = \left( {3 \times 36 + 12} \right)m \cr & = 120\,m \cr} $$
Breadth of rectangular plot = x = 36 m

Let the original length of sides be x
Then, new length :
$$\eqalign{ & = \left( {110\% {\text{ of }}x} \right) \cr & = \frac{{11x}}{{10}} \cr} $$
Original area $${x^2}$$
New area :
$$\eqalign{ & = {\left( {\frac{{11x}}{{10}}} \right)^2} \cr & = \frac{{121{x^2}}}{{100}} \cr} $$
Increase in area :
$$\eqalign{ & = \left( {\frac{{121{x^2}}}{{100}} - {x^2}} \right) \cr & = \frac{{21{x^2}}}{{100}} \cr} $$
∴ Increase % :
$$\eqalign{ & = \left( {\frac{{21{x^2}}}{{100}} \times \frac{1}{{{x^2}}} \times 100} \right)\% \cr & = 21\% \cr} $$

Let a = 3x cm, b = 4x and c = 5x
Then, s = 6x
$$\eqalign{ & A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & \,\,\,\,\,\,\, = \sqrt {6x \times 3x \times 2x \times x} \cr & \,\,\,\,\,\,\, = \left( {6{x^2}} \right)c{m^2} \cr & \therefore 6{x^2} = 216 \cr & \Rightarrow {x^2} = 36 \cr & \Rightarrow x = 6 \cr} $$
So, a = 18 cm, b = 24 cm and c = 30 cm
Perimeter :
= (18 + 24 + 30) cm
= 72 cm

Let the side of the square be a cm

Then, the length of its diagonal = $$\sqrt 2 $$ a cm
Area of equilateral triangle with side :
$$\eqalign{ & = \sqrt 2 a \cr & = \frac{{\sqrt 3 }}{4} \times {\left( {\sqrt 2 a} \right)^2} \cr & = \frac{{\sqrt 3 {a^2}}}{2} \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{\sqrt 3 {a^2}}}{2}:{a^2} \cr & = \sqrt 3 :2 \cr} $$