Area - Aptitude MCQ Questions and Solutions with Explanations

Let the sides of the two squares be x metres and y metres respectively
Then,
$$ \Rightarrow 29{x^2} = 6{y^2} - 1.....(i)$$
And,
$$\eqalign{ & \Rightarrow 9x - 4y = 1 \cr & \Rightarrow 4y = 9x - 1 \cr & \Rightarrow y = \frac{{9x - 1}}{4}.....(ii) \cr} $$
From (i) and (ii), we get :
$$\eqalign{ & \Rightarrow 29{x^2} = 6{\left( {\frac{{9x - 1}}{4}} \right)^2} - 1 \cr & \Rightarrow 29{x^2} = 6\left( {\frac{{81{x^2} + 1 - 18x}}{{16}}} \right) - 1 \cr & \Rightarrow 243{x^2} + 3 - 54x - 8 = 232{x^2} \cr & \Rightarrow 11{x^2} - 54x - 5 = 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {11x + 1} \right) = 0 \cr & \Rightarrow x = 5\,\,m \cr & \therefore y = \frac{{9x - 1}}{4} = \frac{{9 \times 5\,\,m - 1}}{4} = 11\,\,m \cr & \text{Required difference} \cr & = \left(11-5\right) m \cr & = 6\,m } $$

$$\eqalign{ & {\text{Area of }}\Delta {\text{ }}PQR \cr & = \frac{1}{2} \times QR \times PQ \cr & {\text{Given}},\,PQ = 3\,cm \cr & QR = \sqrt {{{\left( {PR} \right)}^2} - {{\left( {PQ} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {25} \right)}^2} - {3^2}} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {625 - 9} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {616} \,cm \cr & \,\,\,\,\,\,\,\,\,\, = 2\sqrt {154} \,cm \cr & \therefore {\text{Area}}\,{\text{of}}\,\Delta \,PQR \cr & = \left( {\frac{1}{2} \times 2\sqrt {154} \times 3} \right)c{m^2} \cr & = 3\sqrt {154} {\mkern 1mu} {\mkern 1mu} c{m^2} \cr} $$

$$\eqalign{ & {\text{Area of the traingle :}} \cr & = \frac{1}{2}ab\sin \theta \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \sin {{45}^ \circ }} \right)c{m^2} \cr & = \left( {\frac{1}{2} \times 10 \times 10 \times \frac{1}{{\sqrt 2 }}} \right)c{m^2} \cr & = \left( {\frac{{50}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}} \right)c{m^2} \cr & = 25\sqrt 2 \,c{m^2} \cr} $$

Let the altitude of the triangle be h₁ and base of each be b.
Then,
$$\frac{1}{2}$$ × b₁ × h₁ = b × h₂, where h₂ = 100 m
⇔ h₁ = 2 h₂
⇔ h₁ = (2 × 100) m
⇔ h₁ = 200 m

$$\eqalign{ & \because \pi {R^2} = 24.64 \cr & \Leftrightarrow {R^2} = \left( {\frac{{24.64}}{{22}} \times 7} \right) \cr & \Leftrightarrow {R^2} = 7.84 \cr & \Leftrightarrow R = \sqrt {7.84} \cr & \Leftrightarrow R = 2.8\,m \cr & \therefore {\text{Circumference :}} \cr & = \left( {2 \times \frac{{22}}{7} \times 2.8} \right)m \cr & = 17.60\,m \cr} $$

Let the radii of the two circle be 3r and 2r respectively
Then, required ratio :
$$\eqalign{ & = \frac{{2\pi \left( {3r} \right)}}{{2\pi \left( {2r} \right)}} \cr & = \frac{3}{2} \cr & = 3:2 \cr} $$

Radius of the plot = 21 m
Area of the path :
$$\eqalign{ & = \pi \left[ {{{\left( {24.5} \right)}^2} - {{\left( {21} \right)}^2}} \right]{m^2} \cr & = \left[ {\pi \left( {24.5 + 21} \right)\left( {24.5 - 21} \right)} \right]{m^2} \cr & = \left( {\frac{{22}}{7} \times 45.5 \times 3.5} \right){m^2} \cr & = 500.5\,{m^2} \cr} $$
∴ Cost of gravelling :
= Rs. (500.5 × 4)
= Rs. 2002

Required area = Area of the quadrant with radius 14 m :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 14 \times 14 \times \frac{{90}}{{360}}} \right){m^2} \cr & = 154\,{m^2} \cr} $$

$$\eqalign{ & s = \frac{{13 + 14 + 15}}{2} = \frac{{42}}{2} = 21 \cr & \therefore \,\,\vartriangle = \sqrt {21 \times 8 \times 7 \times 6} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times 2 \times 3 \times 7} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 84\,c{m^2} \cr} $$
Radius of circle :
$$\eqalign{ & = \left( {\frac{{13 \times 14 \times 15}}{{4 \times 84}}} \right)c{m^2} \cr & = \left( {\frac{{65}}{8}} \right)cm \cr & = 8.125\,\,cm \cr} $$

Distance mode by the skater in 44 sec
= Circumference of the circle
= $$\left( {2 \times \frac{{22}}{7} \times 28} \right)$$     m
= 176 m
Speed of skater :
= $$\left( {\frac{{176}}{{44}}} \right)$$   m/sec
= 4 m/sec
Perimeter of hexagon :
= (6 × 48) m
= 288 m
∴ Required difference :
= $$\left( {\frac{{288}}{{4}}} \right)$$   sec
= 72 sec