Area - Aptitude MCQ Questions and Solutions with Explanations

Let the required distance be x cm
Then,
$$\eqalign{ & \Rightarrow \frac{1}{2} \times \left( {6 + 10} \right) \times x = 32 \cr & \Rightarrow x = 4\,cm \cr} $$

Let the height of the triangle be x cm
Then, radius of the circle = (120% of x) cm = $$\left( {\frac{{6x}}{5}} \right)$$ cm
$$\eqalign{ & \therefore \frac{1}{2} \times 36 \times x = \frac{{22}}{7} \times \frac{{6x}}{5} \times \frac{{6x}}{5} \cr & \Rightarrow x = \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)cm \cr} $$
So, radius of the circle :
$$\eqalign{ & = \left[ {\frac{6}{5} \times \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)} \right]cm \cr & = \left( {\frac{{105}}{{22}}} \right)cm \cr} $$
∴ Area of the circle :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}}} \right)c{m^2} \cr & = \left( {\frac{{1575}}{{22}}} \right)c{m^2} \cr & = 71.6\,c{m^2} \approx 72\,c{m^2} \cr} $$

Distance covered in 5 sec :
$$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times 70 \times 10} \right)cm \cr & = 4400\,cm \cr & = 44\,m \cr} $$
Distance covered in 1 sec :
$$\eqalign{ & = \left( {\frac{{44}}{5}} \right)m \cr & = 8.8\,m \cr} $$
∴ Speed :
$$\eqalign{ & = 8.8\,m/\sec \cr & = \left( {8.8 \times \frac{{18}}{5}} \right)km/hr \cr & = 31.68\,km/hr \cr} $$

$$\eqalign{ & \frac{{2\pi {R_1}}}{{2\pi {R_2}}} = \frac{{23}}{{22}} \cr & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{23}}{{22}} \cr & \Rightarrow {R_1} = \frac{{23{R_2}}}{{22}} \cr & {\text{Also, }}{R_1} - {R_2} = 5\,m \cr & \Rightarrow \frac{{23{R_2}}}{{22}} - {R_2} = 5 \cr & \Rightarrow {R_2} = 110 \cr} $$
∴ Diameter of inner circle :
= (2 × 110) m
= 220 m

$$\eqalign{ & 2\pi R = 100 \cr & R = \frac{{100}}{{2\pi }} = \frac{{50}}{\pi } \cr & R = \frac{1}{2} \times {\text{diagonal}} \cr & \Rightarrow {\text{Diagonal}} = 2R \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 50}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{\pi } \cr} $$
$$\eqalign{ & \therefore {\text{Area of the square}} = \frac{1}{2} \times {\left( {{\text{diagonal}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times {\left( {\frac{{100}}{\pi }} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 2 }} \times \frac{{100}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 2 }}{\pi }cm \cr} $$

Let the original radius be R
New radius = (100 + 200)% of R = 300% of R = 3R
Original area = $$\pi $$R²
New area = $$\pi $$ × (3R)² = 9$$\pi $$R²
Increase in area :
$$\eqalign{ & = \left( {9\pi {{\text{R}}^2} - \pi {{\text{R}}^2}} \right) \cr & = 8\pi {{\text{R}}^2} \cr} $$
∴ Increase %:
$$\eqalign{ & = \left( {\frac{{8\pi {{\text{R}}^2}}}{{\pi {{\text{R}}^2}}} \times 100} \right)\% \cr & = 800\% \cr} $$

Area of square = 4 sq.units
Side of square = $$\sqrt 4 $$ = 2 units
Diagonal of square = $$2\sqrt 2 $$ units
Radius of the circle = $$2\sqrt 2 $$ units
Radius of the circle :
$$\eqalign{ & = \pi {r^2} \cr & = \pi \times {\left( {2\sqrt 2 } \right)^2} \cr & = 8\pi \,\text{sq. units} \cr} $$

Let the radius of circle be r cm and side of square be a cm
Then circumference of circle = $$2\pi r$$ and
Perimeter of square = 4a
According to the question,
$$\eqalign{ & 2\pi r = 4a \times \frac{{110}}{{100}} \cr & \Rightarrow 2\pi r = \frac{{44a}}{{10}} \cr & \Rightarrow r = \frac{{44a}}{{2\pi \times 10}} \cr & \Rightarrow r = \frac{{11a}}{{5\pi }} \cr & \Rightarrow a = \frac{{5\pi r}}{{11 }} . . . . .(i) \cr & {\text{Also, }}\pi {r^2} - {a^2} \Rightarrow 216 \cr} $$
$$ \Rightarrow \pi {r^2} - \frac{{25\pi {r^2}}}{{121}} = 216$$     $$\left[ {{\text{from equation (i)}}} \right]$$
$$\eqalign{ & \Rightarrow \frac{{121\pi {r^2} - 25\pi {r^2}}}{{121}} = 216 \cr & \Rightarrow {r^2}\left[ {121\pi - 25{\pi ^2}} \right] = 26136 \cr} $$
$$ \Rightarrow {r^2}\left[ {121 \times \frac{{22}}{7} - 25 \times \frac{{22}}{7} \times \frac{{22}}{7}} \right]$$      $$ = 26136$$
$$\eqalign{ & \Rightarrow {r^2}\left[ {\frac{{2662}}{7} - \frac{{12100}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2}\left[ {\frac{{6534}}{{49}}} \right] = 26136 \cr & \Rightarrow {r^2} = \frac{{26136 \times 49}}{{6534}} \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14\,cm \cr & \therefore a = \frac{{5\pi r}}{{11}} = 5 \times \frac{{22}}{7} \times \frac{{14}}{{11}} = 20\,cm \cr} $$
Hence, diagonal of square $$ = \sqrt 2 a = 20\sqrt 2 \,cm$$

Given length of the piece of wire = 84 cm
Length of the piece of wire = Circumference of circle
$$\eqalign{ & = 2\pi r \cr & = 2 \times \frac{{22}}{7} \times 84 \cr & = 528\,cm \cr} $$
Length of each side of square = a
∴ Perimeter of square = 4a = 528 cm
∴ Side of square = $$\frac{{528}}{4}$$ = 132 cm

Let the width of the table be x feet.
Then, length of the table = (x + 4) ft
$$\eqalign{ & \therefore x\left( {x + 4} \right) = 45 \cr & \Rightarrow {x^2} + 4x - 45 = 0 \cr & \Rightarrow {x^2} + 9x - 5x - 45 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 5\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr} $$
Hence, length of the table = (5 + 4) = 9 feet