Area - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & 2\left( {14x + 11x} \right) = 100 \cr & \Rightarrow 25x = 50 \cr & \Rightarrow x = 2 \cr} $$
So, length and breadth of the rectangular field are 28 m and 22 m respectively
Area of the circle = Area of the rectangular field
= (28 × 22) m²
= 616 m²
Let the radius of the circle be r metres
Then,
$$\eqalign{ & \frac{{22}}{7} \times {r^2} = 616 \cr & \Rightarrow {r^2} = \frac{{616 \times 7}}{{22}} \cr & \Rightarrow {r^2} = 196 \cr & \Rightarrow r = 14\,m \cr} $$
∴ Diameter :
$$\eqalign{ & = \left( {2 \times 14} \right)\,m \cr & = 28\,m \cr} $$

$$\eqalign{ & \frac{{\pi R_1^2}}{{\pi R_2^2}} = \frac{{16}}{{49}} \cr & \Rightarrow \frac{{R_1^2}}{{\left( {14 \times 14} \right)}} = \frac{{16}}{{49}} \cr & \Rightarrow R_1^2 = \frac{{14 \times 14 \times 16}}{{49}} \cr & \Rightarrow R_1^2 = \frac{{14 \times 4}}{7} = 8\,m \cr} $$

Since the diameter is the longest chord of a circle
So, maximum possible distance = (10 + 5) m = 15 m

Angle traced by the minute hand in 5 minutes :
$$\eqalign{ & = {\left( {\frac{{360}}{{60}} \times 5} \right)^ \circ } \cr & = {30^ \circ } \cr} $$
Area of the sector :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 7 \times 7 \times \frac{{30}}{{360}}} \right)c{m^2} \cr & = 12.83\,c{m^2} \cr} $$

Let A, B and C denote the portions of the three boys
Then, AB = BC = AC
So, ΔABC is equilateral
Let the side of ΔABC be a
Then,
$$\eqalign{ & \frac{a}{{\sqrt 3 }} = 5 \cr & \Rightarrow a = 5\sqrt 3 \cr} $$
∴ Required shortest distance = $$5\sqrt 3 \,m$$

When 3 congruent circles touch each other externally, the triangle formed with their centers is an equilateral triangle. Hence when a circle is surrounded by identical circles, centers of two consecutive circles make an angle of 60° with the central circle. Thus, six identical circles a surround a circle of equal radius.

Let the radius of each circle be r cm

Then the side of the square will be 2r cm
Area covered by the four circle in the square
$$\eqalign{ & = 4 \times \frac{1}{4} \times \pi {r^2} \cr & = \pi {r^2}c{m^2} \cr} $$
Area of the square :
$$\eqalign{ & = {\left( {2r} \right)^2} \cr & = 4{r^2}c{m^2} \cr} $$
Now, according to the question,
Remaining area of the square
$$\eqalign{ & 4{r^2} - \pi {r^2} = 168 \cr & \Rightarrow {r^2}\left( {4 - \frac{{22}}{7}} \right) = 168 \cr & \Rightarrow {r^2} \times \left( {28 - 22} \right) = 168 \times 7 \cr & \Rightarrow {r^2} = \frac{{168 \times 7}}{6} \cr & \Rightarrow {r^2} = 28 \times 7 = 7 \times 4 \times 7 \cr & \Rightarrow r = \sqrt {7 \times 7 \times 2 \times 2} \cr & \Rightarrow r = 7 \times 2 \cr & \Rightarrow r = 14\,cm \cr} $$

Given length and breadth of courtyard is 25 m long and 16 m respectively
Area of courtyard :
$$\eqalign{ & = \left( {25 \times 16} \right)sq.{m^2} \cr & = 400\,sq.{m^2} \cr} $$
Dimensions of bricks 20 cm by 10 cm
Area of the surface of brick $$ = \left( {\frac{{20 \times 10}}{{10000}}} \right){m^2}$$
∴ Number of bricks :
$$\eqalign{ & = \frac{{400}}{{\frac{{20 \times 10}}{{10000}}}} \cr & = \frac{{400 \times 10000}}{{20 \times 10}} \cr & = 20000\,{\text{ bricks}} \cr} $$