Area - Aptitude MCQ Questions and Solutions with Explanations

Let the sides of the triangle be 5x, 5x and 4x cm respectively
Then,
5x + 5x = 4x + 14
⇒ 14 x = 14
⇒ x = 1
So, a = 5 cm, b = 5 cm, c = 4 cm
$$\eqalign{ & s = \frac{{{\text{a + b + c}}}}{2} \cr & \,\,\,\,\,\,\,\, = \left( {\frac{{14}}{2}} \right)cm \cr & \,\,\,\,\,\,\,\, = 7cm \cr} $$
(s - a) = 2 cm, (s - b) = 2 cm, (s - c) = 3 cm
∴ Area of the triangle :
$$\eqalign{ & = \sqrt {7 \times 2 \times 2 \times 3} \cr & = 2\sqrt {21} \,c{m^2} \cr} $$

Let the length of sides of the two triangles be a₁ and a₂ respectively and their altitudes be h₁ and h₂ respectively.
Then,
$$\eqalign{ & \Leftrightarrow \frac{{\frac{{\sqrt 3 }}{4}a_1^2}}{{\frac{{\sqrt 3 }}{4}a_2^2}} = \frac{{25}}{{36}} \cr & \Rightarrow {\left( {\frac{{{a_1}}}{{{a_2}}}} \right)^2} = {\left( {\frac{5}{6}} \right)^2} \cr & \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{5}{6} \cr} $$
And
$$\eqalign{ & \Leftrightarrow \frac{{\frac{1}{2} \times {a_1} \times {h_1}}}{{\frac{1}{2} \times {a_2} \times {h_2}}} = \frac{{25}}{{36}} \cr & \Rightarrow \frac{5}{6} \times \frac{{{h_1}}}{{{h_2}}} = \frac{{25}}{{36}} \cr & \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{{25}}{{36}} \times \frac{6}{5} \cr & \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{5}{6} \cr} $$

$$\eqalign{ & \frac{1}{2} \times 6 \times d = 24 \cr & \Rightarrow d = \frac{{24}}{3} \cr & \Rightarrow d = 8\,cm \cr & OA = 4\,cm\,{\text{and}}\,OB = 3\,cm \cr & \therefore AB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {OB} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{4^2} + {3^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 5\,cm \cr} $$

$$\eqalign{ & \pi {R^2} = 7 \times \left( {2\pi R} \right) \cr & \Rightarrow R = 14 \cr & \therefore {\text{ Circumference :}} \cr & {\text{= }}\left( {2 \times \frac{{22}}{7} \times 14} \right)\text{units} \cr & = 88\,\text{units} \cr} $$

Since weight of the disc is proportional to its area, we have :
$$\eqalign{ & \pi \left( {{R^2} - {r^2}} \right) = \frac{{24}}{{25}}\pi {R^2} \cr & \Rightarrow {R^2} - {r^2} = \frac{{24}}{{25}}{R^2} \cr & \Rightarrow {r^2} = \frac{1}{{25}}{R^2} \cr & \Rightarrow {R^2} = 25{r^2} \cr & \Rightarrow R = 5r \cr} $$

Radius of the required circle :
$$\eqalign{ & = \left( {\frac{1}{2} \times 14} \right)cm \cr & = 7\,cm \cr} $$
Area of the circle :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 7 \times 7} \right)\,c{m^2} \cr & = 154\,c{m^2} \cr} $$

Let original radius be R cm
Then, original circumference = $$2\pi r$$ cm
New radius :
$$\eqalign{ & = \left( {175\% {\text{ of }}R} \right)cm \cr & = \left( {\frac{{175}}{{100}} \times R} \right)cm \cr & = \frac{{7R}}{4}\,cm \cr} $$
New circumference :
$$\eqalign{ & = \left( {2\pi \times \frac{{7R}}{4}} \right)cm \cr & = \frac{{7\pi R}}{2}\,cm \cr} $$
Increase in circumference :
$$\eqalign{ & = \left( {\frac{{7\pi R}}{2} - 2\pi R} \right)cm \cr & = \frac{{3\pi R}}{2}\,cm \cr} $$
Increase % :
$$\eqalign{ & = \left( {\frac{{3\pi R}}{2} \times \frac{1}{{2\pi R}} \times 100} \right)\% \cr & = 75\% \cr} $$

Given length and width of a square base plate of brass is x cm and 1 mm
Volume of the plate of square base = Area of base × height
$$\eqalign{ & = {x^2} \times \frac{1}{{10}} \cr & = \frac{{{x^2}}}{{10}}\,cu.cm. \cr} $$
According to the question,
$$\eqalign{ & \Rightarrow \frac{{{x^2}}}{{10}} \times 8.4 = 4725 \cr & \Rightarrow {x^2} = \frac{{4725 \times 10}}{{8.4}} \cr & \Rightarrow {x^2} = 5625 \cr & \Rightarrow x = \sqrt {5625} = 75\,cm \cr} $$

Radius of circle = 7cm
Given area of rectangle :
= Area of circle
= $${\frac{22}{7} \times 7 \times 7}$$
= 154 cm²

Let the breadth of the rectangle be x metres
Then, length of the rectangle = 2x metres
⇒ 2(2x + x) = 60
⇒ 6x = 60
⇒ x = 10
So, length = 20 m, breadth = 10 m
∴ Area = (20 × 10) m² = 200 m²