Area - Aptitude MCQ Questions and Solutions with Explanations

Length of the fence =$$4\pi R$$
Where, $$R = \frac{{21}}{2}m$$

$$\eqalign{ & 4\pi R \cr & = \left( {4 \times \frac{{22}}{7} \times \frac{{21}}{2}} \right)m \cr & = 132\,m \cr} $$
Cost of fencing :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {132 \times \frac{{25}}{2}} \right) \cr & = {\text{Rs}}{\text{.1650}} \cr} $$

We have :
$$\eqalign{ & 2b + l = 30 \cr & \Rightarrow l = 30 - 2b \cr} $$
$$\eqalign{ & {\text{Area}} = {\text{100 }}{m^2} \cr & \Rightarrow l \times b = 100 \cr & \Rightarrow b\left( {30 - 2b} \right) = 100 \cr & \Rightarrow {b^2} - 15b + 50 = 0 \cr & \Rightarrow \left( {b - 10} \right)\left( {b - 5} \right) = 0 \cr & \Rightarrow b = 10{\text{ or }}b = 5 \cr} $$
When, b = 10, $$l$$ = 10 and when b = 5, $$l$$ = 20
Since the garden is rectangular, so its dimension is 20 m × 5 m

Let the actual length and width of the rectangle be $$l$$ and b respectively.
Then, measured length :
$$ = 100\% {\text{ of }}l = \frac{{11l}}{{10}}$$
Measured width :
$$ = 95\% {\text{ of }}b = \frac{{19b}}{{20}}$$
Actual area = $$lb$$
Measured area :
$$\eqalign{ & = \left( {\frac{{11l}}{{10}} \times \frac{{19b}}{{20}}} \right) \cr & = \frac{{209lb}}{{200}} \cr} $$
Error in measurement :
$$\eqalign{ & = \left( {\frac{{209lb}}{{200}} - lb} \right) \cr & = \frac{{9lb}}{{200}} \cr} $$
∴ Error % :
$$\eqalign{ & = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 4.5\% \cr} $$

Length of rectangle = 25 cm
Breadth of rectangle = 15 cm
Area of rectangle :
$$\eqalign{ & = \left( {25 \times 15} \right)c{m^2} \cr & = 375\,cm \cr} $$
Area of square :
$$\eqalign{ & = \left( {\frac{3}{5} \times 375} \right)c{m^2} \cr & = 225\,c{m^2} \cr} $$
Side of square :
$$\eqalign{ & = \sqrt {225} \,cm \cr & = 15\,cm \cr} $$
Perimeter of square :
$$\eqalign{ & = \left( {4 \times 15} \right)cm \cr & = 60\,cm \cr} $$

Let the side of the square be x cm
Then, its area = x² cm²
Area of the rectangle = 3x² cm²
∴ 40 × $$\frac{3}{2}$$ × x = 3x²
⇔ x = 20 cm

Let the length and breadth of the rectangle be $$l$$ and b respectively
$$\eqalign{ & l - 10 = b + 5 \cr & \Rightarrow l - b = 15.....(i) \cr & {\text{And,}} \cr & \Rightarrow lb - \left( {l - 10} \right)\left( {b + 5} \right) = 210 \cr & \Rightarrow lb - \left( {lb + 5l - 10b - 50} \right) = 210 \cr & \Rightarrow - 5l + 10b = 160 \cr & \Rightarrow - l + 2b = 32.....(ii) \cr} $$
Adding (i) and (ii), we get : b = 47
Putting b = 47 in equation (i), we get $$l$$ = 62
Hence, area of the rectangle :
= $$lb$$
= (62 × 47) sq.units
= 2914 sq.units
Clearly, 2925 > A > 2900

Required ratio :
$$\eqalign{ & = \left( {6 \times 6} \right):\left( {\frac{{\sqrt 3 }}{4} \times 6 \times 6} \right) \cr & = 4:\sqrt 3 \cr} $$

Let the sides be a cm and 2a cm

Then,
$$\eqalign{ & {a^2} + {\left( {2a} \right)^2} = {\left( {10} \right)^2} \cr & \Rightarrow 5{a^2} = 100 \cr & \Rightarrow {a^2} = 20 \cr} $$
∴ Area :
$$\eqalign{ & = \left( {\frac{1}{2} \times a \times 2a} \right) \cr & = {a^2} \cr & = 20\,c{m^2} \cr} $$

Let the length of the other side containing the right angle be x cm
Then,
$$\eqalign{ & \frac{1}{2} \times 4 \times x = 20 \cr & \Rightarrow x = 10{\text{ cm}} \cr} $$
Hypotenuse :
$$\eqalign{ & = \sqrt {{{10}^2} + {4^2}} {\text{ cm}} \cr & = \sqrt {116} {\text{ cm}} \cr & = 2\sqrt {29} \,{\text{cm}} \cr} $$
Let the altitude on the hypotenuse be h cm
Then,
$$\eqalign{ & \frac{1}{2} \times 2\sqrt {29} \times h = 20 \cr & \Rightarrow h = \frac{{20}}{{\sqrt {29} }}{\text{ cm}} \cr} $$

$$\eqalign{ & {{\text{A}}_{\text{1}}} = \frac{{\sqrt 3 }}{2}{a^2}{\text{ and }} \cr & {{\text{A}}_{\text{2}}} = \frac{{\sqrt 3 }}{2}{\left( {2a} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\, = 4 \times \frac{{\sqrt 3 }}{2}{a^2} \cr & \,\,\,\,\,\,\,\,\,\, = 4{{\text{A}}_{\text{1}}} \cr & \therefore {\text{K}} = 4 \cr} $$