Area - Aptitude MCQ Questions and Solutions with Explanations

Side of a rhombus = 13 cm
Diagonal of rhombus = 10 cm
In ΔAOB
$$\eqalign{ & AO = \sqrt {{{13}^2} - {5^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {169 - 25} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {144} \cr & \,\,\,\,\,\,\,\,\,\, = 12\,cm \cr} $$
$$\eqalign{ & \Rightarrow AC = 24 \cr & {\text{Area of rhombus :}} \cr & = \frac{1}{2} \times {d_1} \times {d_2} \cr & = \frac{1}{2} \times 24 \times 10 \cr & = 120\,sq.cm \cr} $$

Let the breadth = x cm
Then, length = (x + 8) m
$$\eqalign{ & \therefore \left( {x + 8} \right)x = \left( {x + 15} \right)\left( {x - 4} \right) \cr & \Rightarrow {x^2} + 8x = {x^2} + 11x - 60 \cr & \Rightarrow x = 20 \cr} $$
So, length = 28 m and breadth = 20 m

Area of the roads :
= (80 × 10 + 60 × 10 - 10 × 10) m²
= 1300 m²
∴ Cost of gravelling :
= Rs. (1300 × 30)
= Rs. 39000

Maximum possible size of a flower bed :
= (H.C.F. of 110, 130, 190) sq. m
= 10 sq. m
∴ Maximum possible length = $$\left( {\frac{{10}}{2}} \right)$$ m = 5 m

$$\eqalign{ & {A_1} = {x^2}{\text{ and }}{A_2} = {\left( {2x} \right)^2} = 4{x^2} \cr & {\text{Increase in area}} = \left( {4{x^2} - {x^2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3{x^2} \cr & {\text{Increase % }} = \left( {\frac{{3{x^2}}}{{{x^2}}} \times 100} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 300\% \cr} $$

Let the height of the room be x metres
Then, breadth of the room = (2x) metres
Area of 4 walls :
$$\eqalign{ & = \left[ {2\left( {16 + 2x} \right) \times x} \right]{m^2} \cr & = \left( {32x + 4{x^2}} \right){m^2} \cr} $$
$$\eqalign{ & \therefore \left( {32x + 4{x^2}} \right) = 168 \times 2 \cr & \Rightarrow {x^2} + 8x - 84 = 0 \cr & \Rightarrow {x^2} + 14x - 6x - 84 = 0 \cr & \Rightarrow x\left( {x + 14} \right) - 6\left( {x + 14} \right) = 0 \cr & \Rightarrow \left( {x + 14} \right)\left( {x - 6} \right) = 0 \cr & \Rightarrow x = 6 \cr} $$
Area of the floor :
= (16 × 12) m²
= 192 m²

Note : The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
Since, XY || BC, we have :

$$\eqalign{ & \angle AXY = \angle ABC{\text{ and }} \cr & \angle AYX = \angle ACB \cr & {\text{Aslo, }}\angle A = \angle A\,\,{\text{(common)}} \cr & {\text{So, }}\vartriangle AXY \sim \,\vartriangle ABC \cr & {\text{Let area (}}\vartriangle ABC{\text{)}} = x\,\text{sq. units} \cr & {\text{Then,}} \cr & {\text{Area (}}\vartriangle AXY{\text{) = }}\frac{x}{2}\text{sq. units} \cr & \frac{{{{\left( {AB} \right)}^2}}}{{{{\left( {AX} \right)}^2}}} = \frac{x}{{\left( {\frac{x}{2}} \right)}} \cr & \Rightarrow \frac{{AB}}{{AX}} = \sqrt 2 \cr & \Rightarrow \frac{{AX + BX}}{{AX}} = \sqrt 2 \cr & \Rightarrow 1 + \frac{{BX}}{{AX}} = \sqrt 2 \cr & \Rightarrow \frac{{BX}}{{AX}} = \left( {\sqrt 2 - 1} \right) \cr & \Rightarrow \frac{{AX}}{{BX}} = \frac{1}{{\left( {\sqrt 2 - 1} \right)}}{\text{ Or 1}}:\left( {\sqrt 2 - 1} \right) \cr} $$

Let the side of the triangle be a cm and each side of the square be b cm
Then,
$$\eqalign{ & X = \frac{{\sqrt 3 }}{4}{a^2}{\text{ and }}Y = {b^2} \cr & {\text{Where }}3a = 4b,i,e.,b = \frac{{3a}}{4} \cr} $$
$$\therefore X = \frac{{\sqrt 3 {a^2}}}{4}{\text{ and }}Y = \frac{{9{a^2}}}{{16}}$$     $$\left[ {\because b = \frac{{3a}}{4}} \right]$$
$$\eqalign{ & {\text{Now,}} \cr & \frac{{\sqrt 3 {a^2}}}{4} = \frac{{1.732{a^2}}}{4} = 0.433{a^2} \cr & {\text{And }}\frac{{9{a^2}}}{{16}} = 0.5625{a^2} \cr & \therefore X < Y \cr} $$

Length of hall = 50 m
Breadth of hall = 45 m
Area of hall = (50 × 45) m
Maximum length of a square tiles = HCF of 50 m and 45 m = 5 metres
Area of tiles = 5 × 5 = 25 sq. m
∴ Number of tiles :
$$\eqalign{ & = \frac{{50 \times 45}}{{25}} \cr & = 90{\text{ tiles}} \cr} $$

Length :
$$\eqalign{ & = \left( {\frac{{1200}}{{30}}} \right)m \cr & = 40\,m \cr} $$
Diagonal :
$$\eqalign{ & = \left( {\sqrt {{{\left( {40} \right)}^2} + {{\left( {30} \right)}^2}} } \right)m \cr & = 50\,m \cr} $$
Length to be fenced :
$$\eqalign{ & = \left( {40 + 30 + 50} \right)m \cr & = 120\,m \cr} $$
∴ Cost of fencing :
$$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {120 \times 100} \right) \cr & {\text{ = Rs}}.12000 \cr} $$