Area - Aptitude MCQ Questions and Solutions with Explanations

We have :
$$\eqalign{ & AE \bot BC{\text{ and }} \cr & AD = BD = CD = r \cr & AE = AD + DE = r + DE \cr} $$
In Δ BDC,
$$\eqalign{ & BE = \sqrt {{{\left( {BD} \right)}^2} - {{\left( {DE} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{r^2} - {{\left( {DE} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \sqrt {\left( {r - DE} \right)\left( {r + DE} \right)} \cr} $$
∴ Area of the triangle :
$$\eqalign{ & = \frac{1}{2} \times BC \times AE \cr & = \frac{1}{2} \times 2BE \times AE \cr & = BE \times AE \cr & = \sqrt {\left( {r - DE} \right)\left( {r + DE} \right)} \left( {r + DE} \right) \cr & = {\left( {r - DE} \right)^{\frac{1}{2}}}{\left( {r + DE} \right)^{\frac{3}{2}}} \cr} $$

Clearly, the longer diagonal of the kite is the diameter of the circle
Also,
∠ABC = 90° (angle in a semi-circle)
Let AB = AD = 3x and BC = CD = 4x
Then,
$$AC = \sqrt {A{B^2} + B{C^2}} = 5x$$
Area of the kite = 2 × area (ΔABC)
$$\eqalign{ & = 2 \times {\text{Area (}}\vartriangle {\text{ABC)}} \cr & = 2 \times \frac{1}{2} \times BC \times AB \cr & = 3x \times 4x \cr & = 12{x^2} \cr} $$
Area of the circle :
$$\eqalign{ & = \pi {r^2} \cr & = \left( {\frac{{22}}{7} \times \frac{{5x}}{2} \times \frac{{5x}}{2}} \right) \cr & = \frac{{275{x^2}}}{{14}} \cr} $$
Area wasted :
$$\eqalign{ & = \left( {\frac{{275{x^2}}}{{14}} - 12{x^2}} \right) \cr & = \frac{{107{x^2}}}{{14}} \cr} $$
Required percentage :
$$\eqalign{ & = \left( {\frac{{107}}{{14}} \times \frac{{14}}{{275}} \times 100} \right)\% \cr & = 39\% \cr} $$

When folded along breadth, we have :
$$\eqalign{ & 2\left( {\frac{l}{2} \times b} \right) = 34 \cr & or,l + 2b = 34.....(i) \cr} $$
When folded along length, we have :
$$\eqalign{ & 2\left( {l \times \frac{b}{2}} \right) = 38 \cr & or,2l + b = 38.....(ii) \cr} $$
Solving (i) and (ii), we get :
$$l$$ = 14 and b = 10
∴ Area of the paper :
= (14 × 10) cm²
= 140 cm²

Let original length = x and original breadth = y
Then,
Original area = xy
New area :
$$\eqalign{ & = \left[ {\frac{{\left( {100 - r} \right)}}{{100}} \times x} \right]\left[ {\frac{{\left( {105 + r} \right)}}{{100}} \times y} \right] \cr & = \left[ {\left( {\frac{{10500 - 5r - {r^2}}}{{10000}}} \right)xy} \right] \cr & \therefore \left( {\frac{{10500 - 5r - {r^2}}}{{10000}}} \right)xy = xy \cr & \Rightarrow {r^2} + 5r - 500 = 0 \cr & \Rightarrow \left( {r + 25} \right)\left( {r - 20} \right) = 0 \cr & \Rightarrow r = 20 \cr} $$

$$\eqalign{ & {\text{Area}} = \frac{{{\text{Total cost}}}}{{{\text{Rate}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{6165}}{{685}}} \right){\text{hectares}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {9 \times 10000} \right){m^2} \cr} $$
∴ Side of the square :
$$\eqalign{ & = \sqrt {90000} \,m \cr & = 300\,m \cr} $$
Perimeter of the field = (300 × 4) m = 1200 m
Cost of fencing = Rs. (1200 × 48.75) = Rs. 58500

$$\eqalign{ & d = \sqrt 2 \times l \cr & \Rightarrow l = \frac{{20}}{{\sqrt 2 }} \cr} $$
∴ Perimeter :
$$\eqalign{ & = \left( {4l} \right)cm \cr & = \left( {\frac{{4 \times 20}}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}} \right)cm \cr & = 40\sqrt 2 \,cm \cr} $$

Let the length and breadth of the rectangle be $$l$$ cm and n cm respectively
Then,
$$\eqalign{ & 2\left( {l + b} \right) = 10 \cr & \Rightarrow l + b = 5 \cr & \Rightarrow b = \left( {5 - l} \right)cm \cr} $$
Area of the rectangle :
$$\eqalign{ & = l\left( {5 - l} \right)c{m^2} \cr & = \left( {5l - {l^2}} \right)c{m^2} \cr} $$
Area of the square :
$$\eqalign{ & = 2\left( {5l - {l^2}} \right)c{m^2} \cr & = \left( {10l - 2{l^2}} \right)c{m^2} \cr} $$
$$\eqalign{ & \therefore \left( {l + 1} \right)\left( {6 - l} \right) = \left( {10l - 2{l^2}} \right) \cr & \Rightarrow {l^2} - 5l + 6 = 0 \cr & \Rightarrow \left( {l - 3} \right)\left( {l - 2} \right) = 0 \cr & \Rightarrow l = 3 \cr} $$
Area of the square :
$$\eqalign{ & = \left( {10 \times 3 - 2 \times 9} \right)c{m^2} \cr & = 12\,c{m^2} \cr} $$
∴ Side of the square $$ = \sqrt {12} \,cm = 2\sqrt 3 $$

$$\eqalign{ & QR = \sqrt {{{\left( {PR} \right)}^2} - {{\left( {PQ} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{25}^2} + {3^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {625 - 9} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {616} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2\sqrt {154} \,cm \cr} $$

Let the length of each of the sides containing the right angle be x cm
Then,
Hypotenuse :
$$\eqalign{ & = \sqrt {{x^2} + {x^2}} \,cm \cr & = \sqrt {2{x^2}} \,cm \cr & = \sqrt 2 x\,cm \cr} $$
Perimeter of the triangle :
$$\eqalign{ & = \left( {x + x + \sqrt 2 x} \right)cm \cr & = \left( {2x + \sqrt 2 x} \right)cm \cr & = \sqrt 2 x\left( {\sqrt 2 + 1} \right)cm \cr & \therefore \sqrt 2 x\left( {\sqrt 2 + 1} \right) = \left( {4\sqrt 2 + 4} \right) \cr & \Rightarrow \sqrt 2 x\left( {\sqrt 2 + 1} \right) = 4\left( {\sqrt 2 + 1} \right) \cr & \Rightarrow \sqrt 2 x = 4 \cr & \Rightarrow x = 2\sqrt 2 \cr} $$
Hence, hypotenuse :
$$\eqalign{ & = \left( {\sqrt 2 \times 2\sqrt 2 } \right)cm \cr & = 4\,cm \cr} $$

$$\eqalign{ & {\text{Area of the field}} = \left( {{\text{Base}} \times {\text{Height}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {150 \times 80} \right){m^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12000\,{m^2} \cr} $$
∴ Cost of watching :
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {12000 \times 0.50} \right) \cr & = {\text{Rs}}{\text{. 6000}} \cr} $$