Practice MCQ Questions and Answer on Area

31.

The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

• (A) 42$$\sqrt 2 $$ sq. cm
• (B) 42 sq. cm
• (C) 84 sq. cm
• (D) 48 sq. cm

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 Let each equal side of isosceles triangle be x cm Perimeter of an isosceles triangle = 36 cm $$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$ BD = DC = 7cm From ΔABD By using Pythagoras theorem : $$\eqalign{ & AD = \sqrt {A{B^2} - B{D^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{11}^2} - {7^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {121 - 49} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {72} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3 \times 2\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\sqrt 2 \,cm \cr} $$ ∴ Area of ΔABC $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$

32.

The radius of the wheel of a vehicle is 70 cm. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle is :

• (A) 29.46 km/hr
• (B) 31.68 km/hr
• (C) 32.72 km/hr
• (D) 36.25 km/hr

Show Answer

 Distance covered in 5 sec : $$\eqalign{ & = \left( {2 \times \frac{{22}}{7} \times 70 \times 10} \right)cm \cr & = 4400\,cm \cr & = 44\,m \cr} $$ Distance covered in 1 sec : $$\eqalign{ & = \left( {\frac{{44}}{5}} \right)m \cr & = 8.8\,m \cr} $$ ∴ Speed : $$\eqalign{ & = 8.8\,m/\sec \cr & = \left( {8.8 \times \frac{{18}}{5}} \right)km/hr \cr & = 31.68\,km/hr \cr} $$

33.

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $$12\sqrt 2 $$ cm, then the area of the triangle is :

• (A) $$24\sqrt 2 $$ cm2
• (B) $$24\sqrt 3 $$ cm2
• (C) $$48\sqrt 3 $$ cm2
• (D) $$64\sqrt 3 $$ cm2

Show Answer

 Let the side of the square be a cm Then, its diagonal = $$\sqrt 2 $$ a cm Now, $$\sqrt 2 $$ a = $$12\sqrt 2 $$ ⇒ a = 12 cm Perimeter of the square = 4a = 48 cm Perimeter of the equilateral triangle = 48 cm Each side of the triangle = 16 cm Area of the triangle : $$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times 16 \times 16} \right)c{m^2} \cr & = \left( {64\sqrt 3 } \right)c{m^2} \cr} $$

34.

The radius of a circle is 20% more than the height of a right-angled triangle. The base of the triangle is 36 cm. If the area of triangle and circle be equal, what will be area of circle ?

• (A) 72 cm2
• (B) 128 cm2
• (C) 144 cm2
• (D) 216 cm2

Show Answer

 Let the height of the triangle be x cm Then, radius of the circle = (120% of x) cm = $$\left( {\frac{{6x}}{5}} \right)$$ cm $$\eqalign{ & \therefore \frac{1}{2} \times 36 \times x = \frac{{22}}{7} \times \frac{{6x}}{5} \times \frac{{6x}}{5} \cr & \Rightarrow x = \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)cm \cr} $$ So, radius of the circle : $$\eqalign{ & = \left[ {\frac{6}{5} \times \left( {\frac{{18 \times 7 \times 5 \times 5}}{{22 \times 6 \times 6}}} \right)} \right]cm \cr & = \left( {\frac{{105}}{{22}}} \right)cm \cr} $$ ∴ Area of the circle :$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}}} \right)c{m^2} \cr & = \left( {\frac{{1575}}{{22}}} \right)c{m^2} \cr & = 71.6\,c{m^2} \approx 72\,c{m^2} \cr} $$

35.

A rectangular lawn 80 metres by 60 metres has two roads each 10 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling them at Rs. 30 per square metre.

• (A) Rs. 3600
• (B) Rs. 3900
• (C) Rs. 36000
• (D) Rs. 39000

Show Answer

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

36.

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

• (A) 34
• (B) 40
• (C) 68
• (D) 88

Show Answer

 We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

37.

The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

• (A) 40%
• (B) 42%
• (C) 44%
• (D) 46%

Show Answer

 Let original length = x metres and original breadth = y metres $$\eqalign{ & {\text{Original}}\,{\text{are}} = \left( {xy} \right){m^2} \cr & {\text{New}}\,{\text{length}} = \left( {\frac{{120}}{{100}}x} \right)m = \left( {\frac{6}{5}x} \right)m \cr & {\text{New}}\,{\text{breadth}} = \left( {\frac{{120}}{{100}}y} \right)m = \left( {\frac{6}{5}y} \right)m \cr & {\text{New}}\,{\text{area}} = \left( {\frac{6}{5}x \times \frac{6}{5}y} \right){m^2} = \left( {\frac{{36}}{{25}}xy} \right){m^2} \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{the}}\,{\text{original}}\,{\text{area = }}xy \cr & {\text{and}}\,{\text{new}}\,{\text{area}}\,\frac{{36}}{{25}}xy\,{\text{is}} \cr & = \left( {\frac{{36}}{{25}}} \right)xy - xy \cr & = xy\left( {\frac{{36}}{{25}} - 1} \right) \cr & = xy\left( {\frac{{11}}{{25}}} \right)\,or\,\left( {\frac{{11}}{{25}}} \right)xy \cr & \therefore {\text{Increase}}\,\% \cr & = \left( {\frac{{11}}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 44\% \cr} $$

38.

The dimensions of a room are 12.5 metres by 9 metres by 7 metres. There are 2 doors and 4 windows in the room; each door measures 2.5 metres by 1.2 metres and each window 1.5 metres by 1 metre. Find the cost of painting the walls at Rs. 3.50 per square metre.

• (A) Rs. 150.50
• (B) Rs. 1011.50
• (C) Rs. 1101.50
• (D) Cannot be determined

Show Answer

 Area of 4 walls : $$\eqalign{ & = 2\left( {l + b} \right) \times h \cr & = \left[ {2\left( {12.5 + 9} \right) \times 7} \right]{m^2} \cr & = 301\,{m^2} \cr} $$ Area of 2 doors and 4 windows : $$\eqalign{ & = \left[ {2\left( {2.5 \times 1.2} \right) + 4\left( {1.5 \times 1} \right)} \right]{m^2} \cr & = 12\,{m^2} \cr} $$ ∴ Area to be painted : $$\eqalign{ & = \left( {301 - 12} \right){m^2} \cr & = 289\,{m^2} \cr} $$ Cost of painting : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {289 \times 3.50} \right) \cr & = {\text{Rs}}{\text{. }}1011.50 \cr} $$

39.

If each side of a rectangle is increased by 50%, its area will increase by :

• (A) 50%
• (B) 125%
• (C) 150%
• (D) 200%

Show Answer

 Let original length = $$l$$ metres and original breadth = b metres Original area : $$ = \left( {lb} \right){m^2}$$ New length : $$\eqalign{ & = \left( {\frac{{150l}}{{100}}} \right)m \cr & = \left( {\frac{{3l}}{2}} \right)m \cr & \text{New breadth :} \cr & = \left( {\frac{{150b}}{{100}}} \right)m \cr & = \left( {\frac{{3b}}{2}} \right)m \cr & \text{New area :} \cr & = \left( {\frac{{3l}}{2} \times \frac{{3b}}{2}} \right){m^2} \cr & = \left( {\frac{{9lb}}{4}} \right){m^2} \cr & \text{Increase} = 1 - \frac{9lb}{4} = \frac{5lb}{4} \cr & \therefore \text{ Increase % :} \cr & = \left( {\frac{{5lb}}{4} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 125\% \cr} $$

40.

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

• (A) 40m
• (B) 50m
• (C) 120m
• (D) Data inadequate

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Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Let}}\,{\text{breadth}} = x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{length}} = \,\left( {x + 20} \right)\,{\text{metres}} \cr & {\text{Perimeter}} = {\frac{{5300}}{{26.50}}} m = 200m \cr & \therefore 2\left[ {\left( {x + 20} \right) + x} \right] = 200 \cr & \Rightarrow 2x + 20 = 100 \cr & \Rightarrow 2x = 80 \cr & \Rightarrow x = 40 \cr & {\text{Hence,}}\,{\text{length}} = x + 20 = 60m \cr} $$