Practice MCQ Questions and Answer on Area

1.

The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per square metre.

• (A) Rs. 15000
• (B) Rs. 15500
• (C) Rs. 15600
• (D) Rs. 16500

Show Answer

 Area of the floor = (5.5 × 3.75)m2 = 20.625m2 ∴ Cost of paying = Rs. (800 × 20.625) = Rs. 16500

2.

The length of a rectangle is three times of its width. If the length of the diagonal is $$8\sqrt {10} $$ cm, then the perimeter of the rectangle is :

• (A) $$15\sqrt {10} $$  cm
• (B) $$16\sqrt {10} $$  cm
• (C) $$24\sqrt {10} $$  cm
• (D) 64 cm

Show Answer

 Let breadth = x cm Then, length = 3x cm $$\eqalign{ & \Rightarrow {x^2} + {\left( {3x} \right)^2} = {\left( {8\sqrt {10} } \right)^2} \cr & \Rightarrow 10{x^2} = 640 \cr & \Rightarrow {x^2} = 64 \cr & \Rightarrow x = 8 \cr} $$ So, length = 24 cm and breadth = 8 cm ∴ Perimeter = [2(24 + 8)] cm = 64 cm

3.

Two sides of a rectangle were measured. One of the sides (length) was measured 10% more than its actual length and the other side (width) was measured 5% less than its actual length. The percentage error in measure obtained for the area of the rectangle is :

• (A) 4.5%
• (B) 5%
• (C) 7.56%
• (D) 15%

Show Answer

 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 % :$$\eqalign{ & = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 4.5\% \cr} $$

4.

A room 5m × 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is Rs. 18 per sq. meter, the cost of carpeting the room will be :

• (A) Rs. 673.92
• (B) Rs. 682.46
• (C) Rs. 691.80
• (D) Rs. 702.60

Show Answer

 Area of the carpet : = [(5 - 0.20) × (8 - 0.20)] m2 = (4.8 × 7.8) m2 = 37.44 m2 ∴ Cost of carpeting : = Rs. (37.44 × 18) = Rs. 673.92

5.

A hall 50 m long and 45 m broad is to be paved with square tiles. Find the largest tile as well as its number in the given options so that the tiles exactly fit in the hall :

• (A) 36 sq. m and 80 tiles
• (B) 16 sq. m and 80 tiles
• (C) 25 sq. m and 90 tiles
• (D) 36 sq. m and 90 tiles

Show Answer

 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 : $$\eqalign{ & = \frac{{50 \times 45}}{{25}} \cr & = 90{\text{ tiles}} \cr} $$

6.

The sides of a rectangle are 8 cm and 6 cm. The corners of the rectangle lie on a circle. Find the are a of the circle without the rectangle :

• (A) 30.6 cm2
• (B) 39 cm2
• (C) 42.4 cm2
• (D) 65.3 cm2

Show Answer

 Diameter of the circle : $$\eqalign{ & = AC = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BC} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{8^2} + {6^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {100} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10\,cm \cr} $$ Radius = 5 cm Required area = (Area of the circle) - (Area of the rectangle) $$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times 5 \times 5} \right) - \left( {8 \times 6} \right)} \right]c{m^2} \cr & = \left( {\frac{{550}}{7} - 48} \right)c{m^2} \cr & = \left( {\frac{{214}}{7}} \right)c{m^2} \cr & = 30.57\,c{m^2} \approx 30.6\,c{m^2} \cr} $$

7.

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is :

• (A) 1520 m2
• (B) 2420 m2
• (C) 2480 m2
• (D) 2520 m2

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 We have : (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103 Solving the two equations, we get : l = 63 and b = 40 ∴ Area = (l × b) = (63 × 40)m2 = 2520 m2

8.

The ratio of the areas of a square of side 6 cm and an equilateral triangle of side 6 cm is :

• (A) 3 : $$\sqrt 3 $$
• (B) 8 : $$\sqrt 3 $$
• (C) 6 : $$\sqrt 3 $$
• (D) 4 : $$\sqrt 3 $$

Show Answer

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

9.

The area of a rectangular field is 2100 sq. meters. If the field is 60 metres long, what is its perimeter ?

• (A) 180 m
• (B) 210 m
• (C) 240 m
• (D) Cannot be determined

Show Answer

Answer & Solution Answer: Option E Solution: $$\eqalign{ & {\text{Breadth}} = \frac{{{\text{Area}}}}{{{\text{Length}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{2100}}{{60}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 35m \cr & \therefore {\text{Perimeter}} = 2\left( {60 + 35} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 190\,m \cr} $$

10.

Three boys are standing on a circular boundary of a fountain. They are at equal distance from each other. If the radius of the boundary is 5 m, the shortest distance between any two boys is :

• (A) $$\frac{{5\sqrt 3 }}{2}m$$
• (B) $$5\sqrt 3 \,m$$
• (C) $$\frac{{15\sqrt 3 }}{2}m$$
• (D) $$\frac{{10\pi }}{3}m$$

Show Answer

 Let A, B and C denote the portions of the three boys Then, AB = BC = AC So, ΔABC is equilateral Let the side of ΔABC be a Then, $$\eqalign{ & \frac{a}{{\sqrt 3 }} = 5 \cr & \Rightarrow a = 5\sqrt 3 \cr} $$ ∴ Required shortest distance = $$5\sqrt 3 \,m$$