Practice MCQ Questions and Answer on Area

141.

If the circumference of a circle is 100 units, then what will be the length of the arc described by an angle of 20 degrees ?

• (A) 5.55 units
• (B) 4.86 units
• (C) 5.85 units
• (D) None of these

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 $$2\pi r = 100$$ So, length of the arc : $$\eqalign{ & = \frac{{2\pi r\theta }}{{360}} \cr & = \left( {\frac{{100 \times 20}}{{360}}} \right){\text{units}} \cr & = \left( {\frac{{50}}{9}} \right){\text{ units}} \cr & = 5.55{\text{ units}} \cr} $$

142.

Of the two square fields, the area of one is 1 hectare while the other one is broader by 1%. The difference in their areas is :

• (A) 100 m2
• (B) 101 m2
• (C) 200 m2
• (D) 201 m2

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 Area = 1 hect. = 10000 sq. m Side = $$\sqrt {10000} $$   m = 100 m Side of the other square = 101 m Difference in their areas : = [(101)2 - (100)2] m2 = [(101 + 100) (101 - 100)] m2 = 201 m2

143.

If the area of a square increase by 69%, then the side of the square increases by :

• (A) 13%
• (B) 30%
• (C) 39%
• (D) 69%

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 Le original area = 100 cm2 Then, new area = 169 cm2 ⇒ Original side = 10 cm New side = 13 cm Increase on 10 cm = 3 cm Increase % : $$\eqalign{ & = \left( {\frac{3}{{10}} \times 100} \right)\% \cr & = 30\% \cr} $$

144.

A circular pond has area equal to 616 m². A circular stage is made at the centre of the pond whose radius is equal to half the radius of the pond. What is the area where water is present ?

• (A) 454 sq. m
• (B) 462 sq. m
• (C) 532 sq. m
• (D) 564 sq. m

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 Let the radius of the pond be R metres Then, $$\eqalign{ & \pi {R^2} = 616 \cr & \Rightarrow {R^2} = \frac{{616 \times 7}}{{22}} \cr & \Rightarrow {R^2} = 196 \cr & \Rightarrow R = 14\,m \cr} $$ Radius of the stage : $$\eqalign{ & = \left( {\frac{{14}}{2}} \right)m \cr & = 7\,m \cr} $$ ∴ Area where water is present : $$\eqalign{ & = \pi \left( {{{14}^2} - {7^2}} \right){m^2} \cr & = \left( {\frac{{22}}{7} \times 21 \times 7} \right){m^2} \cr & = 462\,{m^2} \cr} $$

145.

The area of a circle of radius 5 is numerically what percent of its circumference ?

• (A) 200%
• (B) 225%
• (C) 240%
• (D) 250%

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 $$\eqalign{ & {\text{Required % }} = \left[ {\frac{{\pi \times {{\left( 5 \right)}^2}}}{{2\pi \times 5}} \times 100} \right]\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 250\% \cr} $$

146.

The diagonal of a rectangle is $$\sqrt{41}$$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be :

• (A) 9 cm
• (B) 18 cm
• (C) 20 cm
• (D) 41 cm

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 $$\eqalign{ & \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr & \Rightarrow {l^2} + {b^2} = 41 \cr & Also,\,\,lb = 20 \cr & {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2lb \cr & {\left( {l + b} \right)^2} = 41 + 40 \cr & {\left( {l + b} \right)^2} = 81 \cr & \left( {l + b} \right) = 9 \cr & \therefore {\text{ Perimeter}} = 2\left( {l + b} \right) = 18\,cm \cr} $$

147.

The perimeter of a rectangle is 60 metres. If its length is twice its breadth, then its area is :

• (A) 160 m2
• (B) 180 m2
• (C) 200 m2
• (D) 220 m2

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 Let the breadth of the rectangle be x metres Then, length of the rectangle = 2x metres ⇒ 2(2x + x) = 60 ⇒ 6x = 60 ⇒ x = 10 So, length = 20 m, breadth = 10 m ∴ Area = (20 × 10) m2 = 200 m2

148.

The area of a right-angled triangle is 20 sq.cm and one of the sides containing the right angle is 4 cm. The altitude on the hypotenuse is :

• (A) $$\frac{41}{\sqrt{34}}\text{ cm}$$
• (B) $$\sqrt{\frac{41}{40}}\text{ cm}$$
• (C) $$\frac{29}{\sqrt{20}}\text{ cm}$$
• (D) $$\frac{20}{\sqrt{29}}\text{ cm}$$

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 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,$$\eqalign{ & \frac{1}{2} \times 2\sqrt {29} \times h = 20 \cr & \Rightarrow h = \frac{{20}}{{\sqrt {29} }}{\text{ cm}} \cr} $$

149.

A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top and bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing ?

• (A) 60%
• (B) 65%
• (C) 70%
• (D) 80%

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 Area of the sheet = (30 × 15) cm2 = 450 cm2 Area used for typing = [(30 - 5) × (15 - 2.5)] cm2 = 312.5 cm2 ∴ Required percentage : $$\eqalign{ & = \left( {\frac{{312.5}}{{450}} \times 100} \right)\% \cr & = 69.4\% \approx 70\% \cr} $$

150.

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is :

• (A) 10%
• (B) 10.08%
• (C) 20%
• (D) 28%

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 Let the original length = x and original breadth = y Decrease in area : $$\eqalign{ & = xy - \left( {\frac{{80}}{{100}}x \times \frac{{90}}{{100}}y} \right) \cr & = \left( {xy - \frac{{18}}{{25}}xy} \right) \cr & = \frac{7}{{25}}xy \cr} $$ ∴ Decrease% : $$\eqalign{ & = \left( {\frac{7}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 28\% \cr} $$