If radius of a circle is 3 cm, what is the area of the circle in sq. cm ?
(A) $$6\pi $$ sq. cm
(B) $$9\pi $$ sq. cm
(C) $$\frac{{3\pi }}{2}$$ sq. cm
(D) $$9{\pi ^2}$$ sq. cm
Solution:
Given: Radius of a circle = 3 cm Area of circle :$$\eqalign{ & = \pi {r^2} \cr & = \pi \times {3^2} \cr & = 9\pi {\text{ sq}}{\text{.}}\,{\text{cm}} \cr} $$
53.
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
Solution:
$$2\pi r = 100$$ So, length of the arc : $$\eqalign{ & = \frac{{2\pi r\theta }}{{360}} \cr & = \left( {\frac{{100 \times 20}}{{360}}} \right){\text{units}} \cr & = \left( {\frac{{50}}{9}} \right){\text{ units}} \cr & = 5.55{\text{ units}} \cr} $$
54.
Perimeter of a rectangular field is 160 metres and the difference between its two adjacent sides is 48 metres. The side of a square field, having the same area as that of the rectangle, is :
(A) 4 metres
(B) 8 metres
(C) 16 metres
(D) 32 metres
Solution:
Let the sides of the rectangle be x metres and (x + 48) metres Then, $$\eqalign{ & 2\left( {x + x + 48} \right) = 160 \cr & \Rightarrow 4x + 96 = 160 \cr & \Rightarrow 4x = 64 \cr & \Rightarrow x = 16 \cr} $$ So, sides of the rectangle are 16 metres and 64 metres Area of the rectangle : = (16 × 64) m2 = 1024 m2 Area of the square = 1024 m2 ∴ Side of the square : = $$\sqrt {1024} $$ m = 32 metres
55.
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
Solution:
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 :$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{105}}{{22}} \times \frac{{105}}{{22}}} \right)c{m^2} \cr & = \left( {\frac{{1575}}{{22}}} \right)c{m^2} \cr & = 71.6\,c{m^2} \approx 72\,c{m^2} \cr} $$
56.
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
Solution:
Area of the roads : = (80 × 10 + 60 × 10 - 10 × 10) m2 = 1300 m2 ∴ Cost of gravelling : = Rs. (1300 × 30) = Rs. 39000
57.
A carpenter is designing a table. The table will be in the form of a rectangle whose length is 4 feet more than its width. How long should the table be if the carpenter wants the area of the table to be 45 sq. ft ?
(A) 6 ft
(B) 9 ft
(C) 11 ft
(D) 13 ft
Solution:
Let the width of the table be x feet. Then, length of the table = (x + 4) ft $$\eqalign{ & \therefore x\left( {x + 4} \right) = 45 \cr & \Rightarrow {x^2} + 4x - 45 = 0 \cr & \Rightarrow {x^2} + 9x - 5x - 45 = 0 \cr & \Rightarrow x\left( {x + 9} \right) - 5\left( {x + 9} \right) = 0 \cr & \Rightarrow \left( {x + 9} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr} $$ Hence, length of the table = (5 + 4) = 9 feet
58.
If the radius of a circle is increased by 75%, then its circumference will increase by :
(A) 25%
(B) 50%
(C) 75%
(D) 100%
Solution:
Let original radius be R cm Then, original circumference = $$2\pi r$$ cm New radius : $$\eqalign{ & = \left( {175\% {\text{ of }}R} \right)cm \cr & = \left( {\frac{{175}}{{100}} \times R} \right)cm \cr & = \frac{{7R}}{4}\,cm \cr} $$ New circumference : $$\eqalign{ & = \left( {2\pi \times \frac{{7R}}{4}} \right)cm \cr & = \frac{{7\pi R}}{2}\,cm \cr} $$ Increase in circumference : $$\eqalign{ & = \left( {\frac{{7\pi R}}{2} - 2\pi R} \right)cm \cr & = \frac{{3\pi R}}{2}\,cm \cr} $$ Increase % : $$\eqalign{ & = \left( {\frac{{3\pi R}}{2} \times \frac{1}{{2\pi R}} \times 100} \right)\% \cr & = 75\% \cr} $$
59.
If the area of the trapezium whose parallel sides are 6 cm and 10 cm is 32 sq. cm, then the distance between the parallel sides is :
(A) 2 cm
(B) 4 cm
(C) 5 cm
(D) 8 cm
Solution:
Let the required distance be x cm Then, $$\eqalign{ & \Rightarrow \frac{1}{2} \times \left( {6 + 10} \right) \times x = 32 \cr & \Rightarrow x = 4\,cm \cr} $$
60.
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: