Q7. A horse is tied to a pole with 28 m long string. Find the area where the horse can graze.
$$ \text {Take } \pi= \frac {22}{7} $$Sol.
Given length of the string =28 m
The area over which the horse can graze = Area of a circle of radius 28 m
$$ =\pi r^{2} =\frac{22}{7} \times 28 \times 28 \text { m}^{2} $$ $$ =2464 \text { m}^{2} $$Q8. A steel wire when bent in the form of a square encloses an area of } 121 sq cm. If the same wire is bent in the form of a circle, find the area of the circle.\begin{array}{l} \text {Sol. Area of the square } =121 \text { cm}^{2} \\ \Rightarrow (\text {Side})^{2}=(11)^{2} \text { cm}^{2} \\ \Rightarrow \text {Side} =11 \text { cm} \\ \therefore \text {The perimeter of the square }=4 \times \text {Side} =(4 \times 11) \text { cm} =44 \text { cm} \\\text {Lets assume that r cm be the radius of the circle created with the steel wire.} \\\Rightarrow \text {Circumference of the circle having radius r cm } = \text {Perimeter of the square} \\ \text { We know that the circumference (C) of a circle having radius r } =2 \pi r \\\Rightarrow 2 \pi r=44 \Rightarrow 2 \times \frac {22}{7} \times r = 44 \text { cm} \\ \Rightarrow r = \frac {44 \times 7}{22 \times 2} \text { cm} \\ \Rightarrow r = 7 \text { cm} \\ \therefore \text {Area of the circle} =\pi r^{2} \\ =\frac{22}{7} \times 7 \times 7 \text { cm}^{2} \\ =154 \text { cm}^{2} \\ \end{array}Q9. A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.
Sol.
We know that the circumference (C) of a circle having radius r =2 π r
Circumference of the circular park =352 m
\begin{array}{l} \Rightarrow 2 \pi r=352 \text { m} \\ \Rightarrow r= \frac {352 \times 7}{22 \times 2} \text { m} \\ \Rightarrow r= 56 \text { m} \\\text {Radius of the path including the 7 m wide road } =56+7=63 \text { m} \\ \therefore \text {Area of the road}=\text {Area of park plus circular road } – \text {Area of circular park} \\ =[\pi(63)^{2}-\pi(56)^{2}] \text { m}^{2} \\ =\pi [(63)^{2}-(56)^{2}] \text { m}^{2} \\ =\pi[(63-56)(63+56)] \text { m}^{2} \quad \quad [\text {Using : } a^{2} – b^{2} = (a-b)(a+b)]\\ =\frac{22}{7} \times 7 \times 119 \text { m}^{2} \\ =2618 \text { m}^{2}\therefore \text {Reqired area of the road is } 2618 \text { m}^{2} \end{array}Q10. Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h(2r + h)
Sol.
Radius of the inner circular region = r
Radius of the circular path of uniform width h surrounding the circular region of radius r = (r + h)
Therefore, Area of the path = (Area of inner circular region plus circular path) – (Area of inner circular region)
$$ =[\pi(r+h)^{2}-\pi r^{2}] $$ $$ =\pi[(r+h)^{2}-r^{2}] $$ $$ =\pi[(r+h-r)(r+h+r)] \quad \quad [\text {Using : } a^{2} – b^{2} = (a-b)(a+b)] $$ $$ =\pi[h(2r+h)] $$ $$ =\pi h(2r+h) $$ Hence proved.
Q11. The perimeter of a circle is 4 π r cm. What is the area of the circle?
\begin{array}{l} \text {Sol. Given that the perimeter of the circle } =4 \pi r \text {cm} \\ \text {We know that the perimeter of a circle having radius } r_{1} \text { cm } =2 \pi r_{1} \\ \Rightarrow 2 \pi r_{1} = 4 \pi r \text { cm} \\ \Rightarrow r_{1} = 2r \text { cm} \\ \therefore \text {Radius of circle is 2r cm.} \\ \text {Hence, area of the circle having radius 2r cm } =\pi(2 r)^{2}=4 \pi r^{2} \text { cm}^{2}\end{array}Q12. A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.
\begin{array}{l} \text {Sol. Area of the square } =121 \text { cm}^{2} \\ \Rightarrow (\text {Side})^{2}=(11)^{2} \text { cm}^{2} \\ \Rightarrow \text {Side} =11 \text { cm} \\ \text {Given that the perimeter of the square }=5024 \text { m} \\ \Rightarrow 4 \times (\text {Side})=5024 \text { m} \\ \Rightarrow \text {Side}=\frac {5024}{4} \text { m} \\ \Rightarrow \text {Side}=1256 \text { m} \\\text {Lets assume that r m be the radius of the circle created with the wire.} \\\Rightarrow \text {Circumference of the circle having radius r m } = \text {Perimeter of the square} \\ \text { We know that the circumference (C) of a circle having radius r } =2 \pi r \\\Rightarrow 2 \pi r=5024 \text { m} \\ \Rightarrow r = \frac {5024}{2 \pi} \text { m} \\ \Rightarrow r = \frac {2512}{\pi} \text { m} \\\text {Area of square} : \text {Area of circle} = \frac {\text {Area of square} }{\text {Area of circle} } \\ = \frac {1256 \times 1256}{\pi \times \frac {2512}{\pi} \times \frac {2512}{\pi}} \\ = \frac {1256 \times 1256}{2512 \times \frac {2512}{\pi}} \\ = \frac {1256 \times 1256 \times \pi}{2512 \times 2512} \\ = \frac {1256 \times 1256 \times 22}{2512 \times 2512 \times 7} \\ = \frac {1256 \times 1256 \times 22}{2 \times 1256 \times 2 \times 1256 \times 7} \\ = \frac {11}{14} \\\therefore \text {Area of square} : \text {Area of circle} = 11 :14 \end{array}
$$ \text {Take } \pi= \frac {22}{7} $$Sol.
Given length of the string =28 m
The area over which the horse can graze = Area of a circle of radius 28 m
$$ =\pi r^{2} =\frac{22}{7} \times 28 \times 28 \text { m}^{2} $$ $$ =2464 \text { m}^{2} $$Q8. A steel wire when bent in the form of a square encloses an area of } 121 sq cm. If the same wire is bent in the form of a circle, find the area of the circle.\begin{array}{l} \text {Sol. Area of the square } =121 \text { cm}^{2} \\ \Rightarrow (\text {Side})^{2}=(11)^{2} \text { cm}^{2} \\ \Rightarrow \text {Side} =11 \text { cm} \\ \therefore \text {The perimeter of the square }=4 \times \text {Side} =(4 \times 11) \text { cm} =44 \text { cm} \\\text {Lets assume that r cm be the radius of the circle created with the steel wire.} \\\Rightarrow \text {Circumference of the circle having radius r cm } = \text {Perimeter of the square} \\ \text { We know that the circumference (C) of a circle having radius r } =2 \pi r \\\Rightarrow 2 \pi r=44 \Rightarrow 2 \times \frac {22}{7} \times r = 44 \text { cm} \\ \Rightarrow r = \frac {44 \times 7}{22 \times 2} \text { cm} \\ \Rightarrow r = 7 \text { cm} \\ \therefore \text {Area of the circle} =\pi r^{2} \\ =\frac{22}{7} \times 7 \times 7 \text { cm}^{2} \\ =154 \text { cm}^{2} \\ \end{array}Q9. A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.
Sol.
We know that the circumference (C) of a circle having radius r =2 π r
Circumference of the circular park =352 m
\begin{array}{l} \Rightarrow 2 \pi r=352 \text { m} \\ \Rightarrow r= \frac {352 \times 7}{22 \times 2} \text { m} \\ \Rightarrow r= 56 \text { m} \\\text {Radius of the path including the 7 m wide road } =56+7=63 \text { m} \\ \therefore \text {Area of the road}=\text {Area of park plus circular road } – \text {Area of circular park} \\ =[\pi(63)^{2}-\pi(56)^{2}] \text { m}^{2} \\ =\pi [(63)^{2}-(56)^{2}] \text { m}^{2} \\ =\pi[(63-56)(63+56)] \text { m}^{2} \quad \quad [\text {Using : } a^{2} – b^{2} = (a-b)(a+b)]\\ =\frac{22}{7} \times 7 \times 119 \text { m}^{2} \\ =2618 \text { m}^{2}\therefore \text {Reqired area of the road is } 2618 \text { m}^{2} \end{array}Q10. Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is π h(2r + h)
Sol.
Radius of the inner circular region = r
Radius of the circular path of uniform width h surrounding the circular region of radius r = (r + h)
Therefore, Area of the path = (Area of inner circular region plus circular path) – (Area of inner circular region)
$$ =[\pi(r+h)^{2}-\pi r^{2}] $$ $$ =\pi[(r+h)^{2}-r^{2}] $$ $$ =\pi[(r+h-r)(r+h+r)] \quad \quad [\text {Using : } a^{2} – b^{2} = (a-b)(a+b)] $$ $$ =\pi[h(2r+h)] $$ $$ =\pi h(2r+h) $$ Hence proved.
Q11. The perimeter of a circle is 4 π r cm. What is the area of the circle?
\begin{array}{l} \text {Sol. Given that the perimeter of the circle } =4 \pi r \text {cm} \\ \text {We know that the perimeter of a circle having radius } r_{1} \text { cm } =2 \pi r_{1} \\ \Rightarrow 2 \pi r_{1} = 4 \pi r \text { cm} \\ \Rightarrow r_{1} = 2r \text { cm} \\ \therefore \text {Radius of circle is 2r cm.} \\ \text {Hence, area of the circle having radius 2r cm } =\pi(2 r)^{2}=4 \pi r^{2} \text { cm}^{2}\end{array}Q12. A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.
\begin{array}{l} \text {Sol. Area of the square } =121 \text { cm}^{2} \\ \Rightarrow (\text {Side})^{2}=(11)^{2} \text { cm}^{2} \\ \Rightarrow \text {Side} =11 \text { cm} \\ \text {Given that the perimeter of the square }=5024 \text { m} \\ \Rightarrow 4 \times (\text {Side})=5024 \text { m} \\ \Rightarrow \text {Side}=\frac {5024}{4} \text { m} \\ \Rightarrow \text {Side}=1256 \text { m} \\\text {Lets assume that r m be the radius of the circle created with the wire.} \\\Rightarrow \text {Circumference of the circle having radius r m } = \text {Perimeter of the square} \\ \text { We know that the circumference (C) of a circle having radius r } =2 \pi r \\\Rightarrow 2 \pi r=5024 \text { m} \\ \Rightarrow r = \frac {5024}{2 \pi} \text { m} \\ \Rightarrow r = \frac {2512}{\pi} \text { m} \\\text {Area of square} : \text {Area of circle} = \frac {\text {Area of square} }{\text {Area of circle} } \\ = \frac {1256 \times 1256}{\pi \times \frac {2512}{\pi} \times \frac {2512}{\pi}} \\ = \frac {1256 \times 1256}{2512 \times \frac {2512}{\pi}} \\ = \frac {1256 \times 1256 \times \pi}{2512 \times 2512} \\ = \frac {1256 \times 1256 \times 22}{2512 \times 2512 \times 7} \\ = \frac {1256 \times 1256 \times 22}{2 \times 1256 \times 2 \times 1256 \times 7} \\ = \frac {11}{14} \\\therefore \text {Area of square} : \text {Area of circle} = 11 :14 \end{array}