CBSE Questions for Class 7 Maths Perimeter And Area Quiz 7 - MCQExams.com

Area of circle of radius $$  \dfrac {1}{2} = \pi { ( \dfrac {1}{2} )}^{ 2 } = \dfrac { \pi }{ 4 } $$

Given that at a rate of $$Rs.\ 12$$ per meter, the cost of fencing the circular field is $$Rs. 1320$$

Hence, circumference of the field $$ = \dfrac {1320}{12} = 110 \  m$$
We know that, circumference of a circle is $$2\pi r$$

$$\therefore \ 2\pi r=110$$

$$\Rightarrow 2 \times \dfrac {22}{7} \times r = 110  $$ 

$$ \Rightarrow r = \dfrac {35}{2} \  m $$

We also know that the area of a circle is $$\pi r^2$$

So, area of the field $$ = \dfrac {22}{7} \times \dfrac {35}{2}  \times \dfrac {35}{2} = \dfrac {1925}{2}  \ {m}^{2}  $$

Given that the cost of ploughing one $$m^2$$ area is $$Rs. \ 2$$.

Hence, cost of ploughing the field at $$ Rs. \ 2 $$ per $$ {m}^{2} = \dfrac {1925}{2} \times 2 = Rs. \ 1925 $$.

Hence, option B is correct.

Let the radius of the quadrant be $$r$$ 

Given, 
Perimeter of the square paper $$ = \dfrac {\pi r}{2} + 2r =25  cm  $$

$$\Rightarrow \dfrac {22}{7} \times \dfrac {r}{2} + 2r = 25 $$

$$ \Rightarrow \dfrac {22r + 28r}{14} = 25 $$

$$ \Rightarrow \dfrac {50r}{14} = 25 $$

$$ \Rightarrow r = 7  cm $$

Area of the square paper quadrant of radius $$r$$

$$ = \dfrac {\pi {r}^{2}}{4} \\ = \dfrac {\dfrac {22}{7} \times 7 \times 7}{4} = \dfrac {77}{2}  {cm}^{2}  $$

$$\implies 2 \times \frac {22}{7}

\times R = 506 \ \mathrm{m} $$

$$ R  = 80.5\ \mathrm{  m} $$

$$\implies 2 \times \frac {22}{7}

\times r = 440 \ \mathrm{m} $$

Let, the radius be$$ r cm$$

Circumference will be,

$$2πr=2\times 3.14r=6.28r$$

 According to question,  $$6.28r-1.r=37$$

$$\Rightarrow 5.28r=37$$

 $$\Rightarrow r=\dfrac{37}{5.28}$$

$$\Rightarrow r=$$approx. $$7 cm$$ 

Area$$=\pi {{r}^{2}}=\dfrac{22}{7} \times 7\times 7=154c{{m}^{2}}$$

Hence, this is the answer.