If the area of A circle is $154c{m^2}$ its perimeter is .$\left[ {use\;\pi = \dfrac{{22}}{7}} \right]$
A) 22 cm.
B) 44 cm
C) 50 cm
D) 56 cm
Answer
604.2k+ views
Hint:
By using formula for area of triangle $(\pi {r^2})$, find the value of radius and then using the radius you can easily find perimeter/circumference $(2\pi r)$ of the circle.
Complete step by step solution:
Let the radius of the circle be r .
Now, we know that area of circle $ = \pi {r^2}$
Given that, area of circle $ = 154c{m^2}$
So, $\pi {r^2} = 154$
$ \Rightarrow {r^2} = \dfrac{{154}}{\pi }$
$ \Rightarrow {r^2} = \dfrac{{154}}{{\dfrac{{22}}{7}}}$ $\left[ {Putting\;\pi = \dfrac{{22}}{7}} \right]$
$ \Rightarrow {r^2} = \dfrac{{154 \times 7}}{{22}}$
$ \Rightarrow {r^2} = 7 \times 7$
$ \Rightarrow {r^2} = {7^2}$
$\therefore r = 7cm$
Now, we know that radius of circle is 7 cm.
Perimeter of circle $ = 2\pi r$
$ = 2 \times \dfrac{{22}}{7} \times 7$
Therefore, the perimeter of the circle = 44 cm.
Note:
These types of questions are too common in geometry and mensuration. We’ll be having some one or two unknowns in the formula, they’ll give all the values except one and we need to find that one. It’s easy to solve an equation for an unknown.
By using formula for area of triangle $(\pi {r^2})$, find the value of radius and then using the radius you can easily find perimeter/circumference $(2\pi r)$ of the circle.
Complete step by step solution:
Let the radius of the circle be r .
Now, we know that area of circle $ = \pi {r^2}$
Given that, area of circle $ = 154c{m^2}$
So, $\pi {r^2} = 154$
$ \Rightarrow {r^2} = \dfrac{{154}}{\pi }$
$ \Rightarrow {r^2} = \dfrac{{154}}{{\dfrac{{22}}{7}}}$ $\left[ {Putting\;\pi = \dfrac{{22}}{7}} \right]$
$ \Rightarrow {r^2} = \dfrac{{154 \times 7}}{{22}}$
$ \Rightarrow {r^2} = 7 \times 7$
$ \Rightarrow {r^2} = {7^2}$
$\therefore r = 7cm$
Now, we know that radius of circle is 7 cm.
Perimeter of circle $ = 2\pi r$
$ = 2 \times \dfrac{{22}}{7} \times 7$
Therefore, the perimeter of the circle = 44 cm.
Note:
These types of questions are too common in geometry and mensuration. We’ll be having some one or two unknowns in the formula, they’ll give all the values except one and we need to find that one. It’s easy to solve an equation for an unknown.
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