Question

The area of the circle is given by the expression πx^2+6πx+9π. Find the radius of the circle. Also find perimeter of a circle.​

Answer 1

The area of the circle is given by the expression πx^2+6πx+9π. Find the radius of the circle. Also find perimeter of a circle.

Answer 2

Given :-

• Area of the circle = $\pi x^{2} + 6 \pi x + 9\pi$

Aim :-

• To find the radius of the circle
• To find the perimeter of the circle

Formula :-

• Area of the circle = $\pi \sf{r}^{2}$
• Perimeter of the circle = $2\pi \sf{r}$

Area :-

$\implies \pi x^{2} + 6\pi x + 9\pi$

Taking $\pi$ as the common factor,

$\implies \pi (x^{2} + 6 x + 9)$

By factorizing,

$\implies \pi (x^{2} + 3x + 3x + 9)$

$\implies \pi [x(x + 3) + 3(x+3)]$

$\implies \pi [(x+3)(x+3)]$

$\implies \pi (x+3)^{2}$

We know that, the area of a circle is $\pi \sf{r}^{2}$

Hence,

$\pi \sf{r}^{2} = \pi (x+3)^{2}$

By bring the $\pi$ to the other side,

$\sf{r}^{2} = \dfrac{\pi (x+3)^{2}}{\pi }$

By cancelling $\pi$,

$\sf{r}^{2} = (x+3)^{2}$

Transposing the power,

$\sf{r} = \sqrt{(x+3)^{2}}$

$\sf{r} = (x+3)$

Now that we know the radius, the perimeter of the circle will be :-

Perimeter :-

$\implies 2\pi \sf{r}$

$\implies 2 \times \pi \times (x+3)$

$\implies \pi \times 2(x+3)$

$\implies \pi \times (2x + 6)$

$\implies 2\pi x + 6\pi$

• r represents radius