Question

A circular swimming pool is surrounded by a circular path which is 4 m wide. If the area of the path is 11/25th part of the area of the swimming pool, then find the radius of the swimming pool in metres​

Answer 1

Answer:

R-20ft

Step-by-step explanation:

radius of pool=r

radius including wall=r+4

π(r+4)2-π r2=11\25π r2

(r+4+r) (r+4-r)=11/25 r2

2(r+2) x 4 =11/25 r2

200(r+2)=11 r2

11 r2-200 r-400 =0

11 r2-220r+ 20 r -400=0

11(r-20)+20(r-20)=0

(r-20) (11r +20) =r=20

r-20 ft ans.

Answer 2

Answer:

The radius of the swimming pool is found to be as 20 m long.

Step-by-step explanation:

Let the radius of the circular swimming pool be x,

The area of the swimming pool will be: $\pi\times x^2$

Now, we are given that the circular path is 4m wide, so the radius of the swimming pool with the circular path will be: $x+4$

Now, the area of the circular path: $\pi\times (x+4)^2-\pi\times x^2$

Now, we are given that the area of the circular path is 11/25th part of the swimming pool, so representing this mathematically, we get:

$\pi\times (x+4)^2-\pi\times x^2=\frac{11}{25}\times \pi x^2$

Taking π common out, we get:

$\pi( (x+4)^2- x^2)=\pi(\frac{11}{25}\times x^2)$

or we can say:

$(x+4)^2- x^2=\frac{11}{25}\times x^2$

Simplifying it, we get:

$x^2+16+8x- x^2=\frac{11}{25}\times x^2$

or we can say:

$16+8x=\frac{11}{25}\times x^2$

Using cross multiplication, we get:

$25(16+8x)=11\times x^2$

$400+200x=11 x^2$

Bringing everything to one side of the sign of equality, we get:

$11 x^2-200x-400=0$

Factorising the equation, we get:

$11 x^2-220x+20x-400=0$

Taking factors common, we get:

$11x(x-20)+20(x-20)=0$

or we can say:

$(11x+20)(x-20)=0$

Finding the two solutions, we get:

$x=-\frac{20}{11}$  and $x=20$

But as x is the radius of the pool and radius cannot be negative, so the negative solution is eliminated.
Thus, the radius of the swimming pool is found to be 20m.