Question

if the radius is increased by x % then its area is increased by​

Answer 1

Let radius be r and new radius be r'.

1st case,

Area of a circle = πr² - - 1)

Area increased by x%,

=) r' = (r + x/100 * r)

= (r + xr/100)

= (100r + xr) /100

Then new area A' = πr'²

= π{(100r + xr)/100}²

= π{100r/100 + xr/100}²

= π{r + xr/100}²

= π(r² + x²r² + 2xr²)/10000 - - 2)

Increase in Area = (A' - A)/A × 100 %

= {π(r² + x²r² + 2xr²)/10000 - πr²} × 100% / πr²

= πr²{1 + x² + 2x / 10000 - 1) × 100% / πr²

= (1 + x² + 2x - 10000) × 100% / 10000

= [{(1+x)² - 100²} / 100] %

Answer 2

Let the old radius be r.

So, old area = πr^2 .

New radius = old radius + x% of old radius

$r + \frac{ x}{100} \times r$

$\implies \: \frac{100r + xr}{100}$

New area = π × new radius^2

=π (100r+xr/100)^2

= π (100 r/100 + xr/100)^2

=π (r +xr/100)^2

= π(r^2 + x^2r^2 +2xr^2 ) /10000

Increase in area = New area - old area

= π (r^2 +x^2r^2+2xr^2)/10000 - πr^2

= πr^2 + πx^2r^2 + π2xr^2 /10000 - πr^2

= πr^2 {(1+ x^2 +2x/10000)-1}

percentage Increase in area = Increase in area /old area ×100

=πr^2 {(1+x^2 +2x/10000)-1} / πr^2 ×100

Now, πr^2 will be cancelled by each other.

={(1+x^2 +2x -10000)100} /10000

= (1+x)^2 -(100)^2 /100 %