Question

114.If the radius of a sphere is incresed 3 times,
its surface area increased by ​

Answer 1

Answer:

let the redius be r

surface area = 4×22/7 × r × r

new radius = 3r

new surface area = 4×22/7× 3r × 3r

its surface area increased = (4×22/7× 3r × 3r)/(4×22/7 × r × r)

= 9

ANS:

Its surface area increased by 9 times

plz mark BRAINLIEST answer.

Answer 2

Answer:

If the radius of a sphere is increased 3 times,

its surface area is increased by 800% or it becomes 9-fold

Step-by-step explanation:

Let the initial radius be r

and the new radius be 3r

Surface area of original sphere,

$A =4\pi {r}^{2}$

Surface area of new sphere

$A_n = 4\pi {(3r)}^{2} = 9(4\pi {r}^{2} )$

In this case, the radius is tripled the surface becomes 9-fold (9-times).

Percentage

$\frac{A_n - A}{A} \times 100 = \frac{9(4\pi {r})^{2} - 4\pi {r}^{2} }{4\pi {r}^{2} } \times 100 \\ = \frac{8(4\pi {r})^{2} }{4\pi {r}^{2} } \times 100 \\ = 8 \times 100 = 800\%$

In other words the surface area will increase by 800%.