Question

if the length of radius of a sphere is increased by 50%, let us write how much percent will be increased of it's curved surface area.​

Answer 1

Step-by-step explanation:

Let the initial radius be r.

After 50% increase, new radius r'= r + 50/100*r

= r + 0.5r= 1.5r.

Initial surface area = 4*π*r^2

Final surface area = 4*π*(1.5r)^2

Percent increases=

$\frac{9 \times \pi \times {r}^{2} - 4 \times \pi \times {r}^{2} }{4 \times \pi \times {r}^{2} } \times 100$

$= \frac{5 \times \pi \times {r}^{2} }{4 \times \pi \times {r}^{2} } \times 100$

= 5*25%

= 125%

Answer :- 125% will be increased of its curved surface area.

Answer 2

Answer:

5×25=125

125 is the answer hope it helped