Question

if the length of radius of a sphere is increased by 50%, let us write how much percent will be increase d of its curved surface area​

Answer 1

Answer:

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

4×π×r

2

9×π×r

2

−4×π×r

2

×100

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

4×π×r

2

5×π×r

2

×100

= 5*25%

= 125%

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

Answer 2

Answer:

There will be 125 % increase in curved surface area of sphere.

Step-by-step-explanation:

Let the radius of the sphere be "r" units.

We have given that,

The radius is increased by 50 %.

New radius ( R ) = r + 50 % of r

⇒ New radius = r + 50 / 100 * r

⇒ New radius = r ( 1 + 50 / 100 )

⇒ New radius = r ( 1 + 1 / 2 )

⇒ New radius = r ( 1 + 0.5 )

⇒ New radius = r * 1.5

⇒ New radius = 1.5 r

Now, we know that,

Curved surface area of sphere = 4 π r²

Now,

New curved surface area of sphere = 4 π R²

⇒ New CSA = 4 π * ( 1.5 r )²

⇒ New CSA = 4 π * 2.25 r²

⇒ New CSA = 4 * 2.25 * π r²

⇒ New CSA = 9 π r²

Now,

Percentage increase in area = ( New area - Original area ) / Original area * 100

⇒ Percentage increase in area = ( 9 π r² - 4 π r² ) / 4 π r² * 100

⇒ Percentage increase in area = 5 π r² / 4 π r² * 100

⇒ Percentage increase in area = 5 / 4 * 100

⇒ Percentage increase in area = 5 * 25

⇒ Percentage increase in area = 125 %

∴ There will be 125 % increase in curved surface area of sphere.