Question

the volume of two spheres are in the ratio 64 :27 find difference of their surface area if the sum of their radii is 7

Answer 1

Here is your solution

Ratio of volumes=64:27

Therefore ratio of radii=4:3

sum of radii =>7=4+3

Therefore  radii of 2 spheres = 4 and 3

difference in surface areas=?
$= 4\pi \: r1 {}^{2} - 4\pi \: r2 {}^{2} \\ = 4\pi(r1 {}^{2} -r2 {}^{2}) \\ = 4 \times \frac{22}{7} (4 {}^{2} - 3 {}^{2} ) \\ \\ = \frac{88}{7} (16 - 9) \\ \\ = \frac{88}{\cancel7} \times \cancel7 \\ \\ = 88unit$
Difference in surface areas is 88 unit

hope it helps you

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\therefore \:\:88\:cm^2$

$\bold{\underline {Given:}}$

Volume of two spheres in the ratio = 64: 27

The sum of their radii is = 7cm

$\bold{\underline {To\:find:}}$

The difference of their surface areas = ?

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

Let $r_1$ and $r_2$ be the radii of the two squares and $V_1$ and $V_2$ with their corresponding volumes.

$\because\:\:\:\:\:\:\: \dfrac{V_1}{V_2}=\dfrac{64}{27}\\ \\ \implies \dfrac{\dfrac{4}{3}\pi r_1^3}{\dfrac{4}{3}\pi r^3_2}=\dfrac{64}{27}\\ \\ \implies \left(\dfrac{r_1} {r_2}\right)^3=\left(\dfrac{4}{3}\right)^3\\ \\ \implies \dfrac{r_1}{r_2}=\dfrac{4}{3}$

$\therefore \:\:\:\:\:\:\:r_1=\dfrac{4}{3}×r^2\:\:\:\:\:\:\:\:\:\:$ .......(1)

But, $r_1+r_2=7 \implies \dfrac{4}{3}×r^2+r^2=7$[Using(1)]

$\implies \dfrac{7}{3}×r_2=7\\ \\ \implies r_2=3cm\\ \\ And, r_1=\dfrac{4}{3}×r_2=\dfrac{4}{3}×3=4cm$

Now, surface area of the first sphere $(S_1)=4\pi r_1^2=4×\pi×4×4=64\pi cm^2$

and surface area of the second sphere $(S_2)=4\pi r_2^2=4×\pi×3×3=36\pi cm^2$

•°•$S_1-S_2=(64-36)\pi=28\pi$

$=28× \dfrac{22}{7}=4×22=88cm^2$

Hence, the difference of their surface areas = 88cm²