# Surface Areas of Spheres | Exercise 13.4 | Q7

###### Class 9 Maths | NCERT Solution to Exercise Questions | Chapter 13

Question 7: The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

## NCERT Solution to Class 10 Maths

### Explanatory Answer | Exrecise 13.4 Question 7

Given Data: The diameter of the moon is approximately one fourth the diameter of the earth.
Because radius is half the diameter, the radius of the moon is also approximately one fourth the radius of the earth.
Let the radius of the earth be r1.
Let the radius of the moon be r2.
Radius of the moon, r2 = $$frac{1}{4}$ x r1 or r1 = 4r2 ……$1)

Surface area of moon = 4 × π × r22
Surface area of earth = 4 × π × r12
Ratio of their surface areas = $$frac{\text{Surface Area of Moon}}{\text{Surface Area of Earth}}$ = $\frac{4 × π × r_2^2}{4 × π × r_1^2}$ = $\frac{r_2^2}{ r_1^2}$ = $\frac{r_2^2}{$4 r_2$^2})
= $\frac{1}{16}$
Ratio of surface areas of the moon and the earth = 1 : 16

###### Free CBSE Online CoachingClass 9 Maths

Register in 2 easy steps and
Start learning in 5 minutes!