Surface Areas of Spheres | Exercise 13.4 | Q7

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

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

Exercise 13.1 | Surface Area of Cubes & Cuboids

1. Class 9 Exercise 13.1 Question 3 | Height of hall ▶
2. Class 9 Exercise 13.1 Question 6 | TSA of Herbarium ▶
3. Class 9 Exercise 13.1 Question 7 | Shanti Sweets ▶

Exercise 13.2 | Surface Area of Cylinders

1. Class 9 Exercise 13.2 Question 3 | CSA TSA of Metal Pipe ▶
2. Class 9 Exercise 13.2 Question 4 | Roller & Playground ▶
3. Class 9 Exercise 13.2 Question 9 | Cylindrical Petrol Tank ▶
4. Class 9 Exercise 13.2 Question 10 | CSA of Decorative Lampshade ▶

Exercise 13.3 | Surface Area of Cones

1. Class 9 Exercise 13.3 Question 5 | Length of Tarpaulin 3 m wide ▶
2. Class 9 Exercise 13.3 Question 8 | Cost of painting conical barricades ▶

Exercise 13.4 | Surface Area of Spheres & Hemispheres

1. Class 9 Exercise 13.4 Question 8 | Outer CSA of hemispherical bowl ▶
2. Class 9 Exercise 13.4 Question 9 | Surface area of sphere & Cylinder ▶