A metallic bucket open at the top of height 24 cm is in the form of the frustum of a cone the radii of whose lower and upper circular ends are 7cm and 14cm respectively. Find the area of the metal sheet used to make the bucket.
Answers
Given : Height Of the Bucket = 24cm
Radius of the upper circular end = R1= 14cm.
Radius of the lower circular end = R2 = 7cm.
To Find: Total Surface Area Of the Bucket=?
Formula: TSA of a Frustum = πl(R1 + R2) + πR1(R1) + πR2(R2).
Answer:
3454 cm²
Step-by-step explanation:
We need to get the height of the larger cone and smaller cone.
The linear scale factor = 14/7 = 2
Let the height of the smaller cone be x.
The height of the larger cone will be = x + 24
(x + 24) / x = 2
x + 24 = 2x
x = 24
The height of the larger cone = 24 + 24 = 48 cm
Height of smaller cone = 24 cm
We need to get the slant heights of the two cones.
We will use Pythagoras theorem.
For the smaller cone :
24² + 7² = 625
l = 25 cm
For the larger cone :
48² + 14² = 2500
l = 50
Curved surface area of a cone = 2 × pie × r × l
The smaller cone's curved surface area :
= 22/7 × 2 × 7 × 25 = 1100 cm²
The larger cone's curved surface area :
= 22/7 × 2 × 14 × 50 = 4400 cm²
The curved surface area of the frustum = curved surface of larger cone - curved surface of smaller cone
= 4400 - 1100 = 3300 cm²
The frustum is closed at the bottom so we need the circular area of the bottom.
Area of bottom = 22/7 × 7² = 154 cm²
Total area of the frustum = 154 + 3300 = 3454 cm²