NCERT Class X Chapter 12: Surface Areas And Volumes Example 1

NCERT Class X Chapter 12: Surface Areas And Volumes Example 1

Question:

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Use \( \pi = \frac{22}{7} \))

Given:

  • Total height of top, \( H = 5 \) cm
  • Diameter of top \( = 3.5 \) cm
  • Radius of top, \( r = \frac{3.5}{2} = 1.75 \) cm
  • \( \pi = \frac{22}{7} \)

To Find:

The total surface area of the top to be coloured.

Formula:

  • Curved surface area of a cone: \( \pi r l \)
  • Curved surface area of a hemisphere: \( 2 \pi r^2 \)
  • Slant height of cone: \( l = \sqrt{r^2 + h^2} \)

Solution:

Step 1: Find the height of the cone part.

The hemisphere sits on top of the cone, so the height of the cone is:

$$ h = H - r = 5 - 1.75 = 3.25 \text{ cm} $$

Step 2: Calculate the slant height (\( l \)) of the cone.

$$ l = \sqrt{r^2 + h^2} = \sqrt{(1.75)^2 + (3.25)^2} = \sqrt{3.0625 + 10.5625} = \sqrt{13.625} \approx 3.69 \text{ cm} $$

Step 3: Find the curved surface area of the cone.

$$ \text{CSA of cone} = \pi r l = \frac{22}{7} \times 1.75 \times 3.69 \approx 17.996 \text{ cm}^2 $$

Step 4: Find the curved surface area of the hemisphere.

$$ \text{CSA of hemisphere} = 2\pi r^2 = 2 \times \frac{22}{7} \times (1.75)^2 = 2 \times \frac{22}{7} \times 3.0625 \approx 19.25 \text{ cm}^2 $$

Step 5: Calculate the total area to be coloured.

$$ \text{Total area} = 17.996 + 19.25 = 37.246 \text{ cm}^2 \approx 37.25 \text{ cm}^2 $$

Result:

The area Rasheed has to colour is approximately \( 37.25 \) cm\(^2\).

Next question solution:

Example 2
© Kaliyuga Ekalavya. All rights reserved.

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