NCERT Class X Chapter 12: Surface Areas And Volumes Example 3
NCERT Class X Chapter 12: Surface Areas And Volumes Example 3
Question:
A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Use \( \pi = 3.14 \))
Given:
Height of rocket, \( H = 26 \) cm
Height of conical part, \( h_c = 6 \) cm
Diameter of conical base, \( d_c = 5 \) cm \( \Rightarrow r_c = 2.5 \) cm
Diameter of cylindrical base, \( d_{cy} = 3 \) cm \( \Rightarrow r_{cy} = 1.5 \) cm
\( \pi = 3.14 \)
To Find:
Area to be painted orange (curved surface area of conical part)
Area to be painted yellow (curved surface area of cylindrical part)
Formula:
Curved surface area of cone: \( \pi r l \)
Curved surface area of cylinder: \( 2\pi r h \)
Slant height of cone: \( l = \sqrt{r^2 + h^2} \)
Solution:
Step 1: Find the height of the cylindrical part.
$$ h_{cy} = H - h_c = 26\,\text{cm} - 6\,\text{cm} = 20\,\text{cm} $$Step 2: Calculate the slant height of the cone.
$$ l_c = \sqrt{r_c^2 + h_c^2} = \sqrt{(2.5)^2 + (6)^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5\,\text{cm} $$Step 3: Find the curved surface area of the cone (to be painted orange).
$$ \text{CSA}_{\text{cone}} = \pi r_c l_c = 3.14 \times 2.5 \times 6.5 = 51.025\,\text{cm}^2 $$Step 4: Find the curved surface area of the cylinder (to be painted yellow).
$$ \text{CSA}_{\text{cylinder}} = 2\pi r_{cy} h_{cy} = 2 \times 3.14 \times 1.5 \times 20 = 188.4\,\text{cm}^2 $$Result:
Area to be painted orange (conical part): 51.025 cm2
Area to be painted yellow (cylindrical part): 188.4 cm2
Comments
Post a Comment