Question

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Answer 1

Answer :

Given:

Radius (R) of upper circular end of frustum part

=18/2 = 9cm

Radius (r) of lower circular end of frustum part=Radius of circular end of cylindrical part

= 8/2=4cm

Explanation :

height (h¹ ) of frustum part =22−10=12 cm

height (h² ) of cylindrical part =10 cm

Slant height (l) of frustum part

= √(R-r)²+h1

= √(9-4)²+12²

=√25+144

=√169

= 13 cm.

Area of tin sheet required = CSA of frustum part +CSA of cylindrical part

=π(R+r)l+2πrh2

=22/7×(9+4)×13+2×22/7×4×10

=22/7[169+80]

= 22×249/7

782•4/7 cm²

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Answer 2

Answer:

Upper Radius(R) of Frustum ABEF = 18/2 = 9cm

Lower Radius(r) of Frustum ABEF = 8/2 = 4cm

Height of Frustum(h_1)(h

1

) = 22 - 10 =12cm

Slant Height of Frustum=

$$\begin{lgathered}\tt \sqrt{(R-r)^2 + {h_1}^2} \\\\ \tt \Rightarrow \sqrt{(9-4)^2+12^2} \\\\ \tt \Rightarrow \sqrt{25+144} \\\\ \tt \Rightarrow \sqrt{169} \\\\ \tt = 13\end{lgathered}$$

Height of Cylinder $$(h_2)$$ = 10cm

Radius of Cylinder(r) = 4cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part =

$$\begin{lgathered}\tt \pi (R+r)l + 2\pi rh_2 \\\\ \tt = \dfrac{22}{7} \times (9+4) \times 13 + 2 \times \dfrac{22}{7} \times 4 \times 10 \\\\ \tt = \dfrac{22}{7}(13\times 13 + 2\times 4 \times 10) \\\\ \tt = \dfrac{22}{7}(169+80) \\\\ \tt = \dfrac{22}{7} \times 249 \\\\ \tt = 782.6cm^2\end{lgathered}$$

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