Question

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.​

Answer 1

Given:

• radius of upper circular end of frustum part= 18/2=9cm
• radius of lower circular end of frustum part=radius of circular end of cylindrical end of cylindrical part=8/2=4cm

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

height(h2) of cylindrical part=10cm

Slant height(1) of frustum part= $\sqrt \s (R-r)^{2} +h1^{2}$

                                                =$\sqrt (9-4)^{2} +12^{2}$

                                                =$\sqrt25+144$

                                                =$\sqrt169$

                                                =13cm

Area of tin sheet required= CSA of frustum part+ CSA of cylindrical part= π(R+r)l + 2πrh(2)

=22/7×(9+4)×13+2×22/7×4×10

=22/7[169+80]

=22×249/7

=782 ×4/7 cm²

Answer 2

Given :-

• An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone.
• The total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm.

To Find :-

• What is the area of the tin sheet required to make the funnel.

Solution :-

First, we have to find the slant height of frustum of cone :

Given :

➳ Height of frustum of cone (h₁) = (22 - 10) cm = 12 cm

➳ Radius of top part of frustum of cone (r₁) = 18/2 cm = 9 cm

➳ Radius of lower part of frustum of cone (r₂) = 8/2 cm = 4 cm

As we know that :

$\clubsuit$ Slant Height of Frustum Of Cone Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{Slant\: Height\: (l) =\: \sqrt{h_1^2 + (r_1 - r_2)^2}}}}\: \: \: \bigstar$

By putting those values we get,

$\implies \sf Slant\: Height =\: \sqrt{(12)^2 + (9 - 4)^2}$

$\implies \sf Slant\: Height =\: \sqrt{144 + (5)^2}$

$\implies \sf Slant\: Height =\: \sqrt{144 + 25}$

$\implies \sf Slant\: Height =\: \sqrt{169}$

$\implies \sf\bold{\blue{Slant\: Height =\: 13\: cm}}$

Now, we have to find the CSA of frustum of cone :

As we know that :

$\clubsuit$ CSA of frustum of cone formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{C.S.A_{(Frustum\: Of\: Cone)} =\: {\pi}l(r_1 + r_2)}}}\: \: \: \bigstar$

By putting those above values we get,

$\implies \sf C.S.A_{(Frustum\: Of\: Cone)} =\: \dfrac{22}{7} \times 13(9 + 4)$

$\implies \sf C.S.A_{(Frustum\: Of\: Cone)} =\: \dfrac{286}{7} \times 13$

$\implies \sf\bold{\purple{C.S.A_{(Frustum\: Of\: Cone)} =\: \dfrac{3718}{7}cm^2}}$

Now, we have to find the CSA of cylinder :

Given :

➳ Height of cylindrical part (h) = 10 cm

➳ Radius of cylindrical part (r₂) = 8/2 cm = 4 cm

As we know that :

$\clubsuit$ CSA of Cylinder Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{C.S.A_{(Cylinder)} =\: 2{\pi}rh}}}\: \: \bigstar$

By putting those values we get,

$\implies \sf C.S.A_{(Cylinder)} =\: 2 \times \dfrac{22}{7} \times 4 \times 10$

$\implies \sf C.S.A_{(Cylinder)} =\: \dfrac{44}{7} \times 40$

$\implies \sf\bold{\purple{C.S.A_{(Cylinder)} =\: \dfrac{1760}{7}\: cm^2}}$

Now, we have to find the area of the tin sheet required to make the funnel :

$\footnotesize \bigstar \: \sf\boxed{\bold{\pink{Area\: of\: tin\: sheet =\: C.S.A_{(Frustum\: Of\: Cone)} + C.S.A_{(Cylinder)}}}}\: \: \bigstar$

Given :

➳ C.S.A of frustum of cone = 3718/7 cm²

➳ C.S.A of Cylinder = 1760/7 cm²

According to the question by using the formula we get,

$\dashrightarrow \bf \dfrac{3718}{7} + \dfrac{1760}{7}$

$\dashrightarrow \sf \dfrac{3718 + 1760}{7}$

$\dashrightarrow \sf \dfrac{5478}{7}$

$\dashrightarrow \sf\bold{\red{782.57\: cm^2}}$

$\therefore$ The area of the tin sheet required to make the funnel is 782.57 cm² .