Question

th-som
4. A cylindrical tank is open at the top. It has the base diameter 20 m and height 7 m. How many
mº of metal sheet is used to build the tank?
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Answer 1

Given :

• A cylindrical tank is open at the top. It has the base diameter 20 m and height 7 m.

To find :

• How many m² of metal sheet is used to build the tank?

Solution :

Base diameter of cylindrical tank = 20m

Radius of cylindrical tank = 20/2 = 10m

Height of cylindrical tank = 7m

According to the formula of total surface area of cylinder

★ Curved surface area of cylinder + 2 × area of circle

★ 2πrh + 2πr² ★

• Focus Zone : A cylinder is open at the top

★ 2πrh + πr² ★

→ πr(2h + r)

• Substitute the values

→ 22/7 × 10(2 × 7 + 10)

→ 220/7(14 + 10)

→ 220/7 × 24

→ 5280/7

→ 754.28 m²

•°• The area of metal sheet used to build the tank is 754.28 m²

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Answer 2

$\large \underline{\underline{\bf GIVEN \: :}}$

• A cylindrical tank is open at the top. It has the base diameter 20 m and height 7 m.

$\large \underline{\underline{\bf TO \: FIND \: :}}$

• How many m² of metal sheet is used to build the tank?

$\large \underline{\underline{\bf SOLUTION \: :}}$

• Height of cylindrical tank, h = 7 m
• Diameter of cylindrical tank, d = 20 m

Hence, Radius of cylindrical tank, r = $\sf \dfrac{d}{2} = \dfrac{20m}{2} = 10 m$

Now, total surface area of cylinder = 2πrh + 2πr²

But we are given that cylinder is open at the top.

Hence, Surface area of given cylinder = 2πrh + 2πr² - πr² = 2πrh + πr² = πr(2h + r)

Now, Substitute the values :

$\sf \implies$ π × 10(2 × 7 + 10)

$\sf \implies$ π × 10(14 + 10)

$\sf \implies$ π × 10(24)

$\sf \implies$ π × 240

Now, substitute value of π = $\sf \dfrac{22}{7}$

$\sf \implies \dfrac{22}{7}$ × 240

$\sf \implies \dfrac{5280}{7}$

$\sf \implies 754.285...$

$\sf \implies 754.28 m^{2}\: (approx.)$

Hence, The area of metal sheet used to build the tank is approximately 754.28 m².