Question

The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer 1

Answer:

Given:

• Radius of penholder, r = 3 cm
• Height of penholder, h = 10.5 cm

To Find:

How much cardboard was required to be bought for the competition?

Solution:

We know that,

Penholder has shape of cylindrical shape.

Cardboard required by 1 competitor = CSA of one penholder +Area of base

⇏ 2πrh + πr²

⇏ 2 × 22/7 × 3 + 10.5 + 22/7 + (3)²

⇏ 198 + 198/7

⇏ 1584/7 cm²

★ Cardboard required for 35 competitors,

⇏ 35 × 1584/7

⇏ 7920 cm²

Hence, 7920 cm² of cardboard was required to be bought for the competition.

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More to know:

Curved Surface Area of cylinder = 2πrh

Total surface area of cylinder = 2πr(r + h)

Volume of cylinder = πr²h

Answer 2

Required answer:-

Question:

❥ The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Solution:

Given,

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Radius of penholder = 3 cm

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Height of penholder = 10.5 cm

To find:

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Cardboard required for the competition

Formulas used:

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Area of base = πr²

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Curved surface area = 2πrh

Concept:

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎❥ Cylinders

Step by step explaination:

First we have to calculate curved surface area of penholder.

That is,

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ = ‎2πrh

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎( Value of π is 22/7 )

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$2 \times \dfrac{22}{7} \times 3 + 10.5$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$2 \times \dfrac{22}{7} \times 3+ \dfrac{105}{10}$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=198

Now, we have to calculate area of base.

That is,

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎= ‎πr²

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$\dfrac{22}{7} + (3) {}^{2}$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$\dfrac{22}{7} \times 9$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$\dfrac{198}{7}$

Thus, cardboard required by one 1 competitor = Curved surface area + Area of penholder.

That is,

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$198 + \dfrac{198}{7}$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=1584 cm²

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Cardboard required to be bought for competition:

According to the question there are 35 competitors.

So,

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=$35 \times \dfrac{1584}{7}$

‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎=7920 cm²