A metal sheet of breadth 10 cm is used to make a cone of radius 14 cm and slant height 20 cm. Then length of sheet if 25% of sheet is wasted is .

1.99 m

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1.77 m

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1.82 m

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1.54 m

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Solution

The correct option is A $1.99 m$
We know that, TSA of cone = πr(r+l)

Now, TSA= $\frac{22}{7} \times 14 \times (14 + 20)$
$\Rightarrow T S A = \frac{22}{7} \times 14 \times 34$
$\Rightarrow T S A = 1496 c m^{2}$

Let, the length of sheet be $l$ .
Given that, breadth of the sheet $= 10 c m$
and 25 $%$ of the sheet is wasted
$\to$ $\frac{25}{100} \times 10 l$

Thus, $10 l = \frac{10 l}{4} + 1496$
$\Rightarrow l = 199.467 c m$
$\Rightarrow l = 1.99 m$

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A metal sheet of breadth 10 cm is used to make a cone of radius 14 cm and slant height 20 cm. Then length of sheet if 25% of sheet is wasted is .

The correct option is A 1.99 mWe know that, TSA of cone = πr(r+l) Now, TSA=227×14×(14+20) ⇒TSA=227×14×34 ⇒TSA=1496 cm2 Let, the length of sheet be l. Given that, breadth of the sheet =10 cm and 25% of the sheet is wasted → 25100×10l Thus, 10l=10l4+1496 ⇒l=199.467 cm ⇒l=1.99 m