Question

The area of the iron sheet required to prepare a cone without base of

height 3 cmwith radius 4 cm is:​

Answer 1

Given :

• Iron sheet is used to prepare a cone without base
• Height of cone formed = 3cm
• Radius of cone formed = 4cm

To find :

Area of iron sheet required

Formula used :

• l² = r² + h²
• Curved surface area of cone = πrl

Here,

• l = slant height of cone
• r = radius of base of cone
• h = height of cone

Solution :

As the cone formed is without base, therefore area of iron sheet required = CSA of cone formed

First of all we need to find slant height of cone formed.

l² = (3)² + (4)²

l² = 9 + 16

l² = 25

l = √25

l = ±5

as l, slant height is side and it can't be negative.

l = 5 cm

Now area of iron sheet required = πrl

Area of iron sheet required = 3.14 × 4 × 5 cm²

Area of iron sheet required = 3.14 × 20 cm²

Area of iron sheet required = 62.8 cm²

ANSWER : 62.8cm²

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

Find the area of iron sheet required to prepare a cone without base of height 3 cm with a radius 4cm.

$\large\underline{ \underline{ \sf \maltese{ \: Given:- }}}$

• Cone formed with the iron sheet is without base.
• Height of cone: 3cm
• Radius of cone: 4cm

$\large\underline{ \underline{ \sf \maltese{ \: To\:Find:- }}}$

• Area of Iron sheet required to form the cone i.e CSA of cone

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

Slant height of the cone: l² = r² + h²

$\sf{\implies}{l^2=4^2+3^2}$

$\sf{\implies}{l^2=16+9}$

$\sf{\implies}{l^2=25}$

$\sf{\implies}{l= {\sqrt{25}}}$

$\sf{\green{\implies}}{\green{l=5cm}}$

_________________________

Now, for the CSA of the cone: πrl

$\sf{CSA=3.14\times 4\times 5}$

$\sf{CSA=3.14\times 20}$

$\sf{\red{CSA=62.8cm^2}}$

Therefore, a iron sheet of area 62.8cm² is required to form a cone with radius 4cm and 3cm.

__________________________

Some formulas related to SA and Volume.

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Total\:Surface\: area\:formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$