Question

Find the Total surface area of a cone if it's slant height is 21m and diameter of it's base is 24m.

Answer 1

Answer :-

• The total surfAce area of cone is 1244.57m².

Step-by-step explanation :-

To find :-

• The total surface area of cone..

Solution :-

Given that ,

• The slant height of the cone = 21m.
• Diameter of it's base = 24m.

♦ Radius is

=> Diameter / 2

=> 24 / 2

=> 12m

As we know that ,

Total surface area of cone = πr ( r + L ) .

Where we know,

• π = 22/7
• r = Radius ( radii )..
• L = Slant height.

According to the question :-

The total surface of cone is,

=> Total surface area = πr { r + L } ..

$= \frac{22}{7} \times 12 \times (12 + 21)$

$= \frac{22}{7} \times 12 \times 33$

$= (22 \times 12 \times 33) \div 7$

$= (726 \times 12) \div 7$

$= 8712 \div 7$

$= 1244.57 {m}^{2}$

• Therefore , The total surface area of cone is 1244.57m².

Hope this helps you :)

Answer 2

1244.57 cm²

Step-by-step explanation:

Given:

• Slant height (l) is 21m
• Diameter (d) is 24m

Hence, radius will be :

➝ diameter/2

➝ 24/2

➝ 12m

$\:$

To Find:

• Total Surface Area (TSA) of the cone.

Solution:

As, we know:

$\star \quad{ \underline{ \green{ \boxed{TSA_{(cone)} = \pi r( l+r ) }}}} \quad\star \quad$

Here,

• π = 22/7
• r = 12m
• l = 21m

$\rightarrow \: \frac{22}{7} \times 12 \: (21 + 12)$

$\rightarrow \: \frac{22}{7} \times 12 \: (33)$

$\rightarrow \: \frac{8712}{7} {cm}^{2}$

Therefore, Total Surface Area of Cone is 8712/7 cm² or 1244.57cm².

_____________________

Additional Information:

$\footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}$