Question

maths experts...here's a task for you​

Answer 1

$\bf {\huge {\underline {\boxed {\sf{\purple {Answer:\:1232 \:m^3}}}}}}$

$\textbf {\underline {\underline {Step-By-Step\:Explanation:-}}}$

$\textbf {\underline {\blue {Given:}}}$

• Area of the canvas = 551 $m^2$
• Area wasted = 1 $m^2$
• Radius of the conical tent = 7m

$\textbf {\underline {\blue {To\:Find}}}$

• The volume of conical tent

$\textbf {\underline {Formula\:used:}}$

• Curved surface area of cone = $\dfrac {22}{7}$r×l
• Volume of cone = $\dfrac {1}{3}$×$\dfrac {22}{7}$${r}^2 h$

$\huge\underline\green{\tt Solution:}$

As mentioned in the question, 1 $m^2$ was wasted.

So, C.S.A of cone = 550 $m^2$

$\pie$rl = 550

➡ $\dfrac {22}{7}$ × 7 × l = 550

➡ 22 × l = 550

➡ l = $\dfrac {550}{22}$

•°• l = 25 m

Now,

°•° $l^2 = r^2 + h^2$

Hence,

➡ Height of the conical tent

= $\sqrt {l^2 - r^2}$

= $\sqrt {{25}^2 - {7}^2}$

= $\sqrt {625 - 49 }$

= $\sqrt {576}$

= 24 m

Finally,

➡ Volume of conical tent

$\dfrac {1}{3}$ × $\dfrac {22}{7}$ × 7 × 7 × 24 [/tex]

= $\dfrac{22 × 7}{3}$ × 8

= 1232 $m^3$

Hence,

Volume of the conical tent is 1232 $m^3$