The circumference of the base of a conical tent is 66 m. If the height of the tent is 36 m, what is the area (in m²) of the canvas used in making the tent? (Take π = $\frac{22}{7}$)

1237.5

We know that,

Curved surface area of conical tent = πrl

Circumference of the base of a conical tent = 2πr

l² = r² + h²

We have,

Height of the conical tent h = 36 m

According to the question,

2 πr = 66

= 2 × \(\frac{22}{7}\) × r = 66

= r = 66 × \(\frac{7}{44}\)

= r = \(\frac{21}{2}\)

As we know,

= l² = (\(\frac{21}{2}\))² + 36²

= l² = \(\frac{441}{4}\) + 1296

= l² = \(\frac{5625}{4}\)

= l = \(\frac{75}{2}\)

Now, Curved surface area of conical tent = \(\frac{22}{7}\) × \(\frac{21}{2}\) × \(\frac{75}{2}\)  = 1237.5 m²