Radius of the heap, r = \(\frac{10.5}{2}\) m = 5.25 m
Height of the heap, h = 3 m
Volume of the heap = \(\frac{1}{3}\pi\)r²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5.25 m × 5.25 m × 3 m
= 86.625 m³
Slant height,\(l = \sqrt{r² + h²}\)
\(= \sqrt{(5.25)² + (3)²}\)
\(= \sqrt{27.5625 + 9}\)
\(= \sqrt{36.5625}\)
= 6.046 m
∴ The area of the canvas required to cover the heap = \(\pi\)rl
= \(\frac{22}{7}\) × 5.25 m × 6.046 m
= 99.825 m²
Therefore, 99.825 m² canvas will be provided to protect the heap from rain.
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.