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.
(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
