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)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
