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.
Find the value of the polynomial 5x – 4x 2 + 3 at
(i) x = 0 (ii) x = –1 (iii) x = 2
Factorise each of the following:
(i) 8a 3 + b 3 + 12a 2b + 6ab2
(ii) 8a 3 – b 3 – 12a 2b + 6ab2
(iii) 27 – 125a 3 – 135a + 225a 2
(iv) 64a 3 – 27b 3 – 144a 2b + 108ab2
(v) 27p 3 – \(\frac{1}{ 216}\) – \(\frac{9 }{ 2}\) p2 + \(\frac{1 }{4}\) p