Given,

Diameter of the conical heap (d) = 10.5 m

Radius of the conical heap (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 m

Height of the conical heap (h) = 3 m

By formula,

Volume of the conical heap (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 27.5625 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 82.6875 \\[1em] = \dfrac{1819.125}{21} \\[1em] = \text{86.625 m}^3$

By formula,

Slant height (l) = $\sqrt{r^2 + h^2}$

Substituting values we get :

$l = \sqrt{(5.25)^2 + (3)^2} \\[1em] = \sqrt{27.5625 + 9} \\[1em] = \sqrt{36.5625} \\[1em] = 6.047 \text{ m}$

The area of the canvas required to cover the heap of wheat (A) = Curved surface area of conical heap = πrl

Substituting values we get :

$A = \dfrac{22}{7} \times 5.25 \times 6.047 \\[1em] = \dfrac{698.4285}{7} \\[1em] = 99.775 \text{ m}^2.$

Hence, the volume of the conical heap is 86.625 m³ and the area of the canvas required is 99.775 m².

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CBSE Class IX | Academic Year 2026-2027

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Surface Areas and Volumes — Interactive Study Guide

Formula Master Table

Solid	CSA	TSA	Volume
Cube (a)	4a²	6a²	a³
Cuboid (l,b,h)	2h(l+b)	2(lb+bh+hl)	lbh
Cylinder (r,h)	2πrh	2πr(r+h)	πr²h
Cone (r,h,l)	πrl	πr(l+r)	⅓πr²h
Sphere (r)	4πr²		&frac43;πr³
Hemisphere (r)	2πr²	3πr²	⅔πr³

Common Traps

Cone CSA uses slant height l, NOT height h! l = √(r²+h²)

Hemisphere TSA = CSA + base circle = 2πr² + πr² = 3πr²

Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check

1. Volume of cube with side 5 cm? (125 cm³)
2. CSA of cylinder with r=7, h=10? (2×22/7×7×10 = 440 cm²)
3. Volume of cone with r=3, h=4? (⅓×π×9×4 = 12π cm³)