Chapter-5 (Area And Volume) Notes

Area and Volume: Notes

Mathematics • Grade 10 • Area and Volume

1. Pyramid

A solid with a polygonal base and triangular lateral faces meeting at one common vertex (apex) is a pyramid. In a right square pyramid the four lateral faces are congruent isosceles triangles.

Part	Symbol	Meaning
Side of base	a	Length of one side of the square base
Vertical height	h	Perpendicular from apex to the centre of the base
Slant height	l	Apex to the midpoint of a base side (height of a triangular face)
Edge	e	Apex to a corner of the base

• l² = h² + (a/2)²
• e² = l² + (a/2)²
• Area of base = a²  |  LSA = 2al (4 faces × ½al)  |  TSA = a² + 2al = a(a + 2l)
• Volume V = ⅓ a²h (3 pyramids fill a prism of the same base and height)

Pairs to remember (Pythagorean): 3-4-5, 5-12-13, 6-8-10, 7-24-25, 8-15-17. A square pyramid with a = 10, l = 13 has h = 12.

Example 1. a = 8 cm, l = 3 cm. Find LSA and TSA.

Base = 64; LSA = 2 × 8 × 3 = 48 cm²; TSA = 64 + 48 = 112 cm²

Example 2. a = 10 cm, edge e = 13 cm. Find TSA.

169 = l² + 25 ⇒ l = 12. TSA = 100 + 2 × 10 × 12 = 340 cm²

Example 3. TSA = 144 cm², l = 5 cm. Find h and e.

a² + 10a = 144 ⇒ (a + 18)(a − 8) = 0 ⇒ a = 8. h² = 25 − 16 = 9 ⇒ h = 3 cm; e² = 25 + 16 = 41 ⇒ e = √41 cm

Example 4. a = 12 cm, LSA = 240 cm². Find l and h.

2 × 12 × l = 240 ⇒ l = 10 cm; h² = 100 − 36 = 64 ⇒ h = 8 cm

Example 5. a = 16 cm, h = 6 cm. Find V.

V = ⅓ × 256 × 6 = 512 cm³

Example 6. V = 384 cm³, a = 12 cm. Find LSA.

⅓ × 144 × h = 384 ⇒ h = 8; l² = 64 + 36 ⇒ l = 10. LSA = 2 × 12 × 10 = 240 cm²

Example 7. TSA = 96 cm², a = 6 cm. Find V.

96 = 36 + 12l ⇒ l = 5; h² = 25 − 9 ⇒ h = 4. V = ⅓ × 36 × 4 = 48 cm³

2. Cone

A cone has a circular base and one vertex. The curved surface opens out into a sector whose arc equals the base circumference 2πr.

• Slant height: l² = h² + r² (l = √(h² + r²))
• CSA = πrl  |  TSA = πr² + πrl = πr(r + l)
• Volume V = ⅓ πr²h (3 cones fill a cylinder of the same base and height)

Use π = 22/7 unless told otherwise. If the diameter is given, halve it to get r.

Example 1. h = 12 cm, l = 13 cm. Find base area and CSA.

r² = 169 − 144 = 25 ⇒ r = 5. Base = 22/7 × 25 = 78.57 cm²; CSA = 22/7 × 5 × 13 = 204.28 cm²

Example 2. d = 12 cm, h = 8 cm. Find TSA.

r = 6; l = √(64 + 36) = 10. TSA = 22/7 × 6 × 16 = 301.71 cm²

Example 3. r + l = 64 cm, TSA = 2816 cm². Find CSA.

2816 = 22/7 × r × 64 ⇒ r = 14; l = 50. CSA = 22/7 × 14 × 50 = 2200 cm²

Example 4. TSA = 1320 cm², CSA = 704 cm². Find circumference of base.

πr² = 616 ⇒ r² = 196 ⇒ r = 14. C = 2πr = 88 cm

Example 5. h = 24 cm, l = 26 cm. Find V.

r = √(676 − 576) = 10. V = ⅓ × 22/7 × 100 × 24 = 2514.28 cm³

Example 6. CSA = 1158.3 cm², l = 19.5 cm. Find V.

r = 1158.3 × 7 ÷ (22 × 19.5) = 18.9 cm (the book prints 18.5 by mistake). h = √(19.5² − 18.9²) = 4.8. V = ⅓ × 22/7 × 18.9² × 4.8 = 1796.25 cm³

Example 7. V = 314.86 cm³, r : h = 5 : 12. Find CSA and TSA.

Let r = 5x, h = 12x: 314.86 × 21 = 22 × 25x² × 12x ⇒ x = 1. r = 5, h = 12, l = 13. CSA = 204.28 cm²; TSA = 22/7 × 5 × 18 = 282.85 cm²

3. Combined Solids

Solid	TSA	Volume
Cylinder + Cone (cone on top)	πr² + 2πrh₁ + πrl	πr²(h₁ + ⅓h₂)
Cone + Hemisphere	πrl + 2πr² = πr(l + 2r)	⅓πr²(h + 2r)
Cylinder + Hemisphere	πr² + 2πrh + 2πr² = 2πrh + 3πr²	πr²(h + ⅔r)
Cone + Cone (common base)	πr(l₁ + l₂)	⅓πr²(h₁ + h₂)
Prism + Pyramid (square base)	a² + P·h₁ + 2al (P = 4a)	a²(h₁ + ⅓h₂)

Think! TSA is not simply the sum of the separate TSAs: the joined faces disappear. Volume, however, is always the sum of the parts.

Example 1 (pencil). r = 7, cylinder h₁ = 39, cone h₂ = 24 cm. Find TSA and V.

l = √(576 + 49) = 25. TSA = 22/7 × 7 × (7 + 78 + 25) = 2420 cm². V = 22/7 × 49 × (39 + 8) = 7238 cm³

Example 2 (ice cream). r = 21 cm, V = 32340 cm³. Find cone height and TSA.

⅓ × 22/7 × 441 × (h + 42) = 32340 ⇒ h + 42 = 70 ⇒ h = 28 cm. l = 35. TSA = 22/7 × 21 × (35 + 42) = 5082 cm²

Example 3 (two cones). d = 6 cm, total height 20 cm. Find V.

V = ⅓ × 22/7 × 9 × 20 = 188.57 cm³

Example 4 (stupa). Prism h₁ = 5 m, pyramid h₂ = 0.5 m, a = 2 m. Find V and TSA.

V = 4 × 5 + ⅓ × 4 × 0.5 = 20.67 m³. l = √(0.25 + 1) = 1.12 m. TSA = 4 + 8 × 5 + 2 × 2 × 1.12 = 48.47 m²

Example 5. Two pyramids of equal height, a = 6 cm, total V = 96 cm³. Find h.

96 = ⅔ × 36 × h ⇒ h = 4 cm

4. Cost Estimation

• Volume of tank = base area × height. 1 m³ = 1000 litres; 1 ft³ = 28.317 litres.
• Area of quadrilateral = ½ × d × (p₁ + p₂) (d = diagonal, p₁, p₂ = perpendiculars to it).
• 4 walls + ceiling = 2h(l + b) + lb. Subtract doors and windows. Carpet area = floor area.
• Pillars with a pyramid on top: painted area = P × h + 2al. The base on the ground is not painted.
• Cost = area × rate. Compare quotations by finding the total of every item.

Example 1. Square tank, inner side 3 m, height 4 m. Water it holds?

V = 9 × 4 = 36 m³ = 36,000 litres

Example 2. Land NICE: NC = 40 m, perpendiculars 10 m and 15 m. Area? Labelling 2 days at Rs. 1500 + grass at Rs. 70/m².

Area = ½ × 40 × 25 = 500 m². Cost = 70 × 500 + 3000 = Rs. 38,000

Example 3. Room 14 × 12 × 10 ft, 2 windows 3 ft square, 2 doors 6 × 2 ft. Paint at Rs. 30/ft².

Walls + ceiling = 520 + 168 − 24 − 18 = 646 ft². Cost = Rs. 19,380. Rate Rs. 40 ⇒ 25,840. Increase = Rs. 6,460

Example 4. Two pillars 4 × 4 ft, h₁ = 10 ft, pyramid h₂ = 2 ft. Paint at Rs. 95/ft².

l = √(4 + 4) = 2.83 ft. One pillar = 160 + 2 × 4 × 2.83 = 182.64 ft². Two = 365.28 ft². Cost = Rs. 34,701.60 (book text says 32 ft by mistake; its working uses 2 ft)

Exam tips: (1) Always find l (or h or r) first by Pythagoras. (2) Write the formula before substituting. (3) Keep units consistent. (4) For tents, tanks open at the top and pillars, check which surfaces are really included.

Area and Volume • Grade 10 Mathematics