CG Board Class 8 Mathematics Chapter 19 Solutions: Mensuration-III (क्षेत्रमिति-3) – 2026-27

(i) The shape of the base of cylinder is ___________**

• (ii) Formula of the volume of cylinder is ______________
• (iii) Radius and height of cylinder is 7 cm each. Volume of cylinder is ________

Area of base = 1386 cm², Height h = 15 cm.

Volume of cylinder = Area of base × Height

V = 1386 × 15 = 20790 cm³

(i) Radius = 12 cm, height = 14 cm.**

(ii) Radius = 2.8 cm, height = 5 cm.

(iii) Diameter = 20 m, height = 21 m.

• (i) Given: Radius r = 12 cm, Height h = 14 cm.

6336 cm3

• (ii) Given: Radius r = 2.8 cm, Height h = 5 cm.

V = π r² h = 22/7 × (2.8)² × 5 = 22/7 × 7.84 × 5 = 22 × 1.12 × 5 = 123.20 cm³

123.20 cm3

• (iii) Given: Diameter = 20 m ⇒ Radius r = 10 m, Height h = 21 m.

V = π r² h = 22/7 × (10)² × 21 = 22 × 100 × 3 = 6600 m³

6600 m3

• Let the initial radius be r and height be h.
• Initial Volume / V₁ = π r² h.
• When diameter is halved, the new radius becomes r’ = r/2.
• New Volume / V₂ = π (r/2)² h = (π r² h)/(4) = (V₁)/(4).
• Ratio of Old Volume to New Volume /

(V₁)/(V₂) = (V₁)/((V₁)/(4)) = 4/1 = 4 : 1

Radius r = 2.8 m, Height h = 3.5 m.

• Capacity / :

V = π r² h = 22/7 × (2.8)² × 3.5 = 22/7 × 7.84 × 3.5 = 22 × 7.84 × 0.5 = 86.24 m³

Diameter = 1.4 cm ⇒ Radius r = 0.7 cm, Height/Length h = 90 cm.

• Volume of iron required / :

V = π r² h = 22/7 × (0.7)² × 90 = 22/7 × 0.49 × 90 = 22 × 0.07 × 90 = 138.6 cm³

### *EXERCISE 19.2

(i) Radius = 7 cm, height = 24 cm**

(ii) Diameter = 20 m, height = 21 m.

(iii) Radius = 10.5 cm, height = 35 cm

(iv) Radius = 14 cm, height = 1 m

• (i) Given: Radius r = 7 cm, Height h = 24 cm.
• CSA = 2 × 22/7 × 7 × 24 = 44 × 24 = 1056 cm²
• TSA = 2 × 22/7 × 7 × (7 + 24) = 44 × 31 = 1364 cm²
• (ii) Given: Diameter = 20 m ⇒ r = 10 m, Height h = 21 m.
• CSA = 2 × 22/7 × 10 × 21 = 2 × 22 × 10 × 3 = 1320 m²
• TSA = 2 × 22/7 × 10 × (10 + 21) = (440 × 31)/(7) = (13640)/(7) = 1948.57 m²
• (iii) Given: Radius r = 10.5 cm, Height h = 35 cm.
• CSA = 2 × 22/7 × 10.5 × 35 = 2 × 22 × 1.5 × 35 = 2310 cm²
• TSA = 2 × 22/7 × 10.5 × (10.5 + 35) = 66 × 45.5 = 3003 cm²
• (iv) Given: Radius r = 14 cm, Height h = 1 m = 100 cm.
• CSA = 2 × 22/7 × 14 × 100 = 8800 cm²
• TSA = 2 × 22/7 × 14 × (14 + 100) = 88 × 114 = 10032 cm²

Circumference of base = 2 π r = 176 cm, Height h = 30 cm.

Curved Surface Area = Circumference of base × Height

CSA = 176 × 30 = 5280 cm²

Volume V = 44 cm³, Radius r = 2 cm.

• Step 1: Find Height / :

V = π r² h ⇒ 44 = 22/7 × 2² × h

44 = 88/7 h ⇒ h = (44 × 7)/(88) = 7/2 = 3.5 cm

• Step 2: Find Total Surface Area / :

TSA = 2 π r (r + h) = 2 × 22/7 × 2 × (2 + 3.5)

TSA = 88/7 × 5.5 = 484/7 = 69.14 cm²

Diameter = 14 m ⇒ Radius r = 7 m, Depth/Height h = 25 m.

• Part 1: Volume of mud / :

V = π r² h = 22/7 × 7² × 25 = 22 × 7 × 25 = 3850 m³

• Part 2: Inner Curved Surface Area / :

CSA = 2 π r h = 2 × 22/7 × 7 × 25 = 44 × 25 = 1100 m²

• Part 3: Cost of plastering / :

Cost = 1100 m² × Rs. 3/m² = Rs. 3300

Circumference of base = 44 m, Height h = 44 m.

CSA = Circumference of base × Height = 44 × 44 = 1936 m²

CSA = 1000 π cm², Diameter = 20 cm ⇒ r = 10 cm.

• Formula & Substitution / :

2 π r h = 1000 π

2 π (10) h = 1000 π

20 π h = 1000 π ⇒ h = (1000 π)/(20 π) = 50 cm

Height of cylinder is 50 cm.