Mathematics Practice Solutions: Geometry and Mensuration

Answer Key

• 1: A
• 2: C
• 3: 50°
• 4: B
• 5: D
• 6: C
• 7: C
• 8: C
• 9: A
• 10: A

11. Congruency Proof

Given: OA = OD and OB = OC.

In ΔOBC, since OB = OC, then ∠OBC = ∠OCB. Consequently, ∠OCD = ∠OBA.

In ΔOAD, since OA = OD, then ∠OAD = ∠ODA.

Since ∠OCD = ∠OBA and ∠OAD = ∠ODA, we find that ∠AOB in ΔOAB is equal to ∠COD in ΔOCD.

Conclusion: By SAS congruency, ΔOAB ≅ ΔOCD. Therefore, AB = CD.

12. Conical Cloth Calculation

Given: r = 7m, h = 24m.

l = √(r² + h²) = √(49 + 576) = √625 = 25m.

Curved Surface Area (CSA) = πrl = (22/7) × 7 × 25 = 550m².

Length of cloth = 550m² / 5m = 110m.

13. Sphere Diameter

Surface area of sphere = 154 cm².

4πr² = 154

r² = 154 / (4π) = (154 / 4) × (7 / 22) = 49 / 4.

r = √(49 / 4) = 7/2 = 3.5 cm.

Diameter (D) = 2r = 2 × 3.5 = 7 cm.

14. Shaded Region Area

Area of triangle ABC = 1/2 × base × height = 1/2 × 14 × 15 = 105 cm².

Area of smaller triangle = 1/2 × 12 × 5 = 30 cm².

Area of shaded region = 105 - 30 = 75 cm².

15. Triangle Altitude

Perimeter (P) = a + b + c = 180. Given a=80, b=18, then c = 180 - 80 - 18 = 82 cm.

Semi-perimeter (s) = 180 / 2 = 90 cm.

Area = √[s(s-a)(s-b)(s-c)] = √[90(10)(72)(8)] = √518400 = 720 cm².

For the shortest side (b=18 cm): 720 = 1/2 × 18 × h. Therefore, h = 80 cm.

16. Cost of Conical Canvas

h = 10m, r = 24m. Slant height (l) = √(10² + 24²) = 26m.

CSA = πrl = (22/7) × 24 × 26 = 13728/7 m².

Total cost = (13728/7) × 70 = Rs. 137,280.

17. Circle Theorem Proof

Given: Circle with center O, arc PQ subtending ∠POQ at center and ∠PAQ at the circumference.

To Prove: ∠POQ = 2∠PAQ.

Construction: Join AO and extend to point B.

Proof:

• In ΔAPO: OP = OA (radii), so ∠OPA = ∠OAP.
• In ΔAQO: OQ = OA (radii), so ∠OQA = ∠OAQ.
• Exterior angle ∠BOP = ∠OPA + ∠OAP = 2∠OAP.
• Exterior angle ∠BOQ = ∠OQA + ∠OAQ = 2∠OAQ.
• Adding both: ∠POQ = 2(∠OAP + ∠OAQ) = 2∠PAQ.

18. Volume Calculation

(i) Volume of cylinder = πr²h = π(6)²(8) = 288π cm³.

(ii) Volume of sphere = 4/3 πR³ = 288π. R³ = 216. R = 6 cm.