Case Study Questions on Mensuration Class 10

Case Study Questions on Mensuration Class 10 | CBSE & NCERT

Importance of Mensuration Case Studies

Case Study Questions on Mensuration Class 10 are designed to strengthen conceptual understanding and problem-solving skills. Students often practice Surface Areas and Volumes Class 10 Case Study and solve Mensuration Case Study Questions Class 10 to apply formulas in real-life contexts. These questions help in exams as well as in practical learning.

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For easy revision, many teachers suggest Class 10 Maths Surface Areas and Volumes Case Study Pdf. It includes Cbse Class 10 Surface Areas and Volumes Case Study and detailed Surface Areas and Volumes Case Study Questions With Solutions. Moreover, practicing a Class 10 Maths Case Study on Surface Areas and Volumes builds accuracy and confidence.

Exam-Oriented Preparation

While preparing for boards, focus on Surface Areas and Volumes Case Study Based Questions Class 10. Additionally, Case Study Surface Areas and Volumes Class 10 With Answers helps students check their progress. Solving Class 10 Case Study Questions Maths Surface Areas and Volumes and Cbse Case Study Questions Maths Class 10 Surface Areas and Volumes regularly ensures strong exam readiness. Therefore, students must practice consistently for success.

Case Study 2: Fountain Design

Case Study 2: Mensuration Class 10

A city garden commission plans a centre-piece fountain for a school campus. The fountain design consists of a solid cylindrical pedestal on the ground which supports a shallow frustum-shaped basin. A small hemispherical ornament sits at the centre of the basin as a decorative finial. The pedestal has radius 2 m and height 1.5 m. The frustum basin has lower radius 3 m, upper radius 1 m and vertical height 0.8 m. The hemisphere has radius 0.5 m. The contractor must estimate the exterior surface area to be painted (the base of the pedestal that rests on the ground is not painted) and compute the water capacity of the basin when filled to its brim. The basin is open at the top; only the exposed external surfaces are to be painted (pedestal curved surface, lateral surface of frustum, the top annulus of the basin and the curved surface of the hemisphere). Use π in symbolic answers and give numerical values correct to three significant figures where requested. This problem tests combining curved surface areas, frustum slant computations and volume of frustum.

1. The curved surface area (CSA) of the cylindrical pedestal is:

Answer: (a)

Solution: CSA of a right circular cylinder = 2πrh. Here r = 2 m, h = 1.5 m so CSA = 2π × 2 × 1.5 = 6π m² ≈ 18.85 m².

2. The slant height l of the frustum basin (with R = 3 m, r = 1 m, h = 0.8 m) is:

Answer: (a)

Solution: For a frustum, slant height l = √[(R-r)²+h²] = √[(3-1)²+0.8²] = √[4+0.64] = √4.64 ≈ 2.154 m.

3. The lateral surface area (LSA) of the frustum basin is (use l from previous):

Answer: (a)

Solution: Lateral surface area of a frustum = π(R+r)l. With R = 3, r = 1 and l ≈ 2.154, LSA = π×4×2.154 ≈ 27.07 m².

4. The external horizontal area at the basin top (the top annulus) which is exposed and to be painted equals:

Answer: (a)

Solution: The exposed top horizontal ring (annulus) area = π(R²-r²) = π(3²-1²) = 8π m² ≈ 25.13 m².

5. The total exterior area to be painted (curved surface of pedestal + lateral frustum + top annulus + hemisphere curved surface) is approximately:

Answer: (a)

Solution: Compute each contribution:
CSA of pedestal = 6π m² ≈ 18.85,
LSA of frustum = π(R+r)l ≈ 27.07,
Top annulus = 8π ≈ 25.13,
CSA of hemisphere = 2π(0.5)² = 0.5π ≈ 1.571.
Summing: 6π + π(R+r)l + 8π + 0.5π ≈ 72.62 m². Thus option (a) is correct.

6. The volume of water the frustum basin can hold when filled to the brim equals:

Answer: (a)

Solution: Volume of a frustum = ⅓πh(R² + r² + Rr). Substitute h = 0.8, R = 3, r = 1:
V = ⅓π × 0.8 × (9 + 1 + 3) = (0.8π)/3 × 13 ≈ 10.891 m³.
Hence option (a) is correct.

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