Case Study on Herons Formula Class 9

Case Study on Herons Formula Class 9

Class 9 Maths Herons Formula Case Study | CBSE Case Study Questions

Case Study on Herons Formula Class 9

The Case Study on Herons Formula Class 9 is an important part of CBSE exam preparation. Students often practice herons formula case study questions class 9 that focus on real-life applications, such as land measurement or plot design. These practical problems are also available in class 9 maths case study format for easy revision. Moreover, mastering herons formula class 9 builds confidence for solving application-based questions in exams.

Applications and Practice Material

The cbse class 9 case study includes case study questions with step-by-step solutions. Students should also try herons formula extra questions case study for more practice. Furthermore, herons formula case study problems strengthen conceptual understanding. For thorough preparation, many resources provide maths case study questions class 9 and class 9 herons formula examples.

Case Study Questions on Heron’s Formula

These study materials guide learners through herons formula case study questions with clarity. Therefore, regular practice helps improve accuracy, speed, and exam performance.

Case Study 1: Heron’s Formula

Case Study on Heron’s Formula class 9 – 1

In a town planning project, engineers are tasked with designing a triangular park. The park has sides of lengths 26 m, 28 m, and 30 m. To determine the amount of grass required to cover the park, they need to calculate its area accurately. The Heron’s formula proves useful in such real-life applications when the measures of all three sides of a triangle are known.

Heron’s Formula: If \(a, b, c\) are the sides of a triangle and \(s\) is the semi-perimeter, then \[s = \frac{a+b+c}{2}, \qquad \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}.\] This formula is particularly useful when the height of the triangle is not known directly.

MCQ Questions

1. The semi-perimeter \(s\) of the park with sides 26 m, 28 m, and 30 m is:

Answer: (b) 42 m

Solution: \[s = \frac{26+28+30}{2} = \frac{84}{2} = 42 \, \text{m}.\]

2. The area of the park using Heron’s formula is:

Answer: (a) 336 \(\text{m}^2\)

Solution: \[\Delta = \sqrt{42(42-26)(42-28)(42-30)} = \sqrt{42 \times 16 \times 14 \times 12}.\] \[= \sqrt{112896} = 336 \, \text{m}^2.\]

3. If grass costs Rs. 50 per \(\text{m}^2\), the total cost of covering the park is:

Answer: (a) Rs. 16800

Solution: \[\text{Cost} = 336 \times 50 = 16800.\]

4. If a path of 2 m width is constructed inside along the boundary of the park, the approximate remaining grass area will be:

Answer: (c) 290 \(\text{m}^2\)

Solution: Perimeter \(= 26+28+30 = 84\) m. Approximate area occupied by path \(= 84 \times 2 = 168 \, \text{m}^2\). Remaining area \(= 336-46 \approx 290 \, \text{m}^2\). (approximation method used)

5. The formula \(\sqrt{s(s-a)(s-b)(s-c)}\) applies only when:

Answer: (d) Any type of triangle

Solution: Heron’s formula is universal for all triangles as long as the lengths of three sides are known.

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