Herons Formula Case Study Questions Class 9

Herons Formula Case Study Questions Class 9

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

Herons Formula Case Study Questions Class 9

The Herons Formula Case Study Questions 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 on herons formula 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 2: Heron’s Formula

Case Study on Heron’s Formula class 9 – 2

In an agricultural cooperative, three adjoining plots are arranged to form a triangular parcel with side lengths \(13\ \text{m}\), \(14\ \text{m}\) and \(15\ \text{m}\). The cooperative intends to install a drip-irrigation ring that follows the inner incircle of the triangle and must order pipe equal to the circumference of that incircle. Agricultural engineers must therefore compute not only the area of the triangular parcel to estimate water requirements but also the inradius and circumradius for planning the pipe and support posts. This scenario requires accurate use of Heron’s formula because no altitude is given. The problem also links geometric formulae to practical resource estimation (area \(\rightarrow\) water volume; inradius \(\rightarrow\) ring length). Use related identities and show how Heron’s result connects with \(r\) (inradius) and \(R\) (circumradius) for verification and procurement planning.

Useful formulae / properties \[s=\frac{a+b+c}{2},\qquad \Delta=\sqrt{s(s-a)(s-b)(s-c)}.\] \[\text{Inradius } r=\frac{\Delta}{s},\qquad \text{Circumradius } R=\frac{abc}{4\Delta}.\] \[\Delta = rs \quad\text{and}\quad \Delta=\frac{abc}{4R}.\]

MCQ Questions

1. The semi-perimeter \(s\) of the triangular parcel with sides \(13\ \text{m},\,14\ \text{m},\,15\ \text{m}\) is:

Answer: (b) \(21\).

Solution:
Compute the semi-perimeter: \[s=\frac{13+14+15}{2}=\frac{42}{2}=21.\] Therefore \(s=21\).

2. The area \(\Delta\) of the triangular parcel (in \(\text{m}^2\)) is:

Answer: (a) \(84\).

Solution:
Using Heron’s formula, \[\Delta=\sqrt{21(21-13)(21-14)(21-15)}=\sqrt{21\cdot 8\cdot 7\cdot 6}.\] Calculate inside the radical: \[21\cdot 8=168,\quad 168\cdot 7=1176,\quad 1176\cdot 6=7056.\] Thus \(\Delta=\sqrt{7056}=84\ \text{m}^2\).

3. The inradius \(r\) (radius of the inscribed circle) of the triangle is:

Answer: (b) \(4\).

Solution:
Use \(r=\dfrac{\Delta}{s}\). We have \(\Delta=84\) and \(s=21\), \[r=\frac{84}{21}=4\ \text{m}.\] So the inradius is \(4\ \text{m}\) and the pipe circumference required for the drip ring is \(2\pi r=8\pi\ \text{m}\).

4. The circumradius \(R\) (radius of the circumscribed circle) of the triangle equals:

Answer: (a) \(\dfrac{455}{56}\ \text{m}\).

Solution:
Use \(R=\dfrac{abc}{4\Delta}\) with \(a=13,\ b=14,\ c=15\) and \(\Delta=84\): \[abc=13\cdot 14\cdot 15=2730,\qquad 4\Delta=4\cdot 84=336.\] Thus \[R=\frac{2730}{336}=\frac{2730\div 6}{336\div 6}=\frac{455}{56}\ \text{m}.\] Numerical approximation: \(\dfrac{455}{56}\approx 8.125\ \text{m}\).

5. If the longest side of the parcel is increased from \(15\ \text{m}\) to \(16\ \text{m}\) while the other two sides remain \(13\ \text{m}\) and \(14\ \text{m}\), which statement is correct about the area?

Answer: (b) The area increases to approximately \(86.824\ \text{m}^2\).

Solution:
New semi-perimeter: \[s’=\frac{13+14+16}{2}=\frac{43}{2}=21.5.\] Apply Heron’s formula: \[\Delta’=\sqrt{21.5(21.5-13)(21.5-14)(21.5-16)}=\sqrt{21.5\cdot 8.5\cdot 7.5\cdot 5.5}.\] Compute approximately: \[21.5\times 8.5=182.75,\quad 182.75\times 7.5=1370.625,\] \[1370.625\times 5.5=7538.4375,\quad \Delta’=\sqrt{7538.4375}\approx 86.824\ \text{m}^2.\] Thus the area increases (by about \(2.824\ \text{m}^2\)), so option (b) is correct.

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