Class9th CBSE MATHS

Past CBSE exam questions** on **Heron’s Formula

# **Problem 1: Direct Application of Heron’s Formula**
📌 **CBSE 2018 Question:**
Find the area of a triangle with sides **7 cm, 8 cm, and 9 cm** using Heron’s formula.

#### **Solution:**
**Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{7+8+9}{2} = \frac{24}{2} = 12 \text{ cm}
\]

**Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
= \sqrt{12(12-7)(12-8)(12-9)}
\]

**Step 3: Simplify the Expression**
\[
= \sqrt{12 \times 5 \times 4 \times 3}
\]

\[
= \sqrt{720}
\]

\[
= 6\sqrt{20} = 6 \times 4.47 \approx 26.82 \text{ cm}^2
\]

✅ **Final Answer: 26.82 cm²**

—

### **Problem 2: Application in Composite Figures**
📌 **CBSE 2022 Question:**
A quadrilateral is divided into two triangles with sides **5 cm, 6 cm, 7 cm** and **8 cm, 10 cm, 6 cm**. Find its total area using Heron’s formula.

#### **Solution:**

✅ **Step 1: Find the Area of Triangle 1 (5 cm, 6 cm, 7 cm)**

**Semi-perimeter:**
\[
s = \frac{5+6+7}{2} = \frac{18}{2} = 9 \text{ cm}
\]

Apply Heron’s Formula:
\[
\text{Area} = \sqrt{9(9-5)(9-6)(9-7)}
\]

\[
= \sqrt{9 \times 4 \times 3 \times 2}
\]

\[
= \sqrt{216} \approx 14.7 \text{ cm}^2
\]

✅ **Step 2: Find the Area of Triangle 2 (8 cm, 10 cm, 6 cm)**

**Semi-perimeter:**
\[
s = \frac{8+10+6}{2} = \frac{24}{2} = 12 \text{ cm}
\]

**Apply Heron’s Formula:**
\[
\text{Area} = \sqrt{12(12-8)(12-10)(12-6)}
\]

\[
= \sqrt{12 \times 4 \times 2 \times 6}
\]

\[
= \sqrt{576} = 24 \text{ cm}^2
\]

✅ **Final Step: Total Area of Quadrilateral**
\[
= 14.7 + 24 = 38.7 \text{ cm}^2
\]

✅ **Final Answer: 38.7 cm²**

—

### **Problem 3: Real-Life Application-Based Question**
📌 **CBSE 2021 Question:**
A triangular plot has sides **50 m, 72 m, and 78 m**. Find the **cost of fencing** it at ₹12 per meter.

#### **Solution:**
✅ **Step 1: Find the Perimeter**
\[
\text{Perimeter} = 50 + 72 + 78 = 200 \text{ m}
\]

✅ **Step 2: Find the Semi-Perimeter (s)**
\[
s = \frac{200}{2} = 100 \text{ m}
\]

✅ **Step 3: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{100(100-50)(100-72)(100-78)}
\]

\[
= \sqrt{100 \times 50 \times 28 \times 22}
\]

\[
= \sqrt{3080000} \approx 1755.69 \text{ m}^2
\]

✅ **Step 4: Find the Cost of Fencing**
Since fencing is charged at ₹12 per meter:

\[
\text{Cost} = \text{Perimeter} \times 12
\]

\[
= 200 \times 12 = 2400
\]

✅ **Final Answer: ₹2400**

—

### **Key Takeaways for Exams:**
✔ **Direct Application** questions appear in every exam.
✔ **Composite Figures** questions are tested in higher-order thinking problems.
✔ **Real-Life Application Questions** appear every **2-3 years**.
✔ **Show Full Steps in Exams** for full marks.

—

Would you like **more CBSE-style practice questions** for Heron’s Formula? COMMENT….