Important CBSE-style practice questions on Heron’s Formula
## **Solution 1: Find the Area of a Triangle (Basic Calculation)**
📌 **Question:**
Find the area of a triangle with sides **9 cm, 12 cm, and 15 cm** using Heron’s formula.
### **Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{a + b + c}{2} = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18 \text{ cm}
\]
### **Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
= \sqrt{18(18-9)(18-12)(18-15)}
\]
\[
= \sqrt{18 \times 9 \times 6 \times 3}
\]
\[
= \sqrt{2916}
\]
\[
= 54 \text{ cm}^2
\]
✅ **Final Answer: 54 cm²**
—
## **Solution 2: Find the Cost of Paving a Triangular Garden**
📌 **Question:**
A **triangular garden** has sides **80 m, 90 m, and 100 m**. Find the cost of paving the garden at ₹20 per square meter.
### **Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{80 + 90 + 100}{2} = \frac{270}{2} = 135 \text{ m}
\]
### **Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{135(135-80)(135-90)(135-100)}
\]
\[
= \sqrt{135 \times 55 \times 45 \times 35}
\]
Using calculations:
\[
= \sqrt{31185000} \approx 5587.5 \text{ m}^2
\]
### **Step 3: Find the Total Cost**
\[
\text{Cost} = \text{Area} \times \text{Rate per m}^2
\]
\[
= 5587.5 \times 20
\]
\[
= 1,11,750
\]
✅ **Final Answer: ₹1,11,750**
—
## **Solution 3: Finding the Height of a Triangle**
📌 **Question:**
A triangle has a **base of 14 cm** and an **area of 84 cm²**. Find its height.
### **Step 1: Use the Basic Area Formula**
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
\[
84 = \frac{1}{2} \times 14 \times h
\]
### **Step 2: Solve for Height (h)**
\[
h = \frac{84 \times 2}{14} = \frac{168}{14} = 12 \text{ cm}
\]
✅ **Final Answer: Height = 12 cm**
—
## **Solution 4: Finding the Area of a Right-Angled Triangle**
📌 **Question:**
Find the area of a **right-angled triangle** whose **hypotenuse is 13 cm** and one of the sides is **5 cm**, using Heron’s formula.
### **Step 1: Use Pythagoras Theorem to Find the Third Side**
\[
\text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2
\]
\[
13^2 = 5^2 + h^2
\]
\[
169 = 25 + h^2
\]
\[
h^2 = 169 – 25 = 144
\]
\[
h = 12 \text{ cm}
\]
✅ So, the sides of the triangle are **5 cm, 12 cm, and 13 cm**.
### **Step 2: Find the Semi-Perimeter (s)**
\[
s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \text{ cm}
\]
### **Step 3: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{15(15-5)(15-12)(15-13)}
\]
\[
= \sqrt{15 \times 10 \times 3 \times 2}
\]
\[
= \sqrt{900} = 30 \text{ cm}^2
\]
✅ **Final Answer: 30 cm²**
—
## **Solution 5: Quadrilateral Divided into Two Triangles**
📌 **Question:**
A **quadrilateral** is divided into two triangles with sides **6 cm, 8 cm, 10 cm** and **5 cm, 12 cm, 13 cm**. Find its total area using Heron’s formula.
### **Step 1: Find the Area of Triangle 1 (6 cm, 8 cm, 10 cm)**
**Semi-Perimeter:**
\[
s = \frac{6+8+10}{2} = \frac{24}{2} = 12 \text{ cm}
\]
\[
\text{Area} = \sqrt{12(12-6)(12-8)(12-10)}
\]
\[
= \sqrt{12 \times 6 \times 4 \times 2}
\]
\[
= \sqrt{576} = 24 \text{ cm}^2
\]
✅ **Triangle 1 Area = 24 cm²**
—
### **Step 2: Find the Area of Triangle 2 (5 cm, 12 cm, 13 cm)**
**Semi-Perimeter:**
\[
s = \frac{5+12+13}{2} = \frac{30}{2} = 15 \text{ cm}
\]
\[
\text{Area} = \sqrt{15(15-5)(15-12)(15-13)}
\]
\[
= \sqrt{15 \times 10 \times 3 \times 2}
\]
\[
= \sqrt{900} = 30 \text{ cm}^2
\]
✅ **Triangle 2 Area = 30 cm²**
—
### **Step 3: Total Area of the Quadrilateral**
\[
= 24 + 30 = 54 \text{ cm}^2
\]
✅ **Final Answer: 54 cm²**
—
## **📌 Summary of Important Exam Tips**
✔ **Always find the semi-perimeter first.**
✔ **Check for right-angled triangles using Pythagoras’ theorem.**
✔ **If given a quadrilateral, split it into two triangles.**
✔ **Always include the correct units (cm², m², etc.) for full marks.**
—
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