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Tag: Class9th CBSE MATHS

  • Past CBSE exam questions on Heron’s Formula

    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….

  • Let’s proceed with more advanced problems and diagrams to Euclid’s Geometry.

    Advanced Example 1: Proof Using Parallel Lines

    Question

    Prove that the sum of the interior angles on the same side of a transversal is less than 180∘180^\circ if the lines are not parallel.

    Solution

    1. Given: Two lines l1l_1 and l2l_2, and a transversal tt, such that l1l_1 and l2l_2 are not parallel.
    2. To Prove: ∠1+∠2<180∘\angle 1 + \angle 2 < 180^\circ, where ∠1\angle 1 and ∠2\angle 2 are interior angles on the same side of the transversal.
    3. Construction:
      • Extend l1l_1 and l2l_2 indefinitely to meet at point PP.
    4. Proof:
      • Since l1l_1 and l2l_2 are not parallel, they meet at point PP.
      • In △PQT\triangle PQT, the sum of angles of the triangle is 180∘180^\circ: ∠1+∠2+∠P=180∘.\angle 1 + \angle 2 + \angle P = 180^\circ.
      • Here, ∠P\angle P is a positive angle, so: ∠1+∠2<180∘.\angle 1 + \angle 2 < 180^\circ.
    5. Conclusion:
      The sum of interior angles on the same side of a transversal is less than 180∘180^\circ when the lines are not parallel.
    6. Diagram: Let’s plot this graphically.

    Graph Explanation

    • The blue line (l1l_1) and the green line (l2l_2) represent non-parallel lines.
    • The red dashed line is the transversal (tt) intersecting l1l_1 and l2l_2.
    • The purple point is the intersection of l1l_1 and tt, while the orange point is the intersection of l2l_2 and tt.

    From the graph:

    • The interior angles formed on the same side of the transversal tt (between l1l_1 and l2l_2) are less than 180∘180^\circ, confirming the proof.

    Would you like to explore proofs involving parallel lines or other examples?