NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.4
Question 2

Question. 2

Given \(\triangle ABC \sim \triangle EDF\) with \(AB=5\,\text{cm}\), \(AC=7\,\text{cm}\), \(DE=12\,\text{cm}\), \(DF=15\,\text{cm}\). Find the remaining sides.

Answer:

\(BC=\dfrac{25}{4}\,\text{cm}=6.25\,\text{cm},\; EF=\dfrac{84}{5}\,\text{cm}=16.8\,\text{cm}.\)

Detailed Answer with Explanation:

Step 1: The triangles are similar: \(\triangle ABC \sim \triangle EDF\).

This means their corresponding sides are in the same ratio. The order of letters tells us the matching parts:

  • \(AB \leftrightarrow DE\)
  • \(AC \leftrightarrow EF\)
  • \(BC \leftrightarrow DF\)

Step 2: Write the proportion for the sides:

\[ \dfrac{AB}{DE} = \dfrac{AC}{EF} = \dfrac{BC}{DF} \]

Step 3: First, find \(EF\).

\[ \dfrac{AB}{DE} = \dfrac{AC}{EF} \]

Substitute the values: \( \dfrac{5}{12} = \dfrac{7}{EF} \).

Cross multiply: \( 5 \times EF = 12 \times 7 = 84 \).

So, \( EF = \dfrac{84}{5} = 16.8\,\text{cm} \).

Step 4: Now, find \(BC\).

\[ \dfrac{AB}{DE} = \dfrac{BC}{DF} \]

Substitute the values: \( \dfrac{5}{12} = \dfrac{BC}{15} \).

Cross multiply: \( 5 \times 15 = 12 \times BC \).

So, \( 75 = 12 \times BC \).

\( BC = \dfrac{75}{12} = \dfrac{25}{4} = 6.25\,\text{cm} \).

Final Answer:

\(EF = 16.8\,\text{cm},\; BC = 6.25\,\text{cm}.\)

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 6: Triangles – Exercise 6.4 | Detailed Answers