NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.1
Question 9

Question.  9

9. Given \(\triangle ABC \sim \triangle DFE\) with \(\angle A=30^\circ\), \(\angle C=50^\circ\), \(AB=5\,\text{cm}\), \(AC=8\,\text{cm}\) and \(DF=7.5\,\text{cm}\). Which is true?

(A)

\(DE=12\,\text{cm},\; \angle F=50^\circ\)

(B)

\(DE=12\,\text{cm},\; \angle F=100^\circ\)

(C)

\(EF=12\,\text{cm},\; \angle D=100^\circ\)

(D)

\(EF=12\,\text{cm},\; \angle D=30^\circ\)

Detailed Answer with Explanation:

Step 1: In similar triangles, the order of letters shows the matching corners.
So, \(A \leftrightarrow D\), \(B \leftrightarrow F\), and \(C \leftrightarrow E\).

Step 2: We are given \(\angle A = 30^\circ\) and \(\angle C = 50^\circ\).
The sum of angles in a triangle is \(180^\circ\).
So, \(\angle B = 180^\circ - (30^\circ + 50^\circ) = 100^\circ\).

Step 3: Since \(B \leftrightarrow F\), we know \(\angle F = 100^\circ\).
Also, \(A \leftrightarrow D\), so \(\angle D = 30^\circ\).

Step 4: Find the scale factor between the triangles.
\(AB = 5\,\text{cm}\) and \(DF = 7.5\,\text{cm}\).
So, scale factor \(k = \dfrac{DF}{AB} = \dfrac{7.5}{5} = 1.5\).

Step 5: To find \(DE\), use the matching side of \(AC\).
\(AC = 8\,\text{cm}\). Multiply by the scale factor:
\(DE = AC \times k = 8 \times 1.5 = 12\,\text{cm}\).

Final Answer: \(DE = 12\,\text{cm}\) and \(\angle F = 100^\circ\).
So, the correct option is B.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 6: Triangles – Exercise 6.1 | Detailed Answers