NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.1Question 4
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.1
Question. 4
4. If in triangles \(ABC\) and \(PQR\),
\(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\), then
\(\triangle PQR \sim \triangle CAB\)
\(\triangle PQR \sim \triangle ABC\)
\(\triangle CBA \sim \triangle PQR\)
\(\triangle BCA \sim \triangle PQR\)
Detailed Answer with Explanation:
Step 1: In the question, we are given that the sides of the two triangles are in the same ratio:
\(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\).
Step 2: For similarity of triangles (by the SSS similarity criterion), the three sides of one triangle must be proportional to the three sides of another triangle.
Step 3: Compare the sides:
- \(AB\) in triangle ABC corresponds to \(QR\) in triangle PQR.
- \(BC\) in triangle ABC corresponds to \(PR\) in triangle PQR.
- \(CA\) in triangle ABC corresponds to \(PQ\) in triangle PQR.
Step 4: Now, write down the order of the vertices according to this correspondence:
\(A \rightarrow Q,\; B \rightarrow R,\; C \rightarrow P\).
Step 5: So the triangle \(BCA\) in the first triangle matches with \(PQR\) in the second triangle.
Final Answer: \(\triangle BCA \sim \triangle PQR\).