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

Question 1

Step 1: We are told that \(\angle A = \angle C\). Also, the angle at point \(P\) is common (vertically opposite). So, by AA similarity criterion, triangles \(\triangle APB\) and \(\triangle CPD\) are similar.

Step 2: For similar triangles, the ratios of their corresponding sides are equal. That means: \[ \dfrac{AB}{CD} = \dfrac{AP}{CP} = \dfrac{BP}{PD} \]

Step 3: Substitute the known values: - \(AB = 6\,\text{cm}\) - \(AP = 12\,\text{cm}\) - \(CP = 4\,\text{cm}\)

So, \[ \dfrac{AB}{CD} = \dfrac{AP}{CP} = \dfrac{12}{4} = 3 \]

Step 4: From this ratio, \[ \dfrac{AB}{CD} = 3 \quad \Rightarrow \quad CD = \dfrac{AB}{3} = \dfrac{6}{3} = 2\,\text{cm} \]

Step 5: Now use the ratio with \(BP\) and \(PD\): \[ \dfrac{BP}{PD} = 3 \]

So, \[ PD = \dfrac{BP}{3} = \dfrac{15}{3} = 5\,\text{cm} \]

Final Answer: \(PD = 5\,\text{cm}, \; CD = 2\,\text{cm}.\)