NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.1
Question 11

Question.  11

11. If \(\triangle ABC \sim \triangle QRP\) and \(\dfrac{\operatorname{ar}(ABC)}{\operatorname{ar}(PQR)}=\dfrac{9}{4}\), with \(AB=18\,\text{cm}\) and \(BC=15\,\text{cm}\), then \(PR\) equals

(A)

10 cm

(B)

12 cm

(C)

\(\dfrac{20}{3}\) cm

(D)

8 cm

Detailed Answer with Explanation:

Step 1: We are told \(\triangle ABC \sim \triangle QRP\). That means the two triangles are similar. In similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

Step 2: The ratio of the areas is given as: \[ \dfrac{\operatorname{ar}(ABC)}{\operatorname{ar}(PQR)} = \dfrac{9}{4} \]

Step 3: Take square root to find the ratio of the sides (scale factor): \[ \dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{CA}{PQ} = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2} \]

Step 4: From the correspondence in the question, side \(BC\) of \(\triangle ABC\) matches with side \(PR\) of \(\triangle QRP\).

Step 5: Use the ratio of sides: \[ \dfrac{BC}{PR} = \dfrac{3}{2} \]

Step 6: Put \(BC = 15\,\text{cm}\). \[ \dfrac{15}{PR} = \dfrac{3}{2} \]

Step 7: Cross-multiply: \[ 15 \times 2 = 3 \times PR \] \[ 30 = 3 \times PR \]

Step 8: Divide both sides by 3: \[ PR = \dfrac{30}{3} = 10\,\text{cm} \]

Final Answer: \(PR = 10\,\text{cm}\). So the correct option is A.

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