8. Given \(\triangle ABC \sim \triangle PQR\) and \(\dfrac{BC}{QR}=\dfrac{1}{3}\), then \(\dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)}\) equals
9
3
\(\dfrac{1}{3}\)
\(\dfrac{1}{9}\)
Step 1: We are told that \(\triangle ABC \sim \triangle PQR\). This means both triangles are similar, so their corresponding sides are in the same ratio and their areas are related by the square of that ratio.
Step 2: From the question, \(\dfrac{BC}{QR} = \dfrac{1}{3}\). This means side \(BC\) of triangle ABC is one-third the length of side \(QR\) of triangle PQR.
Step 3: The rule is: \[ \dfrac{\text{Area of one triangle}}{\text{Area of other triangle}} = \left(\dfrac{\text{corresponding side of one}}{\text{corresponding side of other}}\right)^2 \]
Step 4: Apply the rule: \[ \dfrac{\operatorname{ar}(BCA)}{\operatorname{ar}(PRQ)} = \left(\dfrac{BC}{QR}\right)^2 \]
Step 5: Substitute the values: \[ \dfrac{\operatorname{ar}(BCA)}{\operatorname{ar}(PRQ)} = \left(\dfrac{1}{3}\right)^2 = \dfrac{1}{9} \]
Step 6: Take reciprocal (since question asks for \(\dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)}\)): \[ \dfrac{\operatorname{ar}(PRQ)}{\operatorname{ar}(BCA)} = 9 \]
Final Answer: Option A (9)