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Question
The given diagram shows two isosceles triangles which are similar. In the given diagram, PQ and BC are not parallel; PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ.
Calculate:
- the length of AP,
- the ratio of the areas of triangle APQ and triangle ABC.
Solution
i. Given, ΔAQP ~ ΔACB
`=> (AQ)/(AC) = (AP)/(AB)`
`=> (3)/(4 + AP) = (AP)/(3 + 12)`
`=>` AP2 + 4AP – 45 = 0
`=>` (AP + 9)(AP – 5) = 0
`=>` AP = 5 units ...(As length cannot be negative)
ii. Since, ΔAQP ~ ΔACB
∴ `(ar(ΔAPQ))/(ar(ΔACB)) = (PQ^2)/(BC^2)`
`=> (ar(ΔAPQ))/(ar(ΔABC)) = (AP^2)/(BC^2)` ...(PQ = AP)
`=> (ar(ΔAPQ))/(ar(ΔABC)) = (5/15)^2 = 1/9`
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