Math, asked by rsv2829123, 10 months ago

solve the question please.... ​

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Answered by sanketj
1

In ∆KPQ and ∆KLM;

angle KQP = angle KML ... (each 45°, given)

angle PKQ = angle LKM ... (common angle)

∆KPQ ~ ∆KLM ... (by AA test of similarity)

 \frac{KQ}{KM} = \frac{PQ}{LM}

(corresponding sides of similar triangles are proportional)

 \frac{c}{b + c}  =  \frac{x}{a}  \\  \\ x =  \frac{ac}{b + c}

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