Question

the given figure, AP = BQ, PR = QS. Show that ΔAPS = ΔBQ R ​

Answer 1

Answer:

In ∆ APS & ∆ BQR

<APQ=<BQR=90°

AP=BQ(given)

PS=QR( since PR=QS,

PR+RS=QS+RS(when equals are added to equals the wholes are equal)

PS=QR(because RS is a common region for both triangles.)

Therefore,

∆APS=∆BQR(S.A.S criterion)

Step-by-step explanation:

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