In the figure, AC = 10cm, PC = 15cm, PQ = 12cm, find PB.
A
6cm
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B
7cm
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C
8cm
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D
9cm
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Solution
The correct option is D
9cm
In △ABCand△PQC ∠PQC=∠ABC=90∘ ∠C=∠C (common angle) Therefore, △ABC∼△PQC by AA similarity ACPC=ABPQ⇒1015=AB12⇒AB=8cm In △ABC , AB2+BC2=AC2⇒BC=6cm PB = PC – BC = 15 - 6 = 9 cm