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Question

$$\angle BAC$$ of $$\triangle ABC$$ is obtuse and $$AB = AC$$. $$P$$ is a point on $$BC$$ such that $$PC = 12\ cm$$. $$PQ$$  and $$PR$$ are perpendiculars to sides $$AB$$ and $$AC$$ respectively. If $$PQ= 15\ cm$$ and $$PR = 9\ cm$$, then the length of $$PB\ (cm)$$ is


Solution

Given: $$AB = AC$$, $$PQ \perp AB$$ and $$PR \perp AC$$
Since, $$AB = AC$$
$$\angle ABC = \angle ACB$$...(I) (Isosceles triangle property)
Now, In $$\triangle PBQ$$ and $$\triangle PRC$$
$$\angle PBQ = \angle PCR$$ (From I)
$$\angle PQB = \angle PRC$$ (Each $$90^{\circ}$$)
$$\angle QPB = \angle RPC$$ (Third angle)
Thus, $$\triangle QPB \sim \triangle RPC$$ (AAA rule)
Hence, $$\frac{PQ}{PR} = \frac{PB}{PC}$$
$$\frac{15}{9} = \frac{PB}{12}$$
$$PB = \frac{15 \times 12}{9}$$
$$PB = 20$$ cm

Mathematics

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