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Standard X

Mathematics

Chord Properties of Circles

∠ BAC of AB...

Question

∠BAC of △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

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Solution

Given: AB=AC, PQ⊥AB and PR⊥AC
Since, AB=AC
∠ABC=∠ACB...(I) (Isosceles triangle property)
Now, In △PBQ and △PRC
∠PBQ=∠PCR (From I)
∠PQB=∠PRC (Each 90°)
∠QPB=∠RPC (Third angle)
Thus, △QPB∼△RPC (AAA rule)
Hence, PQ/PR=PB/PC
15/9=PB/12
PB=15×12/9
PB=20 cm

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Q. Hello,
Please help me with the following :
Angle BAC of triangle ABC is obtuse and AB=AC. P is a point in bc such that PC =12cm. PQ & PR are perpendicular to sides AB &AC respectively. If PQ=15cm and PR=9cm; find the length of PB.

regards,
Nikita

Angles BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively.

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and

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are points on sides

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and

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Q. In a

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and

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are points on sides

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∠BAC of △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

Given: AB=AC, PQ⊥AB and PR⊥ACSince, AB=AC∠ABC=∠ACB...(I) (Isosceles triangle property)Now, In △PBQ and △PRC∠PBQ=∠PCR (From I)∠PQB=∠PRC (Each 90°)∠QPB=∠RPC (Third angle)Thus, △QPB∼△RPC (AAA rule)Hence, PQ/PR=PB/PC15/9=PB/12PB=15×12/9PB=20 cm

Given: AB=AC, PQ⊥AB and PR⊥AC
Since, AB=AC
∠ABC=∠ACB...(I) (Isosceles triangle property)
Now, In △PBQ and △PRC
∠PBQ=∠PCR (From I)
∠PQB=∠PRC (Each 90°)
∠QPB=∠RPC (Third angle)
Thus, △QPB∼△RPC (AAA rule)
Hence, PQ/PR=PB/PC
15/9=PB/12
PB=15×12/9
PB=20 cm