In the figure- AC -10 cm - PC -15 cm - PQ -12 cm- find PB-A- 6 cmB- 7 cmC- 8 cmD- 9 cm

The correct option is D 9cm In ABC and PQC ∠ PQC = ∠ ABC =90^∘ ∠ C = ∠ C ex (common angle) Therefore, ABC ∼ PQC by AA similarity AC PC = AB PQex ⇒10 15 = AB 1 ...

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Standard IX

Mathematics

SAS Similarity

In the figure...

Question

In the figure, AC = 10cm, PC = 15cm, PQ = 12cm, find PB.

6cm

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7cm

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8cm

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9cm

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Solution

The correct option is D
9cm

In $△ A B C a n d △ P Q C$
$∠ P Q C = ∠ A B C = 90^{\circ}$
$∠ C = ∠ C$ (common angle)
Therefore, $△ A B C \sim △ P Q C$ by AA similarity
$\frac{A C}{P C} = \frac{A B}{P Q} \Rightarrow \frac{10}{15} = \frac{A B}{12} \Rightarrow A B = 8 c m$
In $△ A B C$ ,
$A B^{2} + B C^{2} = A C^{2} \Rightarrow B C = 6 c m$
PB = PC – BC = 15 - 6 = 9 cm

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In the figure, AC = 10cm, PC = 15cm, PQ = 12cm, find PB.

The correct option is D 9cm In △ABC and △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