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

The correct option is B 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 X

Mathematics

Pythagoras Theorem

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 $△ A B C$ , by using Pythagoras theorem
$A B^{2} + B C^{2} = A C^{2} \Rightarrow B C = 6 c m$
Now, PB = PC –BC = 15-6 = 9cm

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If AC = 10cm, PC=15cm and AB=6cm. Find PB?

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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 , by using Pythagoras theorem AB2+BC2=AC2⇒BC=6cm Now, PB = PC –BC = 15-6 = 9cm