PQ -12 cm- find PB-A- 7 cmB- 6 cmC- 9 cmD- 8 cm

The correct option is C 9 cmIn Δ ABC and Δ PQC, ∠ PQC = ∠ ABC = 90^∘ ∠ C = ∠ C (common angle) Therefore, Δ ABC ∼Δ PQC by AA similarity AC/PC=AB/PQ=BC/QC⇒1 ...

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Standard VII

Mathematics

Equal Angles Subtend Equal Sides

PQ =12 cm, fi...

Question

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

6 cm

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7 cm

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8 cm

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9 cm

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Solution

The correct option is D 9 cm
$I n Δ A B C a n d Δ P Q C, ∠ P Q C = ∠ A B C = 90^{\circ} ∠ C = ∠ C (common angle) T h e r e f o r e, Δ A B C \sim Δ P Q C by AA similarity \frac{A C}{P C} = \frac{A B}{P Q} = \frac{B C}{Q C} \Rightarrow \frac{10}{15} = \frac{A B}{12} = \frac{B C}{Q C} \Rightarrow A B = 8 c m I n Δ A B C, A B^{2} + B C^{2} = A C^{2} \Rightarrow 10^{2} = 8^{2} + B C^{2} \Rightarrow B C = 6 c m P B = P C - B C = 15 - 6 = 9 c m H e n c e, P B = 9 c m .$

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

The correct option is D 9 cmIn ΔABC and ΔPQC,∠PQC=∠ABC=90∘∠C=∠C (common angle)Therefore, ΔABC∼ΔPQC by AA similarityACPC=ABPQ=BCQC⇒1015=AB12=BCQC⇒AB=8cmIn ΔABC,AB2+BC2=AC2⇒102=82+BC2⇒BC=6cmPB=PC−BC=15−6=9cmHence, PB=9cm.