In the figure- AQ -1 cm - PC -15 cm - PQ -12 cm and AB -6 cm- find CB-A- 6 cmB- 7 cmC -8 cmD- 9 cm

The correct option is C 8 cm In Δ PQC, By Pythagoras Theorem PQ^2 + QC^2 = PC^2 ⇒ 15^2 12^2=QC^2 ⇒ QC=9 cm QC+AQ=AC ⇒ AC= 10cm Given (1) In Δ ABC, By Pytha ...

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Standard VIII

Mathematics

Pythagorean Triplets

In the figure...

Question

In the figure, AQ = 1 cm, PC = 15cm, PQ = 12 cm and AB = 6 cm, find CB.

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 C
8 cm

$I n Δ P Q C, By Pythagoras Theorem P Q^{2} + Q C^{2} = P C^{2} \Rightarrow 15^{2} - 12^{2} = Q C^{2} \Rightarrow Q C = 9 c m Q C + A Q = A C \Rightarrow A C = 10 c m Given - (1) I n Δ A B C, By Pythagoras Theorem A B^{2} + B C^{2} = A C^{2} \Rightarrow 10^{2} - 6^{2} = B C^{2} B y (1) \Rightarrow B C = 8 c m$

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In the figure, AQ = 1 cm, PC = 15cm, PQ = 12 cm and AB = 6 cm, find CB.

The correct option is C 8 cm In ΔPQC,By Pythagoras TheoremPQ2+QC2=PC2⇒152−122=QC2⇒QC=9cmQC+AQ=AC⇒AC=10cm Given−(1)In ΔABC,By Pythagoras TheoremAB2+BC2=AC2⇒102−62=BC2 By(1)⇒BC=8cm