PS is perpendicular to RS- thenP Q2 equals-A- 2-QR-SRB- 2-QR-SQC- 2-SQ-SR D- 2 - SR

The correct option is A 2.QR.SR PQ^2=PS^2+SR^2 ( ∵, ∠ PSQ = 90^∘) =(PQ^2 SQ^2)+(SQ+QR)^2 =QR^2 SQ^2+SQ^2+QR^2+2SQ.QR (∵ ...

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Byju's Answer

Standard X

Mathematics

Pythagoras Theorem

PS is perpend...

Question

In the figure, if PQ=QR and PS is perpendicular to RS, then $P Q^{2}$ equals:

2.QR.SQ

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2.QR.SR

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2.SQ.SR

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2.SR

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Solution

The correct option is B
2.QR.SR

$P Q^{2} = P S^{2} + S R^{2} (∵, ∠ P S Q = 90^{\circ}) = (P Q^{2} - S Q^{2}) + (S Q + Q R)^{2} = Q R^{2} - S Q^{2} + S Q^{2} + Q R^{2} + 2 S Q . Q R (∵ P Q = Q R) = 2 Q R^{2} + 2. S Q . Q R = 2 Q R (Q R + S Q) = 2. Q R . S R$

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Similar questions

Q. Question 12
If S is a point on the side PQ of a

Δ

PQR such that PS = QS = RS, then

(A) (P R) (Q R) = R S^{2}

(B) Q S^{2} + R S^{2} = Q R^{2}

(C) P R^{2} + Q R^{2} = P Q^{2}

(D) P S^{2} + R S^{2} = P R^{2}

Find the area of hexagon PQRSTU (in $m^{2}$ ) in which QN is perpendicular to PS, RL is perpendicular to PS, TM is perpendicular to PS, UO is perpendicular to PS such that PO= 4 m, ON=2 m, NM= 6 m, ML=2 m, LS= 6 m, UO= 6 m, TM= 8 m, QN= 2 m and RL=4 m.

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Find the area of hexagon PQRSTU in which QN is perpendicular to PS, RL is perpendicular to PS, TM is perpendicular to PS, UO is perpendicular to PS such that PO= 4m, ON=2m, NM= 6m, ML=2m, LS= 6m, UO= 6m, TM= 8m, QN= 2m, RL=4m.