In the adjoining figure, $P, M, Q, R$ are collinear and $P M = M Q = M S, S R^{2} = P R . Q R$ . Then

∠ S Q R = ∠ S M P

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∠ Q S R = ∠ M S P

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∠ Q S R = ∠ M S Q

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∠ Q S M = ∠ P S M

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Solution

The correct option is A $∠ S Q R = ∠ S M P$

$∴ P M = Q M, ∴ ∠ P S M = ∠ Q S M$

$S R^{2} = P R . Q R$

$\Rightarrow \frac{S R}{P R} = \frac{Q R}{S R} (g i v e n)$

$I n Δ S P R a n d Δ S Q R \Rightarrow \frac{S R}{P R} = \frac{Q R}{S R}$

$H e n c e, ∠ S Q R = ∠ S M P$

Hence, proved.

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Similar questions

In the adjoining figure P,M,Q,R are collinear and

PM = MQ = MS, SR² = PR . QR. Then

Q. In this figure QS and RS are bisectors of exterior angles Q and R. Then

∠ Q S R + ∠ P / 2

is equal to

Q. PQRS is a parallelogram whose diagonals intersect at M.

∠ P M S = 54^{\circ}, ∠ Q S R = 25^{\circ} and ∠ S Q R = 30^{\circ};

(i) ∠ R P S

(ii) ∠ P R S

(iii) ∠ P S R

P Q R S

is a parallelogram whose diagonals intersect at

M

.
If

∠ P M S = 54^{o}, ∠ Q S R = 25^{o}

and

∠ S Q R = 30^{o}

; find:
i)

∠ R P S

ii)

∠ P R S

iii)

∠ P S R

$PQRS is a parallelogram whose diagonals intersect at M. i f ∠ P M S = 54^{\circ}, ∠ Q S R = 25^{\circ} a n d ∠ S Q R = 30^{\circ}; f i n d : (i) ∠ R P S (i i) ∠ P R S (i i i) ∠ P S R .$

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In the adjoining figure, P,M,Q,R are collinear and PM=MQ=MS,SR2=PR.QR. Then

The correct option is A ∠SQR=∠SMP∴PM=QM,∴∠PSM=∠QSMSR2=PR.QR⇒SRPR=QRSR(given)InΔSPRandΔSQR⇒SRPR=QRSRHence,∠SQR=∠SMPHence, proved.

In the adjoining figure, $P, M, Q, R$ are collinear and $P M = M Q = M S, S R^{2} = P R . Q R$ . Then