In a $Δ P R S, ∠ P R S = 120^{o}$ . A point $Q$ is taken of $P R$ such that $P Q = Q S$ and $Q R = R S$ then $∠ Q P S = . . . .$

15^{o}

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30^{o}

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45^{o}

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12^{o}

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Solution

The correct option is A $15^{o}$
$P Q = Q S$
$\Rightarrow ∠ Q P S = ∠ Q S P = x (s a y)$
$\Rightarrow ∠ S Q R = 2 x$ (exterior angle)
$= ∠ R S Q$
From $Δ Q R S$
$120 + 2 x + 2 x = 180$
$\Rightarrow x = 15^{o} = ∠ Q P S$

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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}

In Fig. PQRS is a quadrilateral in which diagonals PR and QS intersect in O. [4 MARKS]

Show that:
(i) PQ + QR + RS + SP $>$ PR + QS
(ii) PQ + QR + RS + SP $<$ 2 (PR + QS)

In Fig. T is a point on side QR of $Δ P Q R$ and S is a point such that RT = ST. Which of the following is true?

In the given figure, T is a point on side QR of $Δ P Q R$ and S is a point such that RT = ST.
Which of the following is true?

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In a ΔPRS,∠PRS=120o. A point Q is taken of PR such that PQ=QS and QR=RS then ∠QPS=....

The correct option is A 15oPQ=QS⇒∠QPS=∠QSP=x(say)⇒∠SQR=2x (exterior angle)=∠RSQFrom ΔQRS120+2x+2x=180⇒x=15o=∠QPS

In a $Δ P R S, ∠ P R S = 120^{o}$ . A point $Q$ is taken of $P R$ such that $P Q = Q S$ and $Q R = R S$ then $∠ Q P S = . . . .$

The correct option is A $15^{o}$
$P Q = Q S$
$\Rightarrow ∠ Q P S = ∠ Q S P = x (s a y)$
$\Rightarrow ∠ S Q R = 2 x$ (exterior angle)
$= ∠ R S Q$
From $Δ Q R S$
$120 + 2 x + 2 x = 180$
$\Rightarrow x = 15^{o} = ∠ Q P S$