In the figure, PQRS is a parallelogram, $P M ⊥ R S a n d R N ⊥ P S$ . If PQ = l2cm, $P M = 6 c m$ and $R N = 8 c m,$ then the length of PS is equal to :

18 c m

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9 c m

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4 c m

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12 c m

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Solution

The correct option is A $9 c m$
Area $∥ g m P Q R S = P Q \times P M = 12 \times 6 = 72 {c m}^{2}$
Again, area $∥ g m P Q R S = P S \times R N = P S \times 8$
$\Rightarrow P S \times 8 = 72 \Rightarrow P S = 9 c m$

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In a parallelogram $P Q R S,$ $P Q = 12 c m$ and $P S = 9 c m .$ The bisector of $∠ P$ meets $S R$ in $M .$ $P M$ and $Q R$ both when produced meet at $T .$ Find the length of $R T .$

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in which

∠ Q P S = ∠ S Q R = 90^{o}, P Q = 12 c m, P S = 9 c m, Q R = 8 c m

and

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P Q R S

is a parallelogram with altitude

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on base

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What is

P S

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In the figure, PQRS is a parallelogram, PM⊥RS and RN⊥PS. If PQ = l2cm, PM=6cm and RN=8cm, then the length of PS is equal to :

The correct option is A 9cmArea ∥gmPQRS=PQ×PM=12×6=72cm2Again, area ∥gmPQRS=PS×RN=PS×8⇒PS×8=72⇒PS=9cm

In the figure, PQRS is a parallelogram, $P M ⊥ R S a n d R N ⊥ P S$ . If PQ = l2cm, $P M = 6 c m$ and $R N = 8 c m,$ then the length of PS is equal to :

The correct option is A $9 c m$
Area $∥ g m P Q R S = P Q \times P M = 12 \times 6 = 72 {c m}^{2}$
Again, area $∥ g m P Q R S = P S \times R N = P S \times 8$
$\Rightarrow P S \times 8 = 72 \Rightarrow P S = 9 c m$