The bisectors of the angles of a parallelogram enclose a:

Rhombus

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Rectangle

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Square

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Kite

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Solution

The correct option is B Rectangle
As $x + y = 180^{o} \Rightarrow \frac{x}{2} + \frac{y}{2} = 90^{o}$
$\Rightarrow ∠ D P A = 90^{o} = ∠ S P Q$ (vertically opp)
$∠ S R Q = 90^{o} = ∠ P S R = ∠ P Q R$
Also In $△ A S B^{'} y^{'} &^{'} B^{'}$
are distinction $∴$ As is d if from BS
hence PQRS is A rectangle.

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Similar questions

Q. The bisectors of the angle of a parallelogram enclose a ?

The bisectors of the angles of a parallelogram enclose a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Q. If the bisectors of angles of a quadrilateral enclose a rectangle, then show that it is a parallelogram.

Q. The bisectors of interior angles of a parallelogram forms a ___________.

$In the following diagram, the bisectors of interior angles of the parallelogram PQRS en-close a quadrilateral ABCD.$

$Show that: (i) ∠ P S B + ∠ S P B = 90^{\circ} (i i) ∠ P B S = 90^{\circ} (i i i) ∠ A B C = 90^{\circ} (i v) ∠ A D C = 90^{\circ} (v) ∠ A = 90^{\circ} (v i) ABCD is a rectangle Thus, the bisectors of the angles of a parallelogram enclose a rectangle.$

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The bisectors of the angles of a parallelogram enclose a:

The correct option is B RectangleAs x+y=180o⇒x2+y2=90o⇒∠DPA=90o=∠SPQ (vertically opp)∠SRQ=90o=∠PSR=∠PQRAlso In △ASB′y′&′B′are distinction ∴ As is d if from BS hence PQRS is A rectangle.