Question

In a kite shaped figure $ABCD$, $AB=AD$ and $CB=CD$. Points $P,Q$ and $R$ are mid-points of sides $AB,BC$ and $CD$ respectively. Then, find $\angle PQR$

• A. ${45}^{\circ }$
• B. ${30}^{\circ }$
• C. ${90}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${90}^{\circ }$

Given: ABCD is a kite shaped figure. $AB=AD$, $BC=CD$, P, Q< R are mid points of AB, BC and CD respectively

To Prove: $\angle PQR={90}^{\circ }$

Construction: Join AC and BD
In $△ABC$

P is mid point of AB and Q is mid point of BC

Then, $PQ\parallel AC$ (By Mid point theorem)
In $△BCD$

Q is mid point of BC and R is mid point of CD

Then, $QR\parallel BD$ (mid point theorem)
We know, diagonals of the Kite shaped figure, bisect at right angles.
thus, $AC\perp BD$

hence, $PQ\perp QR$ (Angle made between two lines is same as the angle between their corresponding parallel sides)
or $\angle PQR={90}^{\circ }$