Question 13 Marks

Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.

Given :
$\square PQRS$ is a rhombus. Diagonals PR and SQ intersect each other at point T.
To prove : $PS ^2+ SR ^2+ QR ^2+ PQ ^2= PR ^2+ QS ^2$
Activity :
Diagonals of a rhombus bisect each other.
In $\Delta PQS$, PT is the median and in $\Delta QRS$, RT is the median.
$\therefore$ By Apollonius theorem,
$PQ^{2} + PS^{2} = ⬜ + 2QT^{2}$ .......... (I)
$QR^{2} + SR^{2} = ⬜ + 2QT^{2}$ .......... (II)
adding (I) and (II),
$PQ^{2}+PS^{2}+QR^{2}+SR^{2} = 2(PT^{2}+ ⬜) + 4QT^{2}$
$= 2(PT^{2}+⬜) + 4QT^{2}$ .......... ($RT=PT$)
$= 4PT^{2} + 4QT^{2}$
= (⬜)² + (2QT)²
$\therefore$ $PQ^{2}+PS^{2}+QR^{2}+SR^{2} = PR^{2} + ⬜.$

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