Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. [3 Marks]

Answer 1

We know that, in a rhombus, diagonals bisect each other at 90 degrees.
Diagonals of a rhombus bisect each other. [1 Mark]
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ∴ by Apollonius’ theorem,
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ${PQ}^2$ + ${PS}^2$ = 2${PT}^2$ + 2${QT}^2$....(1)
${QR}^2$ + ${SR}^2$ = 2${RT}^2$ + 2${QT}^2$....(2) [1 Mark]

Adding (1) and (2), we get
${PQ}^2$ + ${PS}^2$ + ${QR}^2$ + ${SR}^2$ = 2${PT}^2$ + 2${QT}^2$ + 2${RT}^2$ + 2${QT}^2$
= 2(${PT}^2$ + ${PT}^2$) + 4${\left(QT\right)}^2$...............(RT = PT)
= 4${\left(PT\right)}^2$ + 4${\left(QT\right)}^2$
= ${\left(2PT\right)}^2$ + $(2QT{)}^2$
= ${PR}^2$ + ${QS}^2$ [1 Mark]