Question

Prove that the sum of squares of the sides
of a rhombus is equal to the sum of the
squares of its diagonals.​

Answer 1

Answer:

see the proof above

Step-by-step explanation:

sum of squares of the diagonal s of a rhombus is equal to the sum of squares of its sides

Answer 2

Answer:

Step-by-step explanation:

Given:

A rhombus ABCD

Diagonals AC and BD are perpendicular bisectors

To Prove:

AB² + BC² + CD² + DA² = BD² + AC²

Proof:

In a rhombus we know that the diagonals are perpendicular bisectors.

Hence,

OD = OB = 1/2 BC------(1)

OC = OA = 1/2 AC------(2)

Also in a rhombus, all the 4 sides are equal.

Therefore,

AB = BC = CD = DA------(3)

Now consider ΔODC

By Pythagoras theorem,

DC² = OD² + OC²

Substitute value of OD and OC from equations 1 and 2,

DC² = (1/2 BD)² + (1/2 AC)²

DC² = 1/4 BD² + 1/4 AC²

Multiply the whole equation by 4

4DC² = BD² + AC²

DC² + DC² + DC² + DC² = BD² + AC²

Substitute equation 3 in the above equation,

AB² + BC² + CD² + DA² = BD² + AC²

Hence proved.

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf A}\put(-0.5,4.2){\bf D}\put(4.2,-0.5){\bf B}\put(4.2,4.2){\bf C}\qbezier(0,0)(0,0)(4,4)\qbezier(4,0)(4,0)(0,4)\put(1.8,1.5){\bf{O}}\end{picture}