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​

Answer 1

Given:

Rhombus ABCD

AC and BD are Diagonals intersecting at O.

Figure is in the attachment provided below.

To Prove that:

Sum of squares of the sides of a rhombus = Sum of squares of its diagonal.

$AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}$

Proof:

AB = BC = CD = AD (All sides of a rhombus are equal)

Diagonals of a rhombus bisect each other at 90 degree or right angles.

∴ ∠ AOB = 90 Degree

AO = 1 / 2 AC

BD = 1 / 2 BO

Consider ΔAOB,

∠ AOB = 90 Degree

Applying Pythagoras theorem in this triangle we get,

$(\frac{1}{2} AC} )^{2} +(\frac{1}{2}BD)^{2} = AB^{2}$

$\frac{AC^{2}}{4} + \frac{BD^{2} }{4} = AB^{2}$

$\frac{AC^{2} + BD^{2} }{4} = AB^{2}$

$AC^{2} + BD^{2} = 4AB^{2}$

$AB^{2} + AB^{2} + AB^{2} + AB^{2} = AC^{2} + BD^{2}$

This means that,

$AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}$  (∵All sides are equal)

Hence Proved

Answer 2

Answer:

Yea, what the person said

Step-by-step explanation:

that's right