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

Answer:

Step-by-step explanation:

given that √3 = 1.732,

√48 = ? , We know that 48 = 16*3, and we also know that √ab = √a * √b,

√48 = √16*√3,

=> 4√3 = 4*1.732,

=> 6.928,

Therefore √48 = 6.728,

Hope you understand , Have a great day !!

Answer 2

Given :

• A rhombus ABCD with diagonals AC & BD intersecting at O.

To Prove :

• Sum of square of all sides = Sum of the squares of its diagonals = $\rm AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2$

Proof :

Since all sides of a Rhombus are equal AB = BC = CD = AD

We know that,

• Diagonals of rhombus bisects each other at right angle.

Therefore, $\angle$ AOB = $\angle$ BOC = $\angle$ COD = $\angle$ DOA = $\rm 90°$

Also, AO = CO = $\dfrac{1}{2}$ AC $\rm \: . \: . \: . \: . \: . \: (1)$

BO = DO = $\dfrac{1}{2}$ BC $\rm \: . \: . \: . \: . \: . \: (2)$

Now, AOB is a right angled triangle, so by using Pythagoras theorem,

$\bullet \: \: \sf \: (Hypotenuse)^2 \: + \: (Height)^2 \: + \: (Base)^2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\longrightarrow \sf \: (AB)^2 = (OA)^2 + (OB)^2 \\ \\ \\ \longrightarrow \sf \: (AB)^2 = \bigg(\dfrac{1}{2} \: AC \bigg)^2 + \bigg(\dfrac{1}{2} \: BD \bigg)^2 \\ \\ \\ \longrightarrow \sf \: (AB)^2 = \dfrac{AC^2}{4} \: + \: \dfrac{BD^2}{4} \\ \\ \\ \longrightarrow \sf \: (AB)^2 = \: \dfrac{AC^2 + BD^2}{4} \\ \\ \\ \longrightarrow \sf \: 4AB^2 = AC^2 + BD^2 \\ \\ \\ \longrightarrow \sf \: AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\bf \underline{Hence \: , Proved!!}$