How will we prove that :

In a $\large \color{blue}{\text{Parallelogram}}$,

$$\large \color{green}{d_1^2 + d_2^2 = 2(a^2 + b^2)}$$

where $d_1$ and $d_2$ are the Diagonals and $a$ and $b$ are the Adjacent Sides of the Parallelogram..

Question

How will we prove that :

In a $\large \color{blue}{\text{Parallelogram}}$,

$$\large \color{green}{d_1^2 + d_2^2 = 2(a^2 + b^2)}$$

where $d_1$ and $d_2$ are the Diagonals and $a$ and $b$ are the Adjacent Sides of the Parallelogram..

lukecrayonz · Answer

Are you smart and just questioning this to make people wonder, or do you not know? Just wondering for my answer

anonymous · Answer

It will be better if we work on question's solution.. Rather than checking my smartness here..

lukecrayonz · Answer

Just wondering because people do post those questions. You use parallelogram law.

lukecrayonz · Answer

Think of the parallelogram as a rhombus. Use properties of a rhombus to solve:)

lukecrayonz · Answer

anonymous · Answer

The angles are 90 in case of parallelogram too???

ganeshie8 · Answer

we allowed to use cosine law @waterineyes

anonymous · Answer

Wait..

Tell me about sine and cosine laws first..

I read once but now forgot all these stuff..

lukecrayonz · Answer

c^2 = a^2 + b^2 - 2ab cos(C)

ganeshie8 · Answer

^^

lukecrayonz · Answer

Ganeshie isn't it strange I know calculus but can barely do simple geometry? ;P

ganeshie8 · Answer

:) you have spent more time in juicy calculus side :))

ganeshie8 · Answer

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