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PQRS is a rhombus. Prove that PR^2 plus QS^2 is equal to 4PQ^2

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Answer:

HEY mate here is your answer.

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Step-by-step explanation:

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Let PQRS be a rhombus WHOSE diagonals PR and QS INTERSECT at O.

It is the PROPERTY of a rhombus that diagonals perpendicularly bisect each other.

∴ In ΔPOQ,

⇒PQ^2 = OP^2 + OQ^2,

⇒(PR/2)^2+(QS/2)^2=PQ^2,

⇒PR^2/4+QS^2/4=PQ^2,

⇒(PR^2+QS^2)/4=PQ^2,

⇒PR² + QS² = 4PQ².

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Hence Proved.✔✔✔✔✔

Hope this helps you.


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