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Diagonals of Rhombus Abcd Intersect Each Other at Point O. Prove That: Oa2 + Oc2 = 2ad2 - "Bd"^2/2 - Mathematics

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प्रश्न

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`

बेरीज
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उत्तर


Diagonals of the rhombus are perpendicular to each other.

In quadrilateral ABCD, ∠AOD = ∠COD = 90°.
So, ΔAOD and ΔCOD are right-angled triangles.

In ΔAOD using Pythagoras theorem,
AD2 = OA2 + OD2
⇒ OA2 = AD2 - OD                ....(i)

In ΔCOD using Pythagoras theorem,
CD2 = OC2 + OD2
⇒ OC2 = CD2 - OD              ....(ii)

LHS = OA2 + OC2
= AD2 - OD2 + CD2 - OD2      ...[ From(i) and (ii) ]
= AD2 + CD - 2OD2

= AD2 + AD2 - 2`("BD"/2)^2` ...[ AD = CD and OD = `"BD"/2`]

= 2AD2 - `("BD")^2/2`

= RHS.

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पाठ 13: Pythagoras Theorem [Proof and Simple Applications with Converse] - Exercise 13 (B) [पृष्ठ १६४]

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सेलिना Concise Mathematics [English] Class 9 ICSE
पाठ 13 Pythagoras Theorem [Proof and Simple Applications with Converse]
Exercise 13 (B) | Q 10 | पृष्ठ १६४

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