Q) Prove that the sum of the squares of the sides of the rhombus equal to the sum of its squares of its diagonal .
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see the diagram.
In a Rhombus, the diagonals are perpendicular. The sides are all equal.
So the triangles AOB, AOD, BOC, OCD are all right angle triangles. The diagonals bisect each other at O.
AB² = AO² + OB²
BC² = BO² + OC²
CD² = CO² + OD²
AD² = AO² + OD²
Add all equations above:
AB² + BC² + CD² + DA²
= 2 AO² + 2 BO² + 2 CO² + 2 DO²
= 4 AO² + 4 BO² ∵ AO = CO AND BO = DO
= AC² + BD² AS AC = 2 AO AND BD = 2 BO
In a Rhombus, the diagonals are perpendicular. The sides are all equal.
So the triangles AOB, AOD, BOC, OCD are all right angle triangles. The diagonals bisect each other at O.
AB² = AO² + OB²
BC² = BO² + OC²
CD² = CO² + OD²
AD² = AO² + OD²
Add all equations above:
AB² + BC² + CD² + DA²
= 2 AO² + 2 BO² + 2 CO² + 2 DO²
= 4 AO² + 4 BO² ∵ AO = CO AND BO = DO
= AC² + BD² AS AC = 2 AO AND BD = 2 BO
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