1. Prove that the sum of the squares of the sides of a rhombus
is equal to the sum of the squares of its diagonals.
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Answer:
□ABCD is a rhombus with O as point of intersection of diagonals.
In Δ AOB,
∠AOB=90°
(since diagonals are perpendicular in rhombus).
By Pythagoras theorem,
AB^2=AO ^2+OB ^2
Similarly,
BC^2 =OC^2+OB^2 ,DC^2=OD ^2 +OC ^2
DA^2=DO^2+OA ^2
AB ^2 +BC^2 +CD ^2 +DA ^2=2(OA ^2+OB^2 +OC^2+OD^2 =4(AO^2 +DO^2)
Rhombus diagonal bisect each other,
AO=OC,DO=OB
AC=AO+OC
AC^2 =OA^2+OC^2+2AO.OC=4AO ^2
Similarly,
DB^2 =4OD ^2
∴AC^2+DB^2 =4(AO^2+DO^2 )
AB^2+BC^2+CD^2+DA^2 =AC^2 +DB^2
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