Question

prove that the sum of square on the sides of a rhombus is equal to the sum so square of the diagonal

Answer 1

hlo frnd ....



Consider a rhombus ABCD, whose diagonals intersect at O.



We know that diagonals of a rhombus bisect each other at 90°, so triangles AOB, AOD, BOC and COD are right angled triangles.

In ΔAOB,


AO2 + OB2 = AB2 (Using Pythagoras theorem)



⇒ AC2 + BD2 = 4AB2

Similarly, AC2 + BD2 = 4BC2

AC2 + BD2 = 4CD2

AC2 + BD2 = 4AD2

Adding all these equations, we get

4(AB2 + BC2 + CD2 + AD2) = 4(AC2 + BD2)

⇒ AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Answer 2

we know,
diagonals of ROHMBUS , perpendicular to each other . and also intersecting point bisects the diagonals .

e.g in figure ABCD is rohmbus,
so, AC and BD are perpendicular .
then,
OA = OC = AC/2 and OB = OD = BD/2

Let AC and BD intersect at O .
now, for ∆AOB
angle AOB = 90°
so, use Pythagoras theorem,
AB² = AO² + OB² ___________(1)

similarly ,
BC² = OB² + OC²________(2)
CD² = OC² + OD²_________(3)
AD² = OA² + OD²________(4)
add all eqns ,
AB² + BC² + CD² + AD² = 2(OA² + OC²) + 2( OB² + OD²)
= 4OA² ( =OC²) + 4OB²(=OD²)
= (2OA)² + (2OB)²
= AC² + BD²

hence,
AB² + BC² + CD² + AD² = AC² + BD²
proved //