# Prove that the sum of squares of a sides of a rhombus is equal to the sum of the squares of its diagonals......pls give answer.....u'll surely get a ''thumbs up''

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

In Î”AOB,

AO2 + OB2 = AB2 (Using Pythagoras theorem) â‡’ AC2 + BD2 = 4AB2

Similarly, AC2 + BD2 = 4BC2

AC2 + BD2 = 4CD2

Adding all these equations, we get

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

Hence  AB2 + BC2 + CD2 + AD2 = AC2 + BD2

• -2

hello,

here is the solution. remember my thumbs up.....

Given :- A rhombus ABCD with diagonals AC and BD intersecting at point O.

To proove : AB+ BC+ CD+ AD= AC+ BD2

Proof :- Since we know that diagonals of rhombus intersect each other at 900 .

Therefore, ang. AOB = ang. BOC = ang. COD = ang. AOD = 900

By Pythagores Theorem,

in triangles AOB, BOC, COD and AOD, we will get :-

AB= AO+ BO---------(1)

BC= BO+ CO--------(2)

CD= CO+ DO-------(3)

AB+ BC+ CD+ AD= 2 (AO+ BO+ CO2 +DO2)

= 2 ( AC/ 2 + BD2 /2) (DIAGONALS BISECT EACH OTHER. AO = CO = AC / 2 BO = DO = BD / 2 )

AC2 + BD2

HENCE PROOVED. PLZ. THUMBS UP.

PLZ. SOLVE MY MATHS QUES.

Posted by ANURAG55(student), on 29/9/11

• 24 hello,

here is the solution. remember my thumbs up.....

Given :- A rhombus ABCD with diagonals AC and BD intersecting at point O.

To proove : AB+ BC+ CD+ AD= AC+ BD2

Proof :- Since we know that diagonals of rhombus intersect each other at 900 .

Therefore, ang. AOB = ang. BOC = ang. COD = ang. AOD = 900

By Pythagores Theorem,

in triangles AOB, BOC, COD and AOD, we will get :-

AB= AO+ BO---------(1)

BC= BO+ CO--------(2)

CD= CO+ DO-------(3)

AB+ BC+ CD+ AD= 2 (AO+ BO+ CO2 +DO2)

= 2 ( AC/ 2 + BD2 /2) (DIAGONALS BISECT EACH OTHER. AO = CO = AC / 2 BO = DO = BD / 2 )

AC2 + BD2

HENCE PROOVED. PLZ. THUMBS UP.

PLZ. SOLVE MY MATHS QUES.

Posted by ANURAG55(student), on 29/9/11
• 75
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