. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answers
Step-by-step explanation:
□ABCD is a rhombus with O as point of intersection of diagonals.
In ΔAOB,
∠AOB=90
0
(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 biset 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
Hence Proved.
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Step-by-step explanation:
ABCD is a rhombus with O as point of intersection of diagonals.
In ΔAOB,
∠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
Hence Proved.