What is the sum of the squares of the medians
(in sq. cm) of a triangle ABC if the sum of the squares
of its sides is 72 sq. cm?
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Answered by
2
Answer:
Correct option is
A
diagonals
The diagonals of a rhombus bisect each other at 90
∘
so triangle AOB,AOD,BOC and COD are right angled triangle.
In △AOB
AO
2
+OB
2
=AB
2
(using Pythagoras theorem)
⇒(
2
AC
)
2
+(
2
BD
)
2
=AB
2
⇒
4
AC
2
+
4
BD
2
=AB
2
⇒AC
2
+BD
2
=4AB
2
Similarly AC
2
+BD
2
=4BC
2
AC
2
+BD
2
=4CD
2
BD
2
+AC
2
=4AD
2
Adding all these we get
⇒4(AB
2
+BC
2
+CD
2
+AD
2
)=4(AC
2
+BD
2
)
Hence the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonals .
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