.ABCD is a rhombus. Diagonals AC and BD bisect each other at O such that AC = 24 cm and
BD = 32 cm. The perimeter of the rhombus is _________
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Secondary School
Math
5 points
If ABCD is a rhombus, prove that

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by Imzan25.03.2018
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Me · Beginner
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Hay!!
Dear user -
in given ABCD rhombus 4ab²
Than prove that ac²+bd²
we use this figure
so,
OA=1/2AC,OB=1/2BD
and △ AOB=90
in Right △AOB, using the above theorem
AB²=OA²+OB²
=> (1/2AC)²+(1/2BD)²
=> AC²+BD²
so,
Hence 4ab²=ac²+bd²
I hope it's help you


Secondary School
Math
5 points
If ABCD is a rhombus, prove that

Ask for details
Follow
Report
by Imzan25.03.2018
Answers
Me · Beginner
Know the answer? Add it here!
Hay!!
Dear user -
in given ABCD rhombus 4ab²
Than prove that ac²+bd²
we use this figure
so,
OA=1/2AC,OB=1/2BD
and △ AOB=90
in Right △AOB, using the above theorem
AB²=OA²+OB²
=> (1/2AC)²+(1/2BD)²
=> AC²+BD²
so,
Hence 4ab²=ac²+bd²
I hope it's help you

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Secondary School Math 6 points
ABCD is a rhombus in which diagonal AC bisects diagonal BD at ‘O’, if AC = 6cm , BD= 8cm , Find the measure of each side.
Ask for details Follow Report by Estheraash 14.09.2018
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Me · Beginner
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hvsai2 Ambitious
In the above figure
AC bisects BD at O
if AC =6cm
AO=3 cm
OC=3 cm
if BD=8cm
BO=4cm
OD=4cm
According to the Pythagorean theorem
Hypotenuse=√(base)2+(height)2
AB=√AO2+OB2
AB=√9+16
AB=√25
AB=5
We know that In rhombus all sides are equal
each side =5cm
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