Math, asked by anamikaprasad12, 10 months ago

.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

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Answered by mahi778dhruv
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Answer:

THIS IS YOUR BRANLIEST AND CORRECT ANSWER .

80cm

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Answered by bimla7037
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Answer:

Brainly.in

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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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