Question

Prove that area of the rhombus is half the product of the diagonals.
​

Answer 1

Answer:

Given ABCD is a rhombus the diagonal AC and BD cut at point O

Then ∠AOD=∠AOB=∠COD=∠BOC=90  

0

 

The area of rhombus ABCD divided diagonal in four parts  

So area of rhombus ABCD =area of triangle AOD+area of triangle AOB+area of triangle BOC+area of triangle COD

=  

2

1

​  

×AO×OD+  

2

1

​  

×AO×OB+  

2

1

​  

×BO×OC+  

2

1

​  

×OD×OC

=  

2

1

​  

×AO(OD+OB)+  

2

1

​  

OC(BO+OD)

=  

2

1

​  

×AO×BD+  

2

1

​  

×OC×BD

=  

2

1

​  

BD(AO+OC)=  

2

1

​  

×BD×AC

So area of rhombus is equal to half of the product of diagonals

Step-by-step explanation:

please mark this answer as a brainliest.

Answer 2

Answer:Given ABCD is a rhombus the diagonal AC and BD cut at point O

Then ∠AOD=∠AOB=∠COD=∠BOC=90  

0

 

The area of rhombus ABCD divided diagonal in four parts  

So area of rhombus ABCD =area of triangle AOD+area of triangle AOB+area of triangle BOC+area of triangle COD

=  

2

1

​  

×AO×OD+  

2

1

​  

×AO×OB+  

2

1

​  

×BO×OC+  

2

1

​  

×OD×OC

=  

2

1

​  

×AO(OD+OB)+  

2

1

​  

OC(BO+OD)

=  

2

1

​  

×AO×BD+  

2

1

​  

×OC×BD

=  

2

1

​  

BD(AO+OC)=  

2

1

​  

×BD×AC

So area of rhombus is equal to half of the product of diagonals

solution

Step-by-step explanation:I hope you understand it and i dont have enough time so pls like this