answer this ...who is perfect in maths ..let me these those perfectionist...
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Heya guys.....
that is very simpu yrrrr. I mean simple....
Given: area of parallelogram ABCD= 90cm²
1. ar(ABEF)= ar(ABCD)= 90cm² [parallelogram ABCD and ABEF on the same base AB and between same parallel FC and AB.]
2. ar(ABD)= 1/2ar(ABEF) [ parallelogram ABEF and triangle ABD on the same base AB and between the same parallel AB and EF ]
then , ar(ABD)= 90/2= 45cm²
3. ar( BEF)= 1/2 ar(ABEF). [ Diagonal bisect the parallelogram]
so , ar(BEF)= 90/2= 45cm²
hopes it will help u
otherwise u can ask from me
I will sure help u
thanks
that is very simpu yrrrr. I mean simple....
Given: area of parallelogram ABCD= 90cm²
1. ar(ABEF)= ar(ABCD)= 90cm² [parallelogram ABCD and ABEF on the same base AB and between same parallel FC and AB.]
2. ar(ABD)= 1/2ar(ABEF) [ parallelogram ABEF and triangle ABD on the same base AB and between the same parallel AB and EF ]
then , ar(ABD)= 90/2= 45cm²
3. ar( BEF)= 1/2 ar(ABEF). [ Diagonal bisect the parallelogram]
so , ar(BEF)= 90/2= 45cm²
hopes it will help u
otherwise u can ask from me
I will sure help u
thanks
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