find the area of BCED
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the answer should be 45 units
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given
DE//BC ------
AD//BC and AE//BC
then. BCAD and BCAE are parallelograms.
BC =5 u.
Distance from BC and DE = 6 u.
Now ar(ABCD) = base * height
= BC* height
= 5*6
= 30u^2
ar(AECD) = 5*6
= 30u^2
ar(BCDE) = ar(ABCD)+ar(AECD)
= 30+30
= 60u ^2
hope it helps you
DE//BC ------
AD//BC and AE//BC
then. BCAD and BCAE are parallelograms.
BC =5 u.
Distance from BC and DE = 6 u.
Now ar(ABCD) = base * height
= BC* height
= 5*6
= 30u^2
ar(AECD) = 5*6
= 30u^2
ar(BCDE) = ar(ABCD)+ar(AECD)
= 30+30
= 60u ^2
hope it helps you
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