Please do it fast and explain step by step and give a reason too for each step. Take your 30 points and do it
Answers
Since AB and CD are parallel, triangles AED and BEC have same altitude drawn between AB and CD.
Let the altitude be drawn from A to CD at M. So we have,
Area of ∆AED = AM · DE / 2 = 45 cm² → (1)
And,
Area of ∆BEC = AM · EC / 2 = 26 cm² → (2)
Because the altitude drawn from B to EC equals AM.
Well, we can find the area of ∆AEB by subtracting the sum of areas of triangles AED and BEC from that of the parallelogram ABCD, i.e.,
Area of ∆AEB,
= (AM · CD) - [(AM · DE / 2) + (AM · EC / 2)]
= (AM · CD) - AM [(DE + EC) / 2]
= AM [CD - (CD / 2)]
= AM · CD / 2 → (3)
But from (1),
DE = 2 × 45 / AM
DE = 90 / AM → (4)
And from (2),
EC = 2 × 26 / AM
EC = 52 / AM → (5)
Adding (4) and (5),
DE + EC = (90 / AM) + (52 / AM)
CD = 142 / AM → (6)
Substituting (6) in (3),
Area of ∆AEB,
= AM · CD / 2
= AM (142 / AM) / 2
= (142 AM) / (2 AM)
= 71 cm²
Hence the answer is 71 cm².
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Short method...
Sum of areas of triangles AED and BEC,
= (AM · DE / 2) + (AM · EC / 2)
= AM (DE + EC) / 2
= AM · CD / 2
But since ABCD is a parallelogram, AB = CD.
So area of ∆AEB = AM · AB / 2 = AM · CD / 2
That is,
Area of ∆AEB = Area of ∆AED + Area of ∆BEC
Hence the answer is 45 + 26 = 71 cm².
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