What is the ratio of the area of parallelogram ABCD to the sum of areas of $Δ A B E$ and $Δ A B F ?$

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Solution

We can clearly see that

EG = DH = FI --------- (1)

So, the height of the parallelogram ABCD is exactly equal to the heights of $Δ A B E$ and $Δ A B F$ .

Also, the base of all the figures are same i.e. AB.

Area of parallelogram $A B C D = A B \times D H$

Area of $Δ A B E = \frac{1}{2} \times A B \times E G$

Area of $Δ A B F = \frac{1}{2} \times A B \times F I$

Area of $Δ A B E + A r e a o f Δ A B F = [\frac{1}{2} \times A B \times E G)] + [\frac{1}{2} \times A B \times F I]$

$= A B \times D H$ [Using equation 1]

Area of $Δ A B E$ + Area of $Δ A B F$ = Area of parallelogram ABCD

Hence, ratio = 1:1

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What is the ratio of the area of parallelogram ABCD to the sum of areas of ΔABE and ΔABF?

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