Question

What is the ratio of the area of parallelogram ABCD to the sum of areas of $\Delta ABEand\Delta ABF?$

• A. 1:2
• B. 1:1
• C. 2:1
• D. 1:3

Answer 1

The correct option is B

1:1

EG, DH and FI are perpendiculars drawn from E, D and F to GI respectively.

We can clearly see that

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

So, the height of the parallelogram ABCD is exactly equal to the height of $\Delta ABEand\Delta ABF$.

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

Area of parallelogram $ABCD=AB\times DH$

Area of $\Delta ABE=\frac{1}{2}\times AB\times EG$

Area of $\Delta ABF=\frac{1}{2}\times AB\times FI$

Area of $\Delta ABE+Areaof\Delta ABF$
$=(\frac{1}{2}\times AB\times EG)+(\frac{1}{2}\times AB\times FI)$

$=2\times \frac{1}{2}\times AB\times DH$
[Using equation 1]

$=AB\times DH$

Area of $\Delta ABE+Areaof\Delta ABF$ = Area of parallelogram ABCD

Hence, ratio of the area of parallelogram ABCD to the sum of areas of $\Delta ABEand\Delta ABF$ is 1:1.