As shown above, a parallelogram ABCD is partitioned by two lines AF and BE, such that the areas of the red triangle ABG = 27 and the blue triangle EFG=12.

What is the area of the yellow region?

Guest Jan 27, 2021

#1**+1 **

The red and blue triangles are similar, so since their areas are in the ratio 27/12 = 9/4, their side lengths will be in the ratio 3/2.

So, let AB = 3k, then EF will equal 2k, (and notice in passing that DE + FC = 3k - 2k = k).

Run a perpendicular from AB to CD, passing through G and meeting the two sides at P and Q.

Let PG = 3h, then GQ = 2h.

Area of the red triangle = 3k*3h/2 = 27, so hk = 6.

(Similarly, area of the blue triangle = 2k*2h/2 = 12, so hk = 6).

Area of the trapezium ABCF = 5h*(AB + CF)/2 = 27 +Y1.

Area of the trapezium ABDE = 5h*(AB + DE)/2 = 27 + Y2.

Adding, 5h*(2AB + CF + DE)/2 = 54 + Y1 + Y2, (DE + CF = k from earlier.)

so, 54 + Y1 + Y2 = 5h*(6k + k)/2 = 35hk/2 = 105,

so Y1 + Y2 = 51.

Tiggsy Jan 28, 2021

#1**+1 **

Best Answer

The red and blue triangles are similar, so since their areas are in the ratio 27/12 = 9/4, their side lengths will be in the ratio 3/2.

So, let AB = 3k, then EF will equal 2k, (and notice in passing that DE + FC = 3k - 2k = k).

Run a perpendicular from AB to CD, passing through G and meeting the two sides at P and Q.

Let PG = 3h, then GQ = 2h.

Area of the red triangle = 3k*3h/2 = 27, so hk = 6.

(Similarly, area of the blue triangle = 2k*2h/2 = 12, so hk = 6).

Area of the trapezium ABCF = 5h*(AB + CF)/2 = 27 +Y1.

Area of the trapezium ABDE = 5h*(AB + DE)/2 = 27 + Y2.

Adding, 5h*(2AB + CF + DE)/2 = 54 + Y1 + Y2, (DE + CF = k from earlier.)

so, 54 + Y1 + Y2 = 5h*(6k + k)/2 = 35hk/2 = 105,

so Y1 + Y2 = 51.

Tiggsy Jan 28, 2021