pushpitkc wrote:

Attachment:

2017-09-06_1052.png

In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1

The triangle ABC is similar to ADE(AA similarity)

It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\)

(sides opposite equal angles are equal)

Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\)

Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\)

Therefore, are of shaded region is \(\frac{1}{2} - \frac{2}{9} = \frac{(9-4)}{18} = \frac{5}{18}\)

(Option B)(E) 5/6

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2017-09-06_1052.png

pushpitkc , (and the rest of the posters, all of whom used triangles' similarity or assumed right angles at some point), I don't think we know we have two right angles.

We are not told that the two vertical lines are parallel.

If the two vertical lines were parallel, then the transversal which forms the base of both triangles, and which would cut the parallel lines, would create two congruent right angles.

We know from the diagram's mark that one angle is a right angle. That angle and the unmarked angle at the smaller triangle's base would be corresponding angles created by parallel lines cut by a transversal, and would be congruent.

In that case, the small and large triangles would share a right angle and a vertex angle, and by AA property of triangles, the triangles would be similar.

We are not told that the shorter vertical line is perpendicular to the horizontal line. Same result as above, i.e., if short vertical line were perpendicular to base, triangles would be similar.

Nor does the figure display a right angle mark where the shorter vertical line meets the base to create a small triangle with one side of length \(\frac{2}{3}\).

It seems all we know is that the two triangles share one angle, at the vertex on the left. The short vertical line might not be vertical. I'm not sure how we can conclude that the triangles are similar.

Am I missing something?

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123