How can we tell that these are similar triangles and therefore the height of the smaller triangle is y?

Let bigger triangle be ABC right angled at B and smaller Triangle be ADE right angled at D

In Triangle ABC, AB = BC therefore, angle BAC = angle ACB = 45

Similarly, In triangle ADE, angle DAE = 45 and angle ADE = 90

therefore, angle AED = 45, (angle sum property)

Now all corresponding angles of the two triangles are equal (90,45,45)

Hence, triangle ABC similar to triangle ADE

Therefore, AB/AD = BC/DE

```
x/AD = x/y
```

therefore, AD = y

Now, DB = AB-AD = x-y

In trapezium DECB

DE = y, BC = x, and DB = x-y

Area (trap. DECB) = 0.5(DE+BC)DB

= (1/2)(x+y)(x-y)

= (1/2)(x^2 - y^2)

= Option A

Nice solution! Also, you can just do area of big triangle(ABC) - area of small triangle(ADE) once you find AD if you don’t wanna go the trapezium route!

No, need for the similar triangle to solve it. In the bigger triangle let’s say ABC right angle at B as two side are `x`

the angle opposite to those equal side will also be equal thus, 90 +x +x = 180 or x = 45.

Let’s move to small triangle ADE right angle at D. Now, we know angle A is 45 ; given angle D is 90 thus angle E will be 45 as 45 + 90 + angle E = 180. So, in small triangle ADE side DE = AD(side opposite to equal angle) and as DE is `y`

AD will also be `y`

. Now, you’ve height and base of both small and big triangle! Subtract area of big triangle from small triangle to get the shaded area!

You are right, it reduces time and complexity. Good observation.