The height drawn from the apex of an obtuse angle of a right-angled trapezoid divides the trapezoid into a square

The height drawn from the apex of an obtuse angle of a right-angled trapezoid divides the trapezoid into a square about a triangle, the area of the triangle is 16 cm.Find the area of the trapezoid if its angle is 45.

1. A, B, C, D – the tops of the trapezoid. ∠А = ∠В = 90 °. ∠D = 45 °. S – area. CE – height.

2.∠DCE = 180 ° – 90 ° – 45 ° = 45 °. The angles ∠DCE and ∠D are equal.

Therefore, ΔDCE is isosceles. Hence, CE = DE.

3. S ΔDCE = CE x DE / 2 = 16 cm². We replace DE with CE:

CE x CE / 2 = 16.

CE² = 2 x 16.

CE = √2 x 16 = 4√2 cm.

4. CE is the side of the square ABCE. Therefore, AB = BC = CE = AE = 4√2 cm.

5. AD = AE + DE = 4√2 + 4√2 = 8√2 cm.

6. S trapezoid = (BC + AD) / 2 x CE = (4√2 + 8√2) / 2 x 4√2 = 24√2 cm².