Huanggang Wuxue county primary school graduation exam, mathematics final exam additional question how to do the second question, a quadrilateral, two of which are right angles, one right angle has a side length of 2, (I drew it casually, I don't know if the original title is like this)

Because the quadrilateral has two right angles and a 45-degree angle, it is a right-angled trapezoid.

Passing through point C at point E is perpendicular to BD, so the quadrilateral ABEC is a rectangle and CDE is a triangle.

Because the length of the right angle is 2.

So it is discussed in two situations.

(1) When AB is equal to 2.

CE=2, because ∠ D = 45 and ∠ CED = 90.

So ∠ DCE = 45, triangle CED is an isosceles right triangle, and DE=CE=2.

Therefore, BE=BD-DE=5-2=3, AC=BE=3.

So the area of the quadrilateral is1/2 (AC+BD) × ce =1/2 × (3+5 )× 2 = 8.

② When AC is equal to 2.

BE=2, so DE = BD-BE = 5-2 = 3.

So AB=CE=DE=3.

So the area of the quadrilateral is1/2 (AC-BD) × ce =1/2 × (2+5) × 3 = 21/2.

As mentioned above, the area of the quadrilateral is 8 or 2 1/2.