What is the area of a rectangular trapezoid if the smaller base and the smaller lateral side are 5 dm

What is the area of a rectangular trapezoid if the smaller base and the smaller lateral side are 5 dm and one of the corners is 45 °.

1. Let’s designate AB – the smaller base, CD – the larger base of the trapezoid;

2. The ACD angle is straight. angle BDC = 45 °;

3. Let us drop the perpendicular from point B to the base CD and denote the point of intersection E;

4. In the resulting triangle BED, the angle BED is 90 °, and the angle DBE is 45 °, since the sum of the angles in the triangle is 180 ° (180 – 45 = 45);

5. The height of the trapezoid BE is 5 in because it is parallel to the AC side and lies between two parallel bases. For the same reason CE = AB = 5 dm;

Since the triangle BED is isosceles, since the two angles in it are equal to 45 °, then the side ED = BE = 5 inches.

Base CD = CE + ED = 5 + 5 = 10 dm.

The area of ​​a trapezoid is defined as the product of the sum of the bases and the height, divided in half.

(AB + CD) * BE / 2 = (5 + 10) * 5/2 = 37.5 sq. dm;

Answer: The area of ​​the trapezoid is 37.5 square meters. dm