In a rectangular trapezoid ABCD, the lateral side AB is 10 cm, AD is the larger base

In a rectangular trapezoid ABCD, the lateral side AB is 10 cm, AD is the larger base, equal to 18 cm, the angle D is 45. Find the area of this trapezoid.

Since the smaller side of a rectangular trapezoid is perpendicular to its bases, it is equal to the height of this trapezoid. Let’s draw the height of the CH. CH divides the trapezoid ABCD into a right-angled triangle CHD and a rectangle ABCH. Consider the triangle CHD. By condition, the lateral side of the trapezoid AB is 10 cm, then AB = CH = 10 cm. The leg of the triangle CHD is known. Angle D is 45 degrees. Let’s find the unknown angle:
angle HCD = 180 degrees – angle D – angle СНD (according to the theorem on the sum of the angles of a triangle);
angle HCD = 180 – 45 – 90 = 45 (degrees).
In the CHD triangle, two angles (HCD and D) are equal to 45 degrees, therefore the CHD triangle is isosceles, its base is CD, and the sides are CH and HD, therefore HD = 10 cm (the sides of an isosceles triangle are equal).
AD = AH + HD;
18 = AH + 10;
AH = 18 – 10 = 8 (cm).
In rectangle ABCH, side AB is equal to side CH, side AH is equal to side BC. Therefore, BC = 8 cm.
Let’s find the area of ​​the trapezoid using the formula:
S = ((AD + BC) / 2) * AB;
S = ((18 + 8) / 2) * 10 = (26/2) * 10 = 13 * 10 = 130 (cm square).
Answer: S = 130 cm square.