# In an isosceles trapezoid, the angle at the base is 45 °, the sides are 9√2cm

**In an isosceles trapezoid, the angle at the base is 45 °, the sides are 9√2cm, and the diagonal is 15cm. Find the perimeter and area of the trapezoid.**

Let’s finish the height BH of the trapezoid ABCD. In a right-angled triangle ABH one of the angles is 45, then the triangle is isosceles and right-angled, then AH = AB * Cos45 = 9 * √2 * √2 / 2 = 9 cm. Then BH = AH = 9 cm.

In a right-angled triangle BDH, according to the Pythagorean theorem, we determine the length of the leg DH.

DH ^ 2 = BD ^ 2 – BH ^ 2 = 225 – 81 = 144.

DН = 12 cm.

Then AD = AH + DH = 9 + 12 = 21 cm.

Let’s draw the height of the CК. Since the triangles ABН and CKD are equal about the hypotenuse and the acute angle, then DK = AH = 9 cm.Then HK = AD – AH – DK = 21 – 9 – 9 = 3 cm.

Then BC = НK = 3 cm, since BCКН is a rectangle.

Determine the perimeter of the trapezoid. Ravsd = (AB + BC + CD + AD) = (9 * √2 + 3 + 9 * √2 + 21) = 18 * √2 + 24 cm.

Determine the area of the trapezoid. S = (ВС + АD) * ВН / 2 = (3 + 21) * 9/2 = 108 cm2.

Answer: The perimeter of the trapezoid is 18 * √2 + 24 cm, the area of the trapezoid is 108 cm2.