The midline of an isosceles trapezoid is 8, the angle at one of the bases is 135 degrees

The midline of an isosceles trapezoid is 8, the angle at one of the bases is 135 degrees, and the lateral side is 5. Find the area of the trapezoid.

The area of ​​the trapezoid is equal to the product of the half-sum of the bases and the height.

S = (a + b) / 2 ∙ h.

The midline of a trapezoid is a line segment connecting the midpoints of the sides of the trapezoid. It is parallel to the bases, and its length is equal to half the sum of the bases. Accordingly, the area of ​​the trapezoid is the product of the midline and the height:

S = m ∙ h, where:

S is the area of ​​the trapezoid;

m is the middle line of the trapezoid;

h – height.

To do this, you need to find the height BH. Consider the triangle ΔАВН. This triangle is rectangular.

Let’s apply the sine theorem. The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:

sin A = BH / AB;

BH = AB ∙ sin A.

Since the sum of the angles of the trapezoid adjacent to one side is 180 °, then:

∠А = 180º – ∠В;

∠А = 180º – 135º = 45º.

sin 45º = √2 / 2 ≈ 0.7071;

BH = 5 · 0.7071 ≈ 3.5 cm;

S = 8 ∙ 3.5 = 28 cm2.

Answer: the area of ​​the trapezoid is 28 cm2.