In trapezoid ABCD AB = CD, AC = 5cm, CH = 3cm. Find the area of the trapezoid.

We calculate the distance from A to H: triangle ACH is rectangular (since CH is height), according to the Pythagorean theorem:

AH = √ (AC² – CH²) = √ (5² – 3²) = √ (25 – 9) = √16 = 4 (cm).

Let’s draw the second height of the trapezoid BE.

The segment AH consists of two segments: AE + EH.

In an isosceles trapezoid, the segments AE and DH are equal, denote each by x.

Then AH = 4;

AE + EH = 4;

x + EH = 4;

hence EH = 4 – x.

The EBCH quadrangle is a rectangle, which means that the base of the trapezoid is BC = EH = 4 – x.

Base AD = AH + DH = 4 + x.

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 = (4 – x + 4 + x) / 2 * 3 = 8/2 * 3 = 12 (cm²).

Answer: the area of ​​the trapezoid is 12 cm².