ABCD rectangular trapezoid BC = 2 cm AD = 10 cm AB refers to CD as 3: 5 find the perimeter.

Let AB = 3x cm and CD = 5x.
We draw the height of the CК, hence BC = AK = 2cm. Therefore KD = AD-AK = 10-2 = 8 (cm).
AB = CK = 3x (cm). In the triangle CDK, by the Pythagorean theorem, we write the expression for the hypotenuse CD ^ 2 = CK ^ 2 + KD ^ 2, we have
5x ^ 2 = 3x ^ 2 + 8 ^ 2;
25x ^ 2 = 9x ^ 2 + 64;
25x ^ 2-9x ^ 2 = 64;
16x ^ 2 = 64;
x ^ 2 = 64: 16;
x ^ 2 = 4;
x1 = 2;
x2 = -2 – this root is not taken into account, since the length cannot be negative.
We have: AB = 3x = 3 * 2 = 6 (cm);
CD = 5x = 5 * 2 = 10 (cm);
P = AB + BC + CD + AD = 6 + 2 + 10 + 10 = 28 (cm).