The bases of an isosceles trapezoid are 1 and 6. Find the perimeter of the trapezoid if the cosine of its acute angle is 5/7.

ABCD – trapezoid, AB = CD, AD = 6, BC = 1, cosA = cosC = 5/7.
1. From the top B draw the heights BH to the base AD. Point H divides the base AD into two segments, one of which is equal to the half-sum of the bases (DH), and the second is the half-difference of the bases (AH).
Consider the triangle AHB: AB – hypotenuse, AH = (AD – BC) / 2 = (6 – 1) / 2 = 5/2 = 2.5.
2. The cosine of an angle in a right-angled triangle is the ratio of the adjacent leg to the hypotenuse, then:
cosA = AH / AB.
Let’s substitute the known data:
2.5 / AB = 5/7;
AB = 2.5 * 7/5 (in proportion);
AB = 17.5 / 5;
AB = 3.5.
Since AB = CD, then CD = 3.5.
3. The perimeter of a trapezoid is equal to the sum of the lengths of all its sides:
P = AB + BC + CD + AD;
P = 3.5 + 1 + 3.5 + 6 = 14.
Answer: P = 14.