The perimeter of an isosceles trapezoid is 28 in and the ratio of lengths of its bases is 5:3. Find the lengths of the sides of the trapezoid if its diagonal bisects the angle at the longer base.
Accepted Solution
A:
Consider the isosceles trapezoid with base lengths actually 5 and 3. The longer base overlaps the shorter one by 1 unit on either end. If the height of the trapezoid is h, the angle a diagonal makes with the longer base is .. α = arctan(h/4) The angle the slant end of the trapezoid makes with the longer base is .. β = arctan(h/1)
The angles are related by .. tan(β) = tan(2α) = 2×tan(α)/(1 -tan(α)^2) .. h = 2*(h/4)/(1 -(h/4)^2) .. 1 -(h/4)^2 = 1/2 . . . . . . . . . . . multiply by the denominator and divide by h .. 16 -h^2 = 8 . . . . . . . . . . . . . . multiply by 16 .. h = √8
The length of the slant side (s) will be .. s = √(1^2 +h^2) = √(1 +*) = 3 so the perimeter of the trapezoid will be .. 5 + 3 + 2*3 = 14
The perimeter of the trapezoid of interest is twice that, so its side lengths will be twice these.
The lengths of the sides of the trapezoid with perimert 28 are 6 and 6. The base lengths are 10 and 6.