The diagonal of an isosceles trapezoid divides its acute angle in half.

| May 4, 2021 | education

The diagonal of an isosceles trapezoid divides its acute angle in half. The perimeter of the trapezoid is 15m, the larger base is 6. Find the middle line of the trapezoid.

By condition, the diagonal AC is the bisector of angle A, hence the angle BAC = CAD.

Angle CAD = BCA as criss-crossing angles at the intersection of parallel straight lines AD and BC secant AC, then angle BAC = BCD, and triangle ABC is isosceles and AB = BC. Since the trapezoid is, by condition, isosceles, then CD = AB = BC.

Determine the perimeter of the trapezoid.

Let the length BC = X m, then AB = BC = CD = X m.

Then the perimeter of the trapezoid is:

P = AB + BC + SD + AD = 3 * X + 6 = 15 m.

3 * X = 15 – 6 = 9 m.

X = 9/3 = 3 m.

BC = 3 m.

Determine the length of the midline of the trapezoid.

KM = (AD + BC) / 2 = (6 + 3) / 2 = 9/2 = 4.5 m.

Answer: The length of the middle line is 4.5 m.



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