Question

а) Боковая сторона, равная 17 , образует с одним из осно ний трапеции угол 135°. Найдите высоту трапеции.

Answer 1

Ответ:

17√2/2

Пошаговое объяснение:

Угол у основания 45°

Строим прямоугольный треугольник в котором один из углов 45°

По теореме Пифагора находим высоту:

h² + h² = 17²

Высота 17√2/2

Answer 2

I can answer your question in detail.

a) The side of the trapezoid, equal to 17, forms an angle of 135° with one of the bases of the trapezoid. Find the height of the trapezoid.

Solution:

Let ABCD be the trapezoid with AB and CD as the parallel bases, and AD = 17 as the side that forms an angle of 135° with CD. Let E be the point on AB such that DE is perpendicular to AB. Then, the height of the trapezoid is h = DE.

We can use the sine rule to find DE. Since angle ADC is 135°, angle ADE is 45°. Therefore,

sin(45°) = DE / AD

DE = AD * sin(45°)

DE = 17 * sin(45°)

DE = 17 * √2 / 2

DE ≈ 12.02

Hence, the height of the trapezoid is h ≈ 12.02.

You can also find some similar problems and solutions on the web . I hope this helps.