In a rectangular trapezoid ABCD with a base of 17 cm and 25 cm

In a rectangular trapezoid ABCD with a base of 17 cm and 25 cm, the diagonal AC is the bisector of the acute angle. Find the smaller side of the trapezoid.

1. Let us denote the angle by the symbol ∠.

2. ∠BAC = ∠CAD, since the bisector AC divides ∠A into two equal parts.

3. ∠АСВ = ∠САD as internal crosswise lying at parallel sides ВС and АD and diagonals АС crossing them.

4. ∠ВАС = ∠АСВ. Therefore, triangle ABC is isosceles. AB = BC = 17 cm.

5. From the top B we draw the height BH.

6. AH = AD – BC = 25 – 17 = 8 cm.

7. BH = √AB ^ 2 -AH ^ 2 = √17 ^ 2 – 8 ^ 2 = √289 – 64 = √225 = 15 cm.

8.CD = BH = 15 cm.

Answer: CD – the smaller side of the trapezoid is 15 cm.