In trapezoid ABCD, diagonal DB is the bisector of angle D. The bisector of angle C intersects the larger base AD at point K. Find the height of the trapezoid if BD = 24, CK = 18.
By the property of the bisectors of the trapezoid, the bisector of the angle of the trapezoid cuts off the isosceles triangle from the trapezoid, then in the triangle CKD CD = KD, and in the triangle BD, BC = CD.
The bisectors of the angles drawn from the corners of the lateral side intersect at right angles, all the corners of the point O are straight.
In the triangle CKD, the height DO is the bisector and the median, therefore, KO = CO = CK / 2 = 18/2 = 9 cm.
Similarly, in the triangle BCD, BO = DO = BD / 2 = 24/2 = 12 cm.
Then, in a right-angled triangle COD, according to the Pythagorean theorem, the hypotenuse CD will be equal to:
CD ^ 2 = CO ^ 2 + DO ^ 2 = 81 + 144 = 225.
СD = КD = 15 cm.
Let us determine the area of the triangle КСD by Heron’s theorem.
The semi-perimeter of the triangle is: p = (CK + KD + CD) / 2 = (18 + 15 + 15) / 2 = 24 cm.
Sxd = √24 * (24 – 18) * (24 – 15) * (24 – 15) = √11664 = 108 cm2.
The area of the triangle KCD can also be determined through the height of the CH.
Sksd = CH * KD / 2.
108 = CH * 15/2.
CH = 108 * 2/15 = 14.4 cm.
Answer: The height of the trapezoid is 14.4 cm.