In an isosceles trapezoid, the perimeter of which is 200, and the area is 1500, you can inscribe a circle.

| July 23, 2021 | education

In an isosceles trapezoid, the perimeter of which is 200, and the area is 1500, you can inscribe a circle. Find the distance from the point of intersection of the diagonals of the trapezoid to its smaller base.

Since a circle is inscribed in the trapezoid, the sum of its lateral sides is equal to the sum of the lengths of its bases.

AB + CD = BC + AD, and since the perimeter of the trapezoid is 200 cm, then AB + CD = BC + AD = 200/2 = 100 cm.

Since AB = CD, AB = CD = 100/2 = 50 cm.

The area of ​​the trapezoid, by condition, is equal to 1500 cm2, then (ВС + АD) * ВН / 2 = 1500.

100 * BH = 3000.

BH = 3000/100 = 30 cm.

From the right-angled triangle ABH, by the Pythagorean theorem, we determine the length of the segment AH.

AH ^ 2 = AB ^ 2 – BH ^ 2 = 50 ^ 2 – 30 ^ 2 = 2500 – 900 = 1600.

AH = 40 cm.

In an isosceles trapezoid, AH = DE = 40 cm.

Then AD = AH + HE + ED = 40 + HE + 40 = 80 + HE.

Since BC = HE, then AD + BC = 100 = 80 + 2 * BC.

BC = 20/2 = 10 cm, then AD = 40 + 10 + 40 = 90 cm.

The BOS triangle is similar to the AOD triangle in two angles, and its similarity coefficient is:

K = AD / BC = 90/10 = 9.

Then MO / KO = 9.

MO = 9 * KO.

KO + MO = BH = 30 cm.

Then KO + 9 * KO = 30.

10 * KO = 30.

KO = 30/10 = 3 cm.

Answer: The distance from the point of intersection of the diagonals to the smaller base is 3 cm.



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