In an isosceles trapezoid ABCD with bases BC = 4, AD = 10, the height BH was drawn. Find the length of AH.

Let’s draw the second height of the CK.

Quadrangle ВСКH is a rectangle, since the heights of ВН and CK are perpendicular to the bases of the trapezoid. Then КН = ВС = 4 cm.

Since the trapezoid is isosceles, the right-angled triangles ABH and CDK are equal in hypotenuse and acute angle, then AH = DE.

AD = 10 = AH + KH + DK = 2 * AH + 4.

2 * AH = 10 – 4 = 6 cm.

AH = 6/2 = 3 cm.

Answer: The length of AH is 3 cm.

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