# The base of an isosceles trapezoid is 3 and 8, and the angle at the base is 60 degrees. Find the diagonal.

**The base of an isosceles trapezoid is 3 and 8, and the angle at the base is 60 degrees. Find the diagonal. In a rhombus ABСD AB = 13 BD = 24 Find the height of the rhombus.**

1). Let an isosceles trapezoid ABCD be given, in which the height of the VC, the base BC = 3 and AD = 8, and the angle at the base of the BAD is 60 degrees, then:

(8 – 3): 2 = 2.5 – the distance from the top A to the point K (the base of the height VK) by the property of the isosceles trapezoid AK = (AD – BC) / 2;

90 ° – 60 ° = 30 ° – the value of ∠АВК, since ∠ВАК + ∠АВК = 90 ° in a right-angled triangle AVK;

2.5 ∙ 2 = 5 – hypotenuse AB, since AB = 2 ∙ AK by the property of the leg, which lies opposite the angle of 30 °;

√ (5² + 8² – 2 ∙ 5 ∙ 8 ∙ ½) = 7 – the diagonal ABCD, since in ΔABD, according to the cosine theorem, BD² = AB² + AK² – 2 ∙ AB ∙ AK ∙ cos60 °.

Answer: The diagonal of the trapezoid is 7.

2). Let a rhombus ABCD be given, in which side AB = 13, diagonal BD = 24, point O – the point of intersection of diagonals AC and BD, then:

∠АВ = 90 ° – by the property of the rhombus diagonals;

BO = BD: 2 = 24: 2 = 12 – by the property of the rhombus diagonals;

AO = √ (13² – 12²) = 5, since in the right-angled triangle ABO, according to the Pythagorean theorem, the equality AB² = AO² + BO² is fulfilled;

5 ∙ 2 = 10 – AC diagonal;

(10 ∙ 24) / 2 = 120 – the area of the rhombus ABСD, since S (ABСD) = (AC ∙ BD) / 2;

120: 13 = 9 + (3/13) – the height of the rhombus, since h = S (ABСD): AB.

Answer: The height of the rhombus is 9 (3/13).