In an isosceles trapezoid, the larger base is 2.7 cm, the side side is 1 cm, the angle between them is 60

In an isosceles trapezoid, the larger base is 2.7 cm, the side side is 1 cm, the angle between them is 60, find the midline of the trapezoid and the area of the trapezoid.

A trapezoid is a quadrangle in which only two sides are parallel and the sides are not equal.

The area of ​​the trapezoid is equal to the product of its midline by the height:

S = m h.

The midline is the line segment connecting the midpoints of the sides of the trapezoid. It is parallel to the bases and equal to their half-sum:

m = (a + b) / 2.

Isosceles is a trapezoid in which the sides are equal: AB = CD. Thus, AH = KD.

Let’s draw two heights BH and SK.

Since the segment of the larger base, located between the two bases, is equal to the length of the smaller base НК = ВС, then:

ВС = АD – АН – КD.

Let us calculate the length of the segment AH. Consider the triangle ΔАВН.

To calculate, we use the cosine theorem, according to which the cosine of an acute angle of a right-angled triangle is the ratio of the adjacent leg to the hypotenuse:

cos A = AH / AB;

AH = AB · cos A;

cos 60º = 1/2;

AH = 1 1/2 = 0.5 cm.

BC = 2.7 – 0.5 – 0.5 = 1.7 cm.

m = (1.7 + 2.7) / 2 = 4.4 / 2 = 2.2 cm.

Let’s find the height of the HB. To do this, apply the Pythagorean theorem:

AB ^ 2 = BH ^ 2 + AH ^ 2;

BH ^ 2 = AB ^ 2 – AH ^ 2;

BH ^ 2 = 12 – 0.52 = 1 – 0.25 = 0.75;

BH = √0.75 ≈ 0.9.

S = 2.2 · 0.9 = 1.98 cm2.

Answer: The middle line of the trapezoid is 2.2 cm, the area of ​​the trapezoid is 1.98 cm2.