In an isosceles trapezoid ABCD AD || BC, angle A = 30 º, height BK = 1, BC = 2√3. a) find the area of the trapezoid

In an isosceles trapezoid ABCD AD || BC, angle A = 30 º, height BK = 1, BC = 2√3. a) find the area of the trapezoid. b) find the area of the triangle KMD, if M is the midpoint of the segment BD.

Given: ABCD – isosceles trapezoid; AD parallel to the aircraft; angle A = 30 degrees; BK – height = 1; BC = 2√3.
Find: a) S (area) of trapezoid ABCD; b) S of triangle KMF, if m. M is the midpoint of BD.
Decision:
A) Since BK is the height, the angle AKB = angle BKD = 90 degrees.
Angle A = 30 degrees. The leg, opposite to the angle of 30 degrees, is equal to 1/2 (half) of the hypotenuse. In this case, AB is the hypotenuse, which means that AB = 2BK = 2 x 1 = 2.
By the Pythagorean theorem: AK ^ 2 = AB ^ 2 – BK ^ 2; AK ^ 2 = 4 – 1 = 3; AK = √3.
Let’s draw the height of the CH. CH = BK, as the height of one trapezoid. Also, BK is parallel to CH – as perpendiculars to one straight line, which means that KВСН is a rectangle, which means that BC = KH = 2√3.
Triangle ABK = triangle DCH – along the hypotenuse and leg (AB = CD – since the trapezoid is isosceles and CH = BK – since the heights are equal), which means AK + HD = 2√3.
AD = AK + KH + HD = 4√3.
S ABCD (half sum of base times height) = (AD + BC) / 2 x BK = (4√3 + 2√3) / 2 x 1 = 3√3.
B) Let’s draw a perpendicular МР, perpendicular to the base АD.
MP is the middle line of triangle BKD (since M is the middle of BD and P is the middle of KD).
The middle line is equal to half the base of the triangle: MP = BK / 2 = 1/2 = 0.5.
KD = KH + HD = 3√3.
S KMF = 1/2 x KD x MP = 0.5 x 3√3 x 0.5 = 3√3 / 4
Answer: 3√3; 3√3 / 4.