In an isosceles trapezoid, the height drawn from the apex of an angle of 130 degrees divides the large

In an isosceles trapezoid, the height drawn from the apex of an angle of 130 degrees divides the large base into 4 and 12 cm segments. Find the area of the trapezoid.

The area of ​​the trapezoid is equal to the product of half the sum of its bases by the height:

S = (a + b) / 2 h.

Since the height of the trapezoid divides its larger base AD into segments AH and HD, which are respectively equal to 4 cm and 12 cm, then:

AD = AH + HD;

AD = 4 + 12 = 16 cm.

Since the trapezoid is isosceles, the KD segment is equal to the AH segment and is 4 cm.

The length of the smaller base of the BC is equal to the part of the larger base, which is located between the heights:

BC = NK = AD – AH – KD;

BC = NK = 16 – 4 – 4 = 8 cm.

To calculate the height of the VN, consider the triangle ΔAVN. To do this, we will use the tangent of the angle ∠B.

Since the angle АВС is 130, and the angle ∠НВС is straight, since it is formed by the perpendicular BC, then:

∠AVN = ∠ABS – ∠НВС;

∠АВН = 130 ° – 90 ° = 40 °;

tg 40 ° = 0.8390.

The tangent of an acute angle in a right triangle is the ratio of the opposite leg to the adjacent one:

tg B = AH / BH;

BH = AH / tg B;

BH = 4 / 0.8390 ≈ 4.8 cm.

S = (8 + 16) / 2 4.8 = 12 4.8 = 57.6 cm2.

Answer: the area of ​​the trapezoid is 57.6 cm2.