ABCD is a trapezoid with AB parallel to DC. If AB=25, BC=24, CD=50 and AD=7, what is the area of ABCD?
We are going to need some sort of angle.
draw BE || AD, then ABED is a parallelogram and BE = 7
In triangle BCE,
BC = 24
BE = 7
CE = 50-25 = 25
by the cosine law:
7^2 = 24^2 + 25^2 - 2(24)(25)cosC
cosC = 1152/1200
angle C = 16.26... ---- I stored the entire number
we can now find the height:
sin 16.26... = height/24
height = 24sin16.26.. = 6.72 --- that came out exact
area = (1/2)(25+50)(6.72) = 252 square units
To find the area of trapezoid ABCD, we can use the formula:
Area = (1/2) * (sum of parallel sides) * (distance between the parallel sides).
In this case, AB and DC are the parallel sides, and let's denote the distance between them as h.
First, we need to calculate the value of h. We can use the Pythagorean theorem on triangle BCD to find h:
BC^2 = AB^2 + CD^2 - (AD - DC)^2
Substituting the given values:
24^2 = 25^2 + 50^2 - (7 - DC)^2
Simplifying:
576 = 625 + 2500 - (49 - 2DC + DC^2)
Rearranging and simplifying further:
576 = 3125 - 49 + 2DC - DC^2
0 = DC^2 - 2DC - 2564
This is a quadratic equation. Applying the quadratic formula, we find:
DC = (2 ± √(2^2 - 4 * 1 * (-2564))) / (2 * 1)
DC = (2 ± √(4 + 10256)) / 2
DC = (2 ± √10260) / 2
DC = (2 ± 101.29) / 2
DC ≈ 51.29 or -49.29
Since the distance cannot be negative, we take DC to be approximately 51.29.
Now that we have the values of AB, DC, and the distance h, we can calculate the area of trapezoid ABCD:
Area = (1/2) * (AB + DC) * h
Area = (1/2) * (25 + 51.29) * h
Area = (1/2) * (76.29) * h
So, to find the area of ABCD, we still need to know the value of h. The value of h can be calculated by finding the height of triangle BCD using triangle BCD's base and sides. Without that information, we cannot determine the area of ABCD.