How do I find the area of a trapazoid with side lenghts 6 & 7, bottom 17 and top 10.
I know the perimeter is 40.
Now I need the area.
draw the trap ABCD where AD=10,DC = 7, BC=17 and BA = 6
Draw AE parallel to DC, then AE=7 and BE = 7
We now have a parallelogram and a triangle.
Let's find angle AEB which is also angle C , call it Ø
by cosine law in triangle ABE
6^2 = 7^2 + 7^2 - 2(7)(7)cosØ
cosØ = .632653
Ø = 50.75386°
area = triangle + parallelogram
= (1/2)(7)(7)sin 50.75386 + (10)(7)sin50.75386
= 73.184
To find the area of a trapezoid, you can use the formula A = (b1 + b2) * h / 2, where b1 and b2 are the lengths of the parallel sides (bottom and top) and h is the height of the trapezoid.
In this case, the bottom side length is 17, the top side length is 10, and you need to find the height.
To find the height, you can use the Pythagorean theorem since you have a right trapezoid. We can create a right triangle by dropping a perpendicular from one of the parallel sides (in this case, the bottom side) to the other parallel side, forming a right angle. Let's say this perpendicular is the height of the trapezoid (h).
Using the Pythagorean theorem, we can solve for h. The formula is a^2 + b^2 = c^2, where a and b are the lengths of the legs of the right triangle, and c is the length of the hypotenuse. In our case, a = (17 - 10)/2 = 3.5 (half the difference between the top and bottom sides), and b = h. We can let c be the height (h).
Using the formula, we have (3.5)^2 + h^2 = 6^2. Simplifying this equation gives us 12.25 + h^2 = 36. Rearranging the equation yields h^2 =36 - 12.25 = 23.75. Taking the square root of both sides gives us h ≈ 4.87 (rounded to two decimal places).
Now that we have the height, we can plug the values into the area formula A = (b1 + b2) * h / 2. Substituting the known values, we get A = (17 + 10) * 4.87 / 2 = 27 * 4.87 / 2 ≈ 65.79.
Therefore, the area of the trapezoid is approximately 65.79 square units.