Isosceles triangle ABE of area 100 square inches is cut by CD into an isosceles trapezoid and a smaller isosceles triangle. The area of the trapezoid is 75 square inches. If the altitude of triangle ABE from A is 20 inches, what is the number of inches in the length of CD?
since the area of ABE is 100, and the altitude is 20, the length of AB is 10.
Since the area of ABCD is 3/4 the area of ABE, CDE has 1/4 the area. So, CDE~ABE, with a scale factor of 1/2.
So, CD = AB/2 = 5
check: the altitude of ABCD is 1/2 * 20 = 10
the area of ABCD is thus (10+20)/2 * 5 = 75
To find the length of CD, we need to use the area of the trapezoid and the altitude of the triangle.
Let's assume that the length of CD is "x" inches.
First, let's find the area of triangle ABE. The formula for the area of a triangle is given by (base * height) / 2.
The base of triangle ABE is the length of CD, which is "x". The height is given as 20 inches.
So, the area of triangle ABE is (x * 20) / 2 = 10x square inches.
Now, let's find the area of the trapezoid. The formula for the area of a trapezoid is given by (height * (base1 + base2)) / 2.
The height of the trapezoid is also given as 20 inches. The bases are the length of CD, which is "x", and the base of triangle ABE, which is "x" as well.
So, the area of the trapezoid is (20 * (x + x)) / 2 = 20x square inches.
According to the problem, the area of the trapezoid is 75 square inches.
So, we have the equation 20x = 75.
Now, we can solve for x:
20x = 75
x = 75 / 20
x = 3.75 inches
Therefore, the number of inches in the length of CD is 3.75 inches.