This kind of problem has a large right triangle.
The altitude is drawn to the hypotenuse creating to smaller right triangles.
A theorem states that the corresponding sides of all three triangles are in the same ratio.
In problem 3, you see that each triangle has a long leg and a short leg.
The ratio of the short leg to the long leg of each triangle is the same.
The small triangle tot eh left has short leg 20 and long leg x.
The small triangle to the upper right has short leg x and long leg 25.
The large triangle has short leg z and long leg y.
All ratios of short leg to long leg of each triangle are equal.
20/x = x/25 = z/y
20/x = x/25 allows you to solve for x.
x^2 = 20 * 25
x^2 = 500
x = sqrt(500)
x = 10sqrt(5)
You can also use the hypotenuses in the ratios.
Now we use the ratio of the short leg to the hypotenuse of each triangle.
x/y = 20/z = z/(20 + 25)
We now x. x = 10sqrt(5)
20/z = z/(20 + 25)
20/z = z/45
z^2 = 900
z = 30
Now we can find y.
x/y = 20/z
10sqrt(5)/y = 20/30
10sqrt(5)/y = 2/3
2y = 3 * 10sqrt(5)
2y = 30sqrt(5)
y = 15sqrt(5)
