Problem 4
Draw a diagonal to connect the opposite non-right angles as shown below. This forms two right triangles. As you can probably guess, we'll use the pythagorean theorem to find the length of this segment. We'll call it x.
a^2+b^2 = c^2
7^2+24^2 = x^2
49+576 = x^2
x^2 = 625
x = sqrt(625)
x = 25
The hypotenuse of each right triangle is 25 units long.
Move your focus to the upper right triangle. We can use the pythagorean theorem again in a slightly different way to find y.
a^2+b^2 = c^2
15^2+y^2 = 25^2
225+y^2 = 625
y^2 = 625-225
y^2 = 400
y = sqrt(400)
y = 20
Answer: y = 20
====================================================
Problem 6
The largest triangle, ignore segment x for now, is isosceles because we have two congruent sides.
As with any isosceles triangle, the vertex angle is bisected by the perpendicular bisector as shown in the diagram. This produces two identical right triangles. Most importantly, it means that horizontal side 4 is cut in half to 4/2 = 2 units.
In short, each smaller right triangle has a horizontal leg of 2, vertical leg x, and hypotenuse 6. Let's focus on just one of those right triangles.
Like the previous problem, we'll use the pythagorean theorem to find x.
a^2+b^2 = c^2
x^2+2^2 = 6^2
x^2+4 = 36
x^2 = 36-4
x^2 = 32
x = sqrt(32)
x = sqrt(16*2)
x = sqrt(16)*sqrt(2)
x = 4*sqrt(2)
Answer: x = 4*sqrt(2)
