Question

find the value of each variable.
7
8
9

Explanation:

Step1: Use the mid - segment theorem for problem 7

The mid - segment of a triangle is parallel to the third side and half its length.
For \(x\):
The segment with length \(x\) is a mid - segment. The side parallel to it has length \(40\). So \(x=\frac{40}{2}=20\)
For \(y\):
The segment with length \(y\) is a mid - segment. The side parallel to it has length \(30\). So \(y = 15\)
For \(z\):
The segment with length \(z\) is a mid - segment. The side parallel to it has length \(10\). So \(z=5\)

Step2: Use the mid - segment theorem for problem 8

For \(x\):
The segment with length \(2x\) is a mid - segment. The side parallel to it has length \(36\). So \(2x=\frac{36}{2}\), then \(x = 9\)
For \(y\):
The segment with length \(y - 2\) is a mid - segment. The side parallel to it has length \(24\). So \(y-2=\frac{24}{2}\), then \(y=14\)
For \(z\):
The segment with length \(z + 8\) is a mid - segment. The side parallel to it has length \(38\). So \(z + 8=\frac{38}{2}\), then \(z=11\)

Step3: Use the mid - segment theorem for problem 9

For \(x\):
The segment with length \(4x\) is a mid - segment. The side parallel to it has length \(56\). So \(4x=\frac{56}{2}\), then \(x = 7\)
For \(y\):
The segment with length \(4y\) is a mid - segment. The side parallel to it has length \(30\). So \(4y=\frac{30}{2}\), then \(y=\frac{15}{2}=7.5\)
For \(z\):
The segment with length \(6z+2\) is a mid - segment. The side parallel to it has length \(40\). So \(6z + 2=\frac{40}{2}\), then \(6z=18\), \(z = 3\)

Answer:

Problem 7:
\(x = 20\)
\(y=15\)
\(z = 5\)

Problem 8:
\(x = 9\)
\(y=14\)
\(z = 11\)

Problem 9:
\(x = 7\)
\(y = 7.5\)
\(z=3\)