Question

find the missing side of each triangle. leave your answers in simplest radical form.
5)
6)
7)
8)
state if each triangle is a right triangle.
9)
10)

Explanation:

Step1: Apply Pythagorean theorem for right - triangles

For a right - triangle with sides \(a\), \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). If we are finding a leg, say \(a\), then \(a=\sqrt{c^{2}-b^{2}}\); if finding the hypotenuse \(c\), then \(c = \sqrt{a^{2}+b^{2}}\).

Step2: Solve problem 5

Given \(a = 7\) mi and \(c=14\) mi, find the other leg \(x\). Using \(x=\sqrt{c^{2}-a^{2}}\), we have \(x=\sqrt{14^{2}-7^{2}}=\sqrt{(14 + 7)(14 - 7)}=\sqrt{21\times7}=\sqrt{147}=7\sqrt{3}\) mi.

Step3: Solve problem 6

Given \(a = 8\) km and \(c=\sqrt{226}\) km, find the other leg \(x\). Using \(x=\sqrt{c^{2}-a^{2}}\), we have \(x=\sqrt{(\sqrt{226})^{2}-8^{2}}=\sqrt{226 - 64}=\sqrt{162}=9\sqrt{2}\) km.

Step4: Solve problem 7

Given \(a = 4\) yd and \(b = 16\) yd, find the hypotenuse \(x\). Using \(x=\sqrt{a^{2}+b^{2}}\), we have \(x=\sqrt{4^{2}+16^{2}}=\sqrt{16 + 256}=\sqrt{272}=2\sqrt{68}=4\sqrt{17}\) yd.

Step5: Solve problem 8

Given \(a = 9\) cm and \(c = 12\) cm, find the other leg \(x\). Using \(x=\sqrt{c^{2}-a^{2}}\), we have \(x=\sqrt{12^{2}-9^{2}}=\sqrt{(12 + 9)(12 - 9)}=\sqrt{21\times3}=\sqrt{63}=3\sqrt{7}\) cm.

Step6: Check if triangle in problem 9 is a right - triangle

Let \(a = 6\) yd, \(b = 12\) yd and \(c = 15\) yd. Calculate \(a^{2}+b^{2}=6^{2}+12^{2}=36 + 144 = 180\) and \(c^{2}=15^{2}=225\). Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

Step7: Check if triangle in problem 10 is a right - triangle

Let \(a = 3\) ft, \(b = 4\) ft and \(c = 5\) ft. Calculate \(a^{2}+b^{2}=3^{2}+4^{2}=9 + 16 = 25\) and \(c^{2}=5^{2}=25\). Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

Answer:

1. \(7\sqrt{3}\) mi
2. \(9\sqrt{2}\) km
3. \(4\sqrt{17}\) yd
4. \(3\sqrt{7}\) cm
5. Not a right - triangle
6. Right - triangle