Question

pythagorean theorem qs
find the missing side of each triangle. round your answers to the nearest tenth if necessary.
1)
2)
3)
4)
5)
6)
7)
8)

Explanation:

1)

Step1: Apply Pythagorean theorem

For a right - triangle \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse. Here \(c = 15\) in, \(b=12\) in, and \(a=x\).

$$x^{2}+12^{2}=15^{2}$$

Step2: Solve for \(x\)

$$x^{2}=15^{2}-12^{2}=225 - 144=81$$

$$x=\sqrt{81}=9$$

in

2)

Step1: Apply Pythagorean theorem

Here \(c = 5\) m, \(a = 3\) m, and \(b=x\)

$$3^{2}+x^{2}=5^{2}$$

Step2: Solve for \(x\)

$$x^{2}=5^{2}-3^{2}=25 - 9 = 16$$

$$x=\sqrt{16}=4$$

m

3)

Step1: Apply Pythagorean theorem

Here \(c = 10\) in, \(a = 8\) in, and \(b=x\)

$$8^{2}+x^{2}=10^{2}$$

Step2: Solve for \(x\)

$$x^{2}=10^{2}-8^{2}=100 - 64 = 36$$

$$x=\sqrt{36}=6$$

in

4)

Step1: Apply Pythagorean theorem

Here \(a = 12\) in, \(b = 5\) in, and \(c=x\)

$$x^{2}=12^{2}+5^{2}$$

Step2: Solve for \(x\)

$$x^{2}=144 + 25=169$$

$$x=\sqrt{169}=13$$

in

5)

Step1: Apply Pythagorean theorem

Here \(c = 5\) in, \(b = 4\) in, and \(a=x\)

$$x^{2}+4^{2}=5^{2}$$

Step2: Solve for \(x\)

$$x^{2}=5^{2}-4^{2}=25 - 16 = 9$$

$$x=\sqrt{9}=3$$

in

6)

Step1: Apply Pythagorean theorem

Here \(a = 6\) in, \(b = 8\) in, and \(c=x\)

$$x^{2}=6^{2}+8^{2}$$

Step2: Solve for \(x\)

$$x^{2}=36+64 = 100$$

$$x=\sqrt{100}=10$$

in

7)

Step1: Apply Pythagorean theorem

Here \(c = 13\) km, \(b = 12\) km, and \(a=x\)

$$x^{2}+12^{2}=13^{2}$$

Step2: Solve for \(x\)

$$x^{2}=13^{2}-12^{2}=169 - 144=25$$

$$x=\sqrt{25}=5$$

km

8)

Step1: Apply Pythagorean theorem

Here \(a = 12\) km, \(b = 9\) km, and \(c=x\)

$$x^{2}=12^{2}+9^{2}$$

Step2: Solve for \(x\)

$$x^{2}=144+81 = 225$$

$$x=\sqrt{225}=15$$

km

Answer:

1. \(9\) in
2. \(4\) m
3. \(6\) in
4. \(13\) in
5. \(3\) in
6. \(10\) in
7. \(5\) km
8. \(15\) km