Question

the pythagorean theorem: color - by - number
directions: find the missing measure (x) for each triangle.
round your answer to the nearest tenth (when needed).
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.

1. a square has legs of 11, what is

the measure of its hypotenuse?

1. a square has a hypotenuse of 13

and a leg that is 9, what does the
other leg measure?
color bank:
light blue: 10.4
gray: 12.4
light green: 11.2
dark blue: 13.6
yellow: 9.4
black: 39.8
dark purple: 30
pink: 14.8
white: 80
red: 19.2
dark green: 15.6
light purple: 3

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), the Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\). If we need to find a leg, say \(a\), then \(a=\sqrt{c^{2}-b^{2}}\); if we need to find the hypotenuse \(c\), then \(c = \sqrt{a^{2}+b^{2}}\).

Step2: Solve problem 4

In problem 4, the legs are \(a = 20\) and \(b=32\), and we need to find the hypotenuse \(x\).
Using the formula \(x=\sqrt{20^{2}+32^{2}}\).
First, calculate \(20^{2}=400\) and \(32^{2}=1024\).
Then \(20^{2}+32^{2}=400 + 1024=1424\).
So \(x=\sqrt{1424}\approx37.7\).

Step3: Solve problem 6

In problem 6, the hypotenuse \(c = 37\) and one leg \(a = 10\), we need to find the other leg \(x\).
Using the formula \(x=\sqrt{37^{2}-10^{2}}\).
Calculate \(37^{2}=1369\) and \(10^{2}=100\).
Then \(37^{2}-10^{2}=1369 - 100 = 1269\).
So \(x=\sqrt{1269}\approx35.6\).

Step4: Solve problem 8

In problem 8, the hypotenuse \(c = 82\) and one leg \(b = 18\), we need to find the other leg \(x\).
Using the formula \(x=\sqrt{82^{2}-18^{2}}\).
Calculate \(82^{2}=6724\) and \(18^{2}=324\).
Then \(82^{2}-18^{2}=6724-324 = 6400\).
So \(x=\sqrt{6400}=80\).

Step5: Solve problem 10

In problem 10, the legs are \(a = 7.9\) and \(b=x\), and the hypotenuse \(c = 14.7\).
Using the formula \(x=\sqrt{14.7^{2}-7.9^{2}}\).
Calculate \(14.7^{2}=216.09\) and \(7.9^{2}=62.41\).
Then \(14.7^{2}-7.9^{2}=216.09 - 62.41=153.68\).
So \(x=\sqrt{153.68}\approx12.4\).

Answer:

For problem 4, \(x\approx37.7\); for problem 6, \(x\approx35.6\); for problem 8, \(x = 80\); for problem 10, \(x\approx12.4\)