Question

name:
pythagorean theorem - worksheet
for each triangle find the missing length. round your answer to the nearest tenth. then find the area and the perimeter.
1.
triangle with hypotenuse 19, one leg 17, missing leg ?
2.
triangle with leg 5, leg 13, hypotenuse ?
3.
triangle with hypotenuse 20, leg 10, missing leg ?
4.
triangle with leg 5, leg 14, hypotenuse ?
for #5 - 9 c is the hypotenuse of the right triangle abc with sides a, b, c

1. \ta = 12 ; b = 5; c =
2. \ta = 8 ; b = ; c = 10
3. \ta = 15 ; b = ; c = 17
4. \ta = ; b = 40 ; c = 50
5. \ta = ; b = 2 ; c = 4

Explanation:

Step1: Solve Problem 1 (Missing leg)

The Pythagorean theorem is \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse. For the first triangle, \(c = 19\), one leg \(a = 17\), find the other leg \(b\).
\(b=\sqrt{c^{2}-a^{2}}=\sqrt{19^{2}-17^{2}}=\sqrt{(19 + 17)(19 - 17)}=\sqrt{36\times2}=\sqrt{72}\approx8.5\)

Step2: Solve Problem 2 (Missing hypotenuse)

For the second triangle, legs \(a = 5\), \(b = 13\), find hypotenuse \(c\).
\(c=\sqrt{a^{2}+b^{2}}=\sqrt{5^{2}+13^{2}}=\sqrt{25 + 169}=\sqrt{194}\approx13.9\)

Step3: Solve Problem 3 (Missing leg)

For the third triangle, \(c = 20\), one leg \(b = 10\), find the other leg \(a\).
\(a=\sqrt{c^{2}-b^{2}}=\sqrt{20^{2}-10^{2}}=\sqrt{400 - 100}=\sqrt{300}\approx17.3\)

Step4: Solve Problem 4 (Missing hypotenuse)

For the fourth triangle, legs \(a = 5\), \(b = 14\), find hypotenuse \(c\).
\(c=\sqrt{a^{2}+b^{2}}=\sqrt{5^{2}+14^{2}}=\sqrt{25 + 196}=\sqrt{221}\approx14.9\)

Step5: Solve Problem 5 (Missing hypotenuse)

Given \(a = 12\), \(b = 5\), \(c=\sqrt{12^{2}+5^{2}}=\sqrt{144 + 25}=\sqrt{169}=13\)

Step6: Solve Problem 6 (Missing leg)

Given \(a = 8\), \(c = 10\), \(b=\sqrt{c^{2}-a^{2}}=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}=6\)

Step7: Solve Problem 7 (Missing leg)

Given \(a = 15\), \(c = 17\), \(b=\sqrt{c^{2}-a^{2}}=\sqrt{17^{2}-15^{2}}=\sqrt{(17 + 15)(17 - 15)}=\sqrt{32\times2}=\sqrt{64}=8\)

Step8: Solve Problem 8 (Missing leg)

Given \(b = 40\), \(c = 50\), \(a=\sqrt{c^{2}-b^{2}}=\sqrt{50^{2}-40^{2}}=\sqrt{(50 + 40)(50 - 40)}=\sqrt{90\times10}=\sqrt{900}=30\)

Step9: Solve Problem 9 (Missing leg)

Given \(b = 2\), \(c = 4\), \(a=\sqrt{c^{2}-b^{2}}=\sqrt{4^{2}-2^{2}}=\sqrt{16 - 4}=\sqrt{12}\approx3.5\)

Answer:

1. Missing length: \(\approx 8.5\) (Area: \(\frac{1}{2}\times17\times8.5 = 72.25\), Perimeter: \(17 + 8.5 + 19 = 44.5\))
2. Missing length: \(\approx 13.9\) (Area: \(\frac{1}{2}\times5\times13 = 32.5\), Perimeter: \(5 + 13 + 13.9 = 31.9\))
3. Missing length: \(\approx 17.3\) (Area: \(\frac{1}{2}\times17.3\times10 = 86.5\), Perimeter: \(17.3 + 10 + 20 = 47.3\))
4. Missing length: \(\approx 14.9\) (Area: \(\frac{1}{2}\times5\times14 = 35\), Perimeter: \(5 + 14 + 14.9 = 33.9\))
5. \(c = 13\)
6. \(b = 6\)
7. \(b = 8\)
8. \(a = 30\)
9. \(a\approx3.5\)