Question

extra practice
in exercises 1 - 6, find the value of x. then tell whether the side lengths form a
pythagorean triple.
1.
2.
3.
4.
5.
6.

1. from school, you biked 1.2 miles due south and then 0.5 mile due east to your house.

if you had biked home on the street that runs directly diagonal from your school to your
house, how many fewer miles would you have biked?
in exercises 8 and 9, verify that the segment lengths form a triangle. is the triangle
acute, right, or obtuse?

1. 90, 216, and 234
2. 1, 1, and $\sqrt{3}$

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).
For problem 1: \(a = 81\), \(b=108\), then \(x^{2}=81^{2}+108^{2}\).
\(81^{2}=6561\), \(108^{2}=11664\), so \(x^{2}=6561 + 11664=18225\), \(x=\sqrt{18225}=135\).
To check if it is a Pythagorean triple: \(81 = 27\times3\), \(108=27\times4\), \(135=27\times5\). Since \(3^{2}+4^{2}=9 + 16=25=5^{2}\), it is a Pythagorean triple.
For problem 2: \(a=x\), \(b = 4\), \(c = 8\). Then \(x^{2}+4^{2}=8^{2}\), \(x^{2}=64 - 16=48\), \(x=\sqrt{48}=4\sqrt{3}\). Since \(4\sqrt{3}\) is not an integer, it is not a Pythagorean triple.
For problem 3: \(a = 15\), \(b=20\), \(x^{2}=15^{2}+20^{2}\). \(15^{2}=225\), \(20^{2}=400\), \(x^{2}=225+400 = 625\), \(x = 25\). Check Pythagorean triple: \(15=5\times3\), \(20=5\times4\), \(25=5\times5\). Since \(3^{2}+4^{2}=25\), it is a Pythagorean triple.
For problem 4: \(a = 6\), \(b = 10\), \(x^{2}=6^{2}+10^{2}\). \(6^{2}=36\), \(10^{2}=100\), \(x^{2}=36 + 100=136\), \(x=\sqrt{136}=2\sqrt{34}\). Not an integer, so not a Pythagorean triple.
For problem 5: \(a=x\), \(b = 55\), \(c = 77\). Then \(x^{2}+55^{2}=77^{2}\), \(x^{2}=77^{2}-55^{2}=(77 + 55)(77 - 55)=(132)(22)=2904\), \(x=\sqrt{2904}=54\). Check Pythagorean triple: \(54=6\times9\), \(55=6\times\frac{55}{6}\) (not in the form \(k\times n\), \(k\times m\), \(k\times p\) where \(n^{2}+m^{2}=p^{2}\)), so not a Pythagorean triple.
For problem 6: \(a = 48\), \(b=90\), \(x^{2}=48^{2}+90^{2}\). \(48^{2}=2304\), \(90^{2}=8100\), \(x^{2}=2304+8100=10404\), \(x = 102\). Check Pythagorean triple: \(48=6\times8\), \(90=6\times15\), \(102=6\times17\). Since \(8^{2}+15^{2}=64 + 225=289=17^{2}\), it is a Pythagorean triple.
For problem 7: Let the two legs of the right - triangle be \(a = 1.2\) and \(b=0.5\). The length of the diagonal \(d=\sqrt{1.2^{2}+0.5^{2}}=\sqrt{1.44 + 0.25}=\sqrt{1.69}=1.3\). The difference in distance is \((1.2 + 0.5)-1.3=0.4\) miles.
For problem 8: Check triangle inequality: \(90+216>234\), \(90 + 234>216\), \(216+234>90\). Then check the type: \(90^{2}+216^{2}=8100+46656=54756\), \(234^{2}=54756\). So it is a right - triangle.
For problem 9: Check triangle inequality: \(1+1>\sqrt{3}\approx1.732\), \(1+\sqrt{3}>1\), \(1+\sqrt{3}>1\). Then check the type: \(1^{2}+1^{2}=2\), \((\sqrt{3})^{2}=3\). Since \(1^{2}+1^{2}<(\sqrt{3})^{2}\), it is an obtuse - triangle.

Answer:

1. \(x = 135\), it is a Pythagorean triple.
2. \(x = 4\sqrt{3}\), it is not a Pythagorean triple.
3. \(x = 25\), it is a Pythagorean triple.
4. \(x = 2\sqrt{34}\), it is not a Pythagorean triple.
5. \(x = 54\), it is not a Pythagorean triple.
6. \(x = 102\), it is a Pythagorean triple.
7. \(0.4\) miles.
8. It is a right - triangle.
9. It is an obtuse - triangle.