Answer:
1) [tex]9 < x < 31[/tex]
2) [tex]10 < x < 54[/tex]
3) Not a triangle
4) This is a triangle
5) [tex]6\sqrt{2}[/tex]
6) [tex]4\sqrt{6}[/tex]
7) [tex]x=\sqrt{296} \\x=2\sqrt{74} \\x=17.2047[/tex]
8) [tex]x=7[/tex]
9) Obtuse
10) Obtuse
Step-by-step explanation:
1) This one is a bit difficult to explain without a picture, but the third side needs to be longer than the difference of the 2 sides, and shorter than the sum of the 2 sides.
[tex]20+11=31\\20-11=9[/tex]
The third side must be shorter than 31 and longer than 9.
[tex]9 < x < 31[/tex]
2) Same thing here.
[tex]32+22=54\\32-22=10\\10 < x < 54[/tex]
3) Use the same method as the first 2 questions.
[tex]10 + 5 = 15\\10 - 5 = 5[/tex]
15 is greater than 5, but not less than 15. Not a triangle.
4) Same thing again.
[tex]12+2.6=14.6\\12-2.6=9.4[/tex]
12 is greater than 9.4 and less than 14.6. This is a triangle.
5) Split 72 into its prime factors, rewrite any pairs in exponent form, and apply [tex]\sqrt{x^2} =x[/tex]
[tex]\sqrt{72} \\\sqrt{2*2*2*3*3}\\\sqrt{(2*2)*(3*3)*2} \\\sqrt{2^2*3^2*2} \\(2*3)\sqrt{2} \\6\sqrt{2}[/tex]
6) Same thing once again.
[tex]\sqrt{96}\\\sqrt{2*2*2*2*2*3}\\\sqrt{(2*2)*(2*2)*2*3}\\\sqrt{2^2*2^2*2*3} \\(2*2)\sqrt{2*3} \\4\sqrt{6}[/tex]
7) Use the pythagorean theorem ([tex]a^2+b^2=c^2[/tex])
[tex]14^2+10^2=x^2\\196+100=x^2\\x^2=296\\x=\sqrt{296} ,-\sqrt{296}[/tex]
Distances cant be negative, so [tex]x=\sqrt{296} \\x=2\sqrt{74}[/tex]
8) The time to repeat the same method has come.
[tex]5^2+x^2=\sqrt{74} ^2\\x^2+25=74\\x^2=49\\x=7,-7[/tex]
Distances cant be negative, so [tex]x=7[/tex]
9) Check distances with the pythagorean theorem.
[tex]12^2+9^2=17^2\\144+81=289\\225\neq289[/tex]
The third side is too long to be a right triangle, so this triangle is obtuse.
10) Repeat the previous method once more.
[tex]6^2+(2\sqrt{55} )^2=17^2\\36+220=289\\256\neq 289[/tex]
Third side is too long, triangle is obtuse.
