Question

1. determine algebraically if the triangles shown below can be classified as right triangles.
2. which set of the three numbers do not represent the lengths of the sides of a right triangle?

a 9, 12, 15
b 12, 16, 20
c 7, 24, 25
d 7, 7, 14
e 5, 12, 13

Explanation:

Step1: Apply Pythagorean theorem for first triangle

For triangle \(ABC\) with sides \(a = 3\), \(b=4\), \(c = 5\). Check if \(a^{2}+b^{2}=c^{2}\).
\(3^{2}+4^{2}=9 + 16=25\) and \(5^{2}=25\). So, \(3^{2}+4^{2}=5^{2}\).

Step2: Apply Pythagorean theorem for second triangle

For triangle \(FGE\) with sides \(f = 5\), \(g=6\), \(e = 8\). Check if \(f^{2}+g^{2}=e^{2}\).
\(5^{2}+6^{2}=25+36 = 61\) and \(8^{2}=64\). So, \(5^{2}+6^{2}
eq8^{2}\).

Answer:

The first triangle (\(ABC\)) is a right - triangle (since \(3^{2}+4^{2}=5^{2}\)), and the second triangle (\(FGE\)) is not a right - triangle (since \(5^{2}+6^{2}
eq8^{2}\)).

For problem 6:

Step1: Recall Pythagorean theorem

For a set of numbers \(a,b,c\) (\(c\) is the largest number) to be sides of a right - triangle, \(a^{2}+b^{2}=c^{2}\) must hold.

Step2: Check option A

For \(a = 9\), \(b = 12\), \(c=15\). \(9^{2}+12^{2}=81 + 144=225\) and \(15^{2}=225\). So, \(9^{2}+12^{2}=15^{2}\).

Step3: Check option B

For \(a = 12\), \(b = 16\), \(c = 20\). \(12^{2}+16^{2}=144+256 = 400\) and \(20^{2}=400\). So, \(12^{2}+16^{2}=20^{2}\).

Step4: Check option C

For \(a = 7\), \(b = 24\), \(c = 25\). \(7^{2}+24^{2}=49+576 = 625\) and \(25^{2}=625\). So, \(7^{2}+24^{2}=25^{2}\).

Step5: Check option D

For \(a = 7\), \(b = 7\), \(c = 14\). \(7^{2}+7^{2}=49 + 49=98\) and \(14^{2}=196\). So, \(7^{2}+7^{2}
eq14^{2}\).

Step6: Check option E

For \(a = 5\), \(b = 12\), \(c = 13\). \(5^{2}+12^{2}=25+144 = 169\) and \(13^{2}=169\). So, \(5^{2}+12^{2}=13^{2}\).