Question

1. if rs = 40, st = 41, and rt = 11, is △rst a right triangle? explain.

Explanation:

Step1: Recall Pythagorean theorem

For a triangle with side lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), if \(a^{2}+b^{2}=c^{2}\), then the triangle is a right triangle. Here, the side lengths are \(RS = 40\), \(ST=41\), and \(RT = 11\). The longest side is \(ST = 41\). So we check if \(RS^{2}+RT^{2}=ST^{2}\) or \(RT^{2}+ST^{2}=RS^{2}\) or \(RS^{2}+ST^{2}=RT^{2}\). We will check the first one (since \(RS\) and \(RT\) are shorter than \(ST\)).

Step2: Calculate \(RS^{2}+RT^{2}\)

\(RS^{2}=40^{2}=1600\), \(RT^{2}=11^{2} = 121\). Then \(RS^{2}+RT^{2}=1600 + 121=1721\).

Step3: Calculate \(ST^{2}\)

\(ST^{2}=41^{2}=1681\).

Step4: Compare the two results

Since \(1721
eq1681\), we check another combination. Let's check \(RT^{2}+ST^{2}\). \(RT^{2}+ST^{2}=121 + 1681=1802
eq1600=RS^{2}\). And \(RS^{2}+ST^{2}=1600+1681 = 3281
eq121=RT^{2}\). So none of the Pythagorean theorem conditions are met.

Answer:

\(\triangle RST\) is not a right triangle because for the side lengths \(RS = 40\), \(ST=41\), and \(RT = 11\), none of the combinations of the sum of the squares of two sides equals the square of the third side (i.e., \(40^{2}+11^{2}
eq41^{2}\), \(11^{2}+41^{2}
eq40^{2}\), and \(40^{2}+41^{2}
eq11^{2}\)).