Question

sharel wants to build a right triangle. which of these sets of sides would not work? {99,195,168} {17,8,15} {166,144,220} {41,9,40}

Explanation:

Step1: Check the Pythagorean theorem

For a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the longest side.

For the set \(\{99,195,168\}\)

Let \(a = 99\), \(b=168\), \(c = 195\)
\(a^{2}+b^{2}=99^{2}+168^{2}=9801 + 28224=38025\)
\(c^{2}=195^{2}=38025\)
So \(99^{2}+168^{2}=195^{2}\)

For the set \(\{17,8,15\}\)

Let \(a = 8\), \(b = 15\), \(c=17\)
\(a^{2}+b^{2}=8^{2}+15^{2}=64 + 225=289\)
\(c^{2}=17^{2}=289\)
So \(8^{2}+15^{2}=17^{2}\)

For the set \(\{166,144,220\}\)

Let \(a = 144\), \(b = 166\), \(c = 220\)
\(a^{2}+b^{2}=144^{2}+166^{2}=20736+27556 = 48292\)
\(c^{2}=220^{2}=48400\)
Since \(144^{2}+166^{2}
eq220^{2}\)

For the set \(\{41,9,40\}\)

Let \(a = 9\), \(b = 40\), \(c = 41\)
\(a^{2}+b^{2}=9^{2}+40^{2}=81+1600 = 1681\)
\(c^{2}=41^{2}=1681\)
So \(9^{2}+40^{2}=41^{2}\)

Answer:

\(\{166,144,220\}\)