Question

determine whether a triangle with the given side lengths is a right triangle. (a) 5, 12, 13 (b) 15, 36, 39 (c) 10, 11, 15 (d) 10, 26, 28

Explanation:

Step1: Apply Pythagorean theorem

For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), check if \(a^{2}+b^{2}=c^{2}\)

For (a) \(5\), \(12\), \(13\)

Step2: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)

\(a = 5\), \(b=12\), \(c = 13\)
\(a^{2}+b^{2}=5^{2}+12^{2}=25 + 144=169\)
\(c^{2}=13^{2}=169\)
Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

For (b) \(15\), \(36\), \(39\)

Step3: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)

\(a = 15\), \(b = 36\), \(c=39\)
\(a^{2}+b^{2}=15^{2}+36^{2}=225+1296 = 1521\)
\(c^{2}=39^{2}=1521\)
Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

For (c) \(10\), \(11\), \(15\)

Step4: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)

\(a = 10\), \(b = 11\), \(c = 15\)
\(a^{2}+b^{2}=10^{2}+11^{2}=100 + 121=221\)
\(c^{2}=15^{2}=225\)
Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

For (d) \(10\), \(26\), \(28\)

Step5: Calculate \(a^{2}+b^{2}\) and \(c^{2}\)

\(a = 10\), \(b = 26\), \(c = 28\)
\(a^{2}+b^{2}=10^{2}+26^{2}=100+676 = 776\)
\(c^{2}=28^{2}=784\)
Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

Answer:

(a) Right triangle
(b) Right triangle
(c) Not a right triangle
(d) Not a right triangle