Question

determine whether a triangle with the given side lengths is a right triangle.

(a) 20, 21, 29
(b) 6, 9, 11
(c) 4, 8, 9
(d) 13, 20, 26

Explanation:

Step1: Recall the Pythagorean theorem

For a triangle with side lengths \(a\), \(b\), \(c\) (\(c\) is the longest side), if \(a^{2}+b^{2}=c^{2}\), then it is a right - triangle.

Step2: Check part (a)

Let \(a = 20\), \(b=21\), \(c = 29\).
Calculate \(a^{2}+b^{2}=20^{2}+21^{2}=400 + 441=841\).
Calculate \(c^{2}=29^{2}=841\).
Since \(a^{2}+b^{2}=c^{2}\), it is a right - triangle.

Step3: Check part (b)

Let \(a = 6\), \(b = 9\), \(c=11\).
Calculate \(a^{2}+b^{2}=6^{2}+9^{2}=36+81 = 117\).
Calculate \(c^{2}=11^{2}=121\).
Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

Step4: Check part (c)

Let \(a = 4\), \(b = 8\), \(c = 9\).
Calculate \(a^{2}+b^{2}=4^{2}+8^{2}=16 + 64=80\).
Calculate \(c^{2}=9^{2}=81\).
Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

Step5: Check part (d)

Let \(a = 13\), \(b = 20\), \(c = 26\).
Calculate \(a^{2}+b^{2}=13^{2}+20^{2}=169+400 = 569\).
Calculate \(c^{2}=26^{2}=676\).
Since \(a^{2}+b^{2}
eq c^{2}\), it is not a right - triangle.

Answer:

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