Answers:
25. b. 9, 40, 41
26. c. 14, 19, 24
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Explanation for problem 25
We use the converse of the pythagorean theorem.
If a^2+b^2 = c^2 is a true equation, then we have a right triangle with sides a,b,c where c is the longest side.
For part (a) we have a = 9, b = 41, c = 41
Then,
a^2+b^2 = c^2
9^2+41^2 = 41^2
1762 = 1681
The two sides aren't the same value at the end, so we have a false equation when (a,b,c) = (9,41,41). These three sides do NOT form a right triangle.
Repeat these steps for parts (b) through (d)
You should find that part (b) does lead to a true equation
a^2+b^2 = c^2
9^2+40^2 = 41^2
1681 = 1681
Therefore, a right triangle occurs when (a,b,c) = (9,40,41).
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Explanation for problem 26
We'll use the same idea as the previous problem. This time we're looking for non-right triangles.
Choice (a) can be ruled out because 3^2+4^2 = 5^2 leads to 25 = 25. Same goes for choice (d). That has been scaled up by 10.
Choice (b) gives 21^2+28^2=35^2 which boils down to 1225 = 1225, so this is ruled out as well.
Choice (c) on the other hand gives:
a^2+b^2 = c^2
14^2+19^2 = 24^2
557 = 576
Showing that a triangle with sides 14,19,24 is not a right triangle.