Which of the following sets of sides could NOT form a right triangle: a 3 — 4 — 5
b 6 — 8 — 10
c 8 — 14 — 17
d 9 — 40 — 41

Explanation:

Choice A is a non-answer because 3-4-5 does represent a right triangle. We can see so from the Pythagorean theorem

a^2+b^2 = c^2

3^2+4^2 = 5^2

9+16 = 25

25 = 25

We get a true equation, so (a,b,c) = (3,4,5) is a solution to a^2+b^2 = c^2

This confirms a 3-4-5 triangle is a right triangle.

The same steps can be used to show 6-8-10 is also a right triangle (note how each value has been doubled compared to 3-4-5). We can cross choice B off the list.

We can also see that choice D is a non-answer as well.

Choice C is the only thing left. The proper Pythagorean triple is actually 8-15-17.

Let's see what happens when a = 8, b = 14, c = 17

a^2+b^2 = c^2

8^2+14^2 = 17^2

64+196 = 289

260 = 289 false equation

We get a false equation, so a = 8, b = 14, c = 17 does not form a right triangle.

darn right!

step-by-step explanation:

steve kerr's sons name is let that skin in.