8.7k views
1 vote
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

User Hiura
by
9.3k points

1 Answer

5 votes

Answer: C. 8-14-17

===================================================

Step-by-step 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.

User Sam Kingston
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories