M17-04

If the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then:

For a right triangle: \(a^2 +b^2= c^2\).
For an acute (a triangle that has all angles less than 90°) triangle: \(a^2 +b^2>c^2\).
For an obtuse (a triangle that has an angle greater than 90°) triangle: \(a^2 +b^2<c^2\).

As for your other question: you don't need to know any formula. It's enough to know that you CAN find the angles because the triangle will be fixed.

Hi Bunuel,

I did not understand how this condition is true for a right triangle. For right angle triangle it should be AC^2+AB^2=BC^2. Am i missing anything here?

(2) AC^2 + AB^2 \gt BC^2. This condition will hold for equilateral triangle (all angles are 60 degrees) as well as for a right triangle with right angle at B or C (so in this case one angle will be 90 degrees, for example consider a right triangle: AC=5, AB=4 and BC=3). Not sufficient.

Thank you

AC^2+AB^2=BC^2 is true for a right triangle where BC is hypotenuse. Think what would it be if for example, AC is hypotenuse.

We can solve this in another way

1. Lets write all the sides with respect to one side
2AB=3BC=4AC
AB=3/2BC
AC=3/4BC

Now use AB^2 + AC^2 = BC^2 and substitute in this u will get LHS is greater than BC^2 So it is obviously obtuse triangle.

2. AC^2+AB^2>BC^2 As per this statement we have no information about which side is hypotenuse. so This is clearly insufficient.

The explanation says that we can find the angles no matter what they are. I get that - when given the ratios of three sides all the triangles with that ratio are similar. What I don't understand is why we don't need to know the angles. The question asks weather all than angles are less than 90 degrees. Here's an example:

Ratio of three sides 1:1:1

All angles are 60 degrees ANSWER YES

Ratio of three sides 3:4:5

One angle is 90 degrees ANSWER NO

With trigonometry we could find find the actual angles, but without it we couldn't tell decisively if one of the angles is equal to or greater than 90 degrees.
I'm pretty sure that I'm missing an identity somewhere, but don't know what it is!

The ratio of the sides can only be \(AB:BC:AC=6:4:3\). It cannot be 1:1:1 or 3:4:5. Does those ratio satisfy \(2AB = 3BC = 4AC\)? No.

ALL triangles with this ratio (\(AB:BC:AC=6:4:3\)) are similar and have the same fixed angles. No matter whether they are 6, 4, and 3 or 12, 8, and 6, etc They all will have the same angles.

Hi Bunuel, can you please help me understand how you got the ratio to be 6:4:3? I am having difficulty in calculating that.

Given: \(2AB = 3BC = 4AC\).

The least common multiple of 2, 3, and 4 is 12.

Divide by 12: \(AB/6 = BC/4 = AC/3\).

\(AB:BC:AC=6:4:3\).

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.