Is it a Right Triangle?

Let the $3-4-5$ triangle be oriented so that its lower left corner is at the origin of a standard $xy$ -grid, with the side length $4$ oriented along the positive $x$ -axis and the side length $3$ oriented along the positive $y$ -axis.

Now the slope of $BC$ is clearly $-\dfrac{4}{3}.$ To find the slope of $AB,$ note first that the upper-left corner of the box with side lengths $3$ lies at $P(-3,3).$ Next, after noting that the side of length $5$ of the $3-4-5$ triangle has slope $-\dfrac{3}{4},$ we see that the sides of the box of side lengths $5$ perpendicular to this side of the triangle will have slope $\dfrac{4}{3}.$ The uppermost vertex $Q$ of the box with side lengths $5$ will then lie at $(0,3) + (3,4) = (3,7),$ and so the slope of $AB$ will be that of $PQ$ , namely $\dfrac{7 - 3}{3 - (-3)} = \dfrac{2}{3}.$ Since the slopes of $AB$ and $BC$ are not negative reciprocals of one another, we can conclude that $\angle ABC$ is not a right angle.

For completeness we should also calculate the slope of $AC.$ The lower-right vertex $R$ of the box with side lengths $4$ lies at $R(4,-4),$ and the rightmost vertex $S$ of the box of side lengths $5$ lies at $(4,0) + (3,4) = (7,4).$ The slope of $AC$ is then the slope of $RS,$ namely $\dfrac{4 - (-4)}{7 - 4} = \dfrac{8}{3}.$ Again as this value is not the negative reciprocal of the slopes of either of the other two sides of $\Delta ABC$ we can conclude that none of the interior angles of $\Delta ABC$ are $90^{\circ},$ and thus $\Delta ABC$ is not a right-angled triangle.

I did the same Brian....But I also plotted this in Geogebra and here is the accurate image with the angle s

The above diagram refers to where I am in my steps below. 4,8 and 14 refer to lengths of the side of the smaller triangle they are each near, and the rest refer to angles.

When I did this, I used a series of trig identities and triangle properties to find the sides and angles necessary to find the angles of $\triangle ABC$ . And it was done in 18 simple steps: (note: A scientific calculator is needed for this. Having a calculator that can store numbers lets you be more precise here, but rounding to 2 decimal places will still get you the correct answer)

1) $\tan^{-1}\frac{4}{3}=53.13$

2) $\tan^{-1}\frac{3}{4}=36.87$

3) $360-90-90-53.13=126.87$

4) $c^{2}=3^{2}+5^{2}-2(3)(5)(\cos 126.87) \color{#D61F06}{\Rightarrow} c=2\sqrt{13}$

5) $\frac{\sin 126.87}{2\sqrt{13}}=\frac{\sin A}{3} \color{#D61F06}{\Rightarrow} A=19.44$

6) $180-90-19.44=70.56$

The same logic used in 3-6 can be used in 7-10

7) $143.13$

8) $\sqrt{73}$

9) $16.31$

10) $73.69$

11) $180-70.56-73.69=35.75$

12) $180-143.13-16.31=20.56$

13) $180-90-20.56=69.44$

14) $5$ : This side is opposite the vertical angle of a right angle, making this a 3-4-5 right triangle and congruent to the original 3-4-5 right triangle.

15) $\cong (2) =36.87$

16) $180-90-36.87=53.13$

17) $180-69.44-53.13=57.43$

18) $180-35.75-57.43=86.82$

None of 11, 17 or 18 = 90, so the answer is $\color{#D61F06}{No}.\space\space\space\Box$

Firstly, we note that if $\triangle ABC$ were a right-angled triangle, $\angle a$ and $\angle b$ must add up to 90°. Then, we construct 2 triangles (in red) where we can determine the values of $\angle a$ and $\angle b$ .

• $\angle a$ = $\tan ^{ -1 }{ \frac { 4 }{ 6 } }=\quad 33.7°$
• $\angle b$ = $\tan ^{ -1 }{ \frac { 4 }{ 3 } }=\quad 53.1°$

Since $\angle a$ + $\angle b$ =53.1°+33.7°=86.8° $\neq$ 90°, therefore $\triangle ABC$ is not a right-angled triangle.