# How to Find the Length of the Hypotenuse of a Right Triangle?

• MHB
• DrLiangMath
In summary, at triangle ABC, angle BCA is C. Angle CDA is 135 - C. AC.AB=3, DC=5, ∠ CAD=$45^o$, AB ⊥ BC. Find the length of AC.
DrLiangMath
AB=3, DC=5, CAD=$45^o$, AB ⊥ BC. Find the length of AC.

MathTutoringByDrLiang said:
AB=3, DC=5, CAD=$45^o$, AB ⊥ BC. Find the length of AC.

View attachment 11867
It's inefficient but start by defining angle BCA to be C. Then angle CDA = 135 - C. Call x = AC. Then by Law of Sines
$$\displaystyle \dfrac{sin(90)}{x} = \dfrac{sin(45)}{5}$$

$$\displaystyle \dfrac{sin(135 - C)}{x} = \dfrac{sin(45)}{5}$$

This gives a biquadratic in x. But, as I said, it's inefficient. There is probably a better approach.

-Dan

Thank you for your response. But we still need one more equation since we have 2 variables x and C.

MathTutoringByDrLiang said:
Thank you for your response. But we still need one more equation since we have 2 variables x and C.
I gave two equations. The first is the Law of Sines in triangle ABC and the second is the Law of Sines in triangle ADC. Perhaps I should have stated that.

-Dan

Hey MathTutoringByDrLiang,

Are you actually looking for an answer to this problem? Or do you have an elegant answer to share?
If the latter, then we might move this thread to the https://mathhelpboards.com/forums/challenge-questions-and-puzzles.28/ subforum.

Btw, at this time I could provide an answer, but it's rather long winded and I'm still looking for a more elegant solution.

Here's the best I could find.
\begin{tikzpicture}
%preamble \usetikzlibrary {angles,quotes}
\coordinate[label=left:A] (A) at (0,3);
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=right:C] (C) at (6,0);
\coordinate[label=below: D] (D) at (1,0);
\draw (B) rectangle +(0.2,0.2);
\draw[thick] (A) -- (B) -- (C) -- cycle (A) -- (D);
\path (A) -- node[ left ] {3} (B) -- node[below] {$y$} (D) -- node[below] {5} (C) -- node[above right] {$x$} (A);
\pic ["$45^\circ$", draw, angle radius=0.4cm, angle eccentricity=2.2,pic text options={shift={(0.1,0)}}] {angle = D--A--C};
\pic ["$135^\circ-C$", draw, angle radius=0.3cm, angle eccentricity=2.2,pic text options={shift={(0.5,0)}}] {angle = C--D--A};
\end{tikzpicture}
Let $x=AC$, which is what we want to find.
Let $y=BD$.

From the law of sines we have:
$$\frac 5{\sin 45}=\frac x{\sin(135-C)} \implies x=5\cdot\frac{\sin(135-C)}{\sin 45}=5\cdot\frac{\sin 135\cos C -\cos 135\sin C}{\sin 45}=5\cdot\frac{\sin 45\cos C +\cos 45\sin C}{\sin 45}=5(\cos C+\sin C) \tag 1$$
From the definitions of cosine and sine:
$$\cos C=\frac{BD+5}{x}=\frac{y+5}{x} \tag 2$$
$$\sin C=\frac 3{x} \tag 3$$
From Pythagoras:
$$x^2 = (y + 5)^2 + 3^2 \tag 4$$

Substitute (2) and (3) into (1) to find:
$$x=5\left(\frac{y+5}x+\frac 3x\right) \implies x^2=5(y+8)\tag 5$$
Substitute in (4):
$$5(y+8) = (y + 5)^2 + 3^2 \implies y^2+5y-6= (y-1)(y+6) = 0 \implies y =1 \tag 6$$
Substitute back into (5):
$$x^2=5(1+8)=45 \implies x=3\sqrt 5$$

Last edited:
Klaas van Aarsen said:
Hey MathTutoringByDrLiang,

Are you actually looking for an answer to this problem? Or do you have an elegant answer to share?
If the latter, then we might move this thread to the https://mathhelpboards.com/forums/challenge-questions-and-puzzles.28/ subforum.

Btw, at this time I could provide an answer, but it's rather long winded and I'm still looking for a more elegant solution.

I do have a solution to share. Different form the solution given by Klaas van Aarsen, my solution is based on creating new triangles. If you don't mind, I would like post it here for your reference:

Just for fun, I've created a TikZ diagram that corresponds to your video.
TikZ is natively supported on this site similar to formulas. Click on the picture to see the $\LaTeX$ code.
\begin{tikzpicture}[scale=2]
%preamble \usepackage{tkz-euclide}
\tkzDefPoint(0,3){A}
\tkzDefPoint(0,0){B}
\tkzDefPoint(6,0){C}
\tkzDefPoint(1,0){D}

\tkzDrawSegments[ultra thick](A,B B,C C,A A,D)

\tkzDefPointWith[orthogonal,K=-1](D,A)
\tkzGetPoint{E}
\tkzDefPointBy[projection=onto B--C](E)
\tkzGetPoint{F}
\tkzDrawSegments[help lines](D,E E,F)

\tkzLabelPoints[ left ](A,B)
\tkzLabelPoints[ right ](C)
\tkzLabelPoints[ below ](D)
\tkzLabelPoints[ below,help lines ](F)
\tkzLabelPoints[ above right,help lines ](E)
\tkzLabelSegment(B,A){3}
\tkzLabelSegment(C,D){5}
\tkzLabelSegment(D,B){$x$}
\tkzLabelSegment[help lines](E,F){$x$}
\tkzLabelSegment[help lines](D,F){3}
\tkzLabelSegment[help lines](F,C){2}

\tkzLabelAngles[pos=1](D,A,C){$45^\circ$}
\tkzMarkRightAngles(A,B,C)
\tkzMarkAngles[size=0.7,mark=none](D,A,C)

\tkzLabelAngles[pos=1,help lines](A,E,D){$45^\circ$}
\tkzMarkAngles[size=0.7,mark=none,help lines](A,E,D)
\tkzMarkAngles[size=0.6,mark=||,help lines](B,A,D F,D,E)
\tkzMarkRightAngles[help lines](A,D,E E,F,D)
\tkzMarkSegments[mark=|,help lines](A,D D,E)
\end{tikzpicture}

Wow, it looks so nice! Thank you!

## 1. What is the Pythagorean theorem and how does it relate to finding the length of the hypotenuse?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This means that if we know the lengths of two sides of a right triangle, we can use the Pythagorean theorem to find the length of the hypotenuse.

## 2. How do I use the Pythagorean theorem to find the length of the hypotenuse?

To use the Pythagorean theorem, you will need to know the lengths of two sides of the right triangle. Once you have these values, you can plug them into the formula a2 + b2 = c2, where c represents the length of the hypotenuse. Then, solve for c by taking the square root of both sides of the equation.

## 3. Can I use the Pythagorean theorem to find the length of the hypotenuse if I only know the length of one side?

No, the Pythagorean theorem can only be used to find the length of the hypotenuse if you know the lengths of two sides of the triangle. If you only know the length of one side, you will need to use other geometric principles or measurements to find the length of the hypotenuse.

## 4. Are there any other methods for finding the length of the hypotenuse of a right triangle?

Yes, there are other methods for finding the length of the hypotenuse, such as using trigonometric functions or applying the Law of Cosines. These methods may be more complex and require additional information about the triangle, but they can also be useful in certain situations.

## 5. Can I use the Pythagorean theorem to find the length of the hypotenuse in a non-right triangle?

No, the Pythagorean theorem only applies to right triangles. In a non-right triangle, you will need to use other methods, such as the Law of Cosines or the Law of Sines, to find the length of the hypotenuse.

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