Euclid had a triangle in mind. Find its third side.

Question

Problem:

Euclid had a triangle in mind. The triangles longest side is $20$ and another side is $10$. Its area is $80$. What is the exact length of the third side ?

My attempt:

I have read up this and this, but they don't help much except for the fact that $AB$ (longest side of length $20$) can be considered the base, and the height be derived as $8$.

Question:

1. Why does such a method, which is correct in its implementation, give an incorrect result ? (i.e. using the above method I got a right-triangle of sides $8$, $20$, and $10$ - which is impossible)
2. How do I further solve this to get the required answer ?

Answer 1

You can answer this using just Pythagorus as illustrated above.

NOTE: Mistake in my illustration above: AD = sqrt(10^2 - h^2)

Answer 2

One of the several forms of Heron's formula is $$ A=\frac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2},$$ hence $$ |a^2+b^2-c^2|=\sqrt{4a^2b^2-16A^2}.$$ With $A=80$, $a=20$, $b=10$, this becomes $$ |500-c^2|=\sqrt{160000-102400}=\sqrt{57600}=240.$$ Then either $c^2=500+240$, but then $c>a$. Or $c^2=500-240=260$, so $$c=2\sqrt{65}.$$

Answer 3

It is correct you draw height and you get that height corresponding to side 20 is equal to 8 now you have a right-triangle with side 8,10 and x(note: x is a part of 20) from this we get x is equal to 6 and that means that the the second part of the side of 20 is 14 from that you can get side $=\sqrt{260}$ and if the height goes out of the triangle you get that it's 20+6 and from that you can get it is equal $=\sqrt{740}$(since the longest side is 20 this can not be the case)

Answer 4

A very simple different approach is to first find the $\sin$ of the angle between the sides of the lengths $10$ and $20$. Then the area formula tells us that ($\alpha$ denotes the angle between those sides) $$A=80=\frac{1}{2}\cdot10\cdot20\cdot \sin\alpha=100\cdot \sin\alpha\implies \sin\alpha=0.8$$

Then by using the fact that $\cos\alpha=\pm\sqrt{1-\sin^2\alpha}$ and that $\alpha$ is acute, since the side of the length $20$ is the longest one, we further have that $$\cos\alpha=\sqrt{1-\sin^2\alpha}=0.6$$

By using the Law of Cosines and the fact that $a$ is positive, where $a$ is the unknown side of the triangle, we have that $$a=\sqrt{10^2+20^2-2\cdot10\cdot20\cdot0.6}=2\sqrt{65}$$

Hence the unknown side is $a=2\sqrt{65}$.

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A proof that the angle $\alpha$ is acute goes as follows:

A well-known fact proven by Euclid shows that a greater angle in any triangle is opposite to a longer side, and vice versa. Hence, in this case, the longest side, that is, the side of length $20$, is opposite to the biggest angle (let it be $\beta$).

Now assume that $\alpha\ge 90^{\circ}$. Then, since $\beta$ is the largest of the bunch, we have that $\beta\ge\alpha\ge 90^{\circ}$ (I assume that 'the longest side' means that the other sides could be of the same length as the largest one, but could not be greater). Call the unnamed angle $\gamma$. Then $\gamma>0^{\circ}$ and we therefore have that $\alpha+\beta+\gamma>180^{\circ}$, which is a contradiction, since all the angles of any triangle in Euclidean geometry always add up to exactly $180^{\circ}$. Hence $\alpha<90^{\circ}$, i.e., $\alpha$ is acute.

This proof may sound obvious, but wanted to point it out anyway.