1
$\begingroup$

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 ?
$\endgroup$
  • $\begingroup$ To get the answer, this might help : en.wikipedia.org/wiki/Heron%27s_formula $\endgroup$ – Amateur May 11 '14 at 19:49
  • $\begingroup$ @Amateur I used that already. But it takes a long time to solve. Given that this problem can come in an exam of 60 questions in 60 mins, I think there must be an alternate method and I am looking for it. Thanks for the hint though :) $\endgroup$ – Gaurang Tandon May 11 '14 at 19:53
  • $\begingroup$ I would go for $2T=bc\sin A$, which gives, $\sin A=4/5$. Get $\cos A = \sqrt{1-(\frac45)^2}=\frac35$. Now cosine rule and be done :) $\endgroup$ – Sawarnik May 11 '14 at 20:06
1
$\begingroup$

enter image description here

You can answer this using just Pythagorus as illustrated above.

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

$\endgroup$
  • $\begingroup$ Note height doesn't have to be in triangle area. $\endgroup$ – kingW3 May 11 '14 at 20:11
  • 1
    $\begingroup$ I assume that this way, using only Pythagoras, was the one expected by the problem poser ... (One has to consider if the height is inside or outside the triangle - but here it must be inside as we are given that $20$ is the longest side) $\endgroup$ – Hagen von Eitzen May 11 '14 at 20:11
  • $\begingroup$ @kingW3 It actually has to as the angle $\angle CAB$ is acute. $AB$ is the longest side $\implies$ $\angle C$ is the biggest angle$\implies$ other angles are acute. $\endgroup$ – user26486 May 11 '14 at 20:22
3
$\begingroup$

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 $$c=2\sqrt{65}.$$

$\endgroup$
1
$\begingroup$

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 $a=2\sqrt{65}$.


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.

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ Except that the height cannot go out of the triangle. $AB$ is the longest side $\implies$ $\angle C$ is the biggest angle$\implies$ other angles are acute. That implies that the height will never go out of the triangle. $\endgroup$ – user26486 May 11 '14 at 20:25
  • $\begingroup$ I kinda missed the point about the longest side $\endgroup$ – kingW3 May 11 '14 at 21:03
  • $\begingroup$ And yet you haven't edited your answer and left the part that claims that the unknown side could be $\sqrt{740}$ given that the height goes out of the triangle. $\endgroup$ – user26486 May 11 '14 at 21:36
  • $\begingroup$ To avoid confusion, you must edit your answer. The unknown side cannot be equal to $\sqrt{740}$. $\endgroup$ – user26486 May 11 '14 at 22:33
  • $\begingroup$ I don't really have to edit my answer,as he is practicing for an exam and on exam he can get that 20 wasn't the longest side prolly he would do it the same way.Those kind of trick questions are common on exams. $\endgroup$ – kingW3 May 12 '14 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.