0
$\begingroup$

An isosceles triangle has a height of $7$m from the base. The perimeter measures $18 + 8\sqrt{2}$. What is the area?

I have tried applying Pythagoras' Theorem, the law of cosines and sines, this law https://sv.m.wikipedia.org/wiki/Areasatsen and base times height = Area with no success.

Thanks in advance.

$\endgroup$
  • $\begingroup$ If its laid isosceles triangle, its complicate solution. $\endgroup$ – Takahiro Waki Apr 17 '18 at 16:01
1
$\begingroup$

As Hagen von Eitzen says, you can split the triangle into 2 right-angled triangles. As it is isosceles, we can say the original triangle will have 2 sides of length $a$, and $1$ side of length $b$. From this, we can work out;

$18+8 \sqrt{2}=2a+b$.

Now, if the base is length $b$, we can use the height to form our right-angled triangles, made up of lengths $\frac{b}{2}$, $a$ and $7$. Given $a^2+b^2=c^2$, $(\frac{b}{2})^2+7^2=a^2 \rightarrow \frac{b^2}{4}+49=a^2 \rightarrow b^2=4a^2-196 \rightarrow b=\sqrt{4a^2-196}$.

By putting this into the first equation, we get;

$18+8\sqrt{2}=2a+\sqrt{4a^2-196}$. I take away $2a$ from both sides of this equation to form;

$18+8\sqrt{2}-2a=\sqrt{4a^2-196}$. I square both sides of the equation to form;

$(18+8\sqrt{2}-2a)^2=4a^2-196$. If I expand the brackets, I get;

$4a^2-72a+452+288\sqrt{2}-32a\sqrt{2}=4a^2-196$. This is the same as;

$72a+32\sqrt{2}(a)=648+288\sqrt{2} \rightarrow a(72+32\sqrt{2})=648+288\sqrt{2}\rightarrow a=\frac{648+288\sqrt{2}}{72+32\sqrt{2}}\rightarrow a=9$.

Now take this back to the original equation, and you get;

$18+8\sqrt{2}=2(9)+b \rightarrow b=8\sqrt{2}$.

We now simply use half base times height to get the area, which is $7\times \frac{8\sqrt{2}}{2}=28\sqrt{2}$.

So, there is your answer - the area is $28\sqrt{2}$.

EDIT

I've just spotted an even simpler way of calculating the value of a. Once you know $b^2=4a^2-196$, you can rearrange this to form;

$b^2-4a^2=-196$. The difference between two squares means this is the same as;

$(b-2a)(b+2a)=-196$. As we know $b+2a=18+8\sqrt{2}$, this means;

$(b-2a)(18+8\sqrt{2})=-196 \rightarrow b = 2a-\frac{196}{18+8\sqrt{2}}$. By putting this back into the original equation, you get;

$18+8\sqrt{2}=2a+(2a-\frac{196}{18+8\sqrt{2}}) \rightarrow 4a = 18+8\sqrt{2}+\frac{196}{18+8\sqrt{2}} \rightarrow 4a=36 \rightarrow a=9$.

$\endgroup$
  • $\begingroup$ What enabled you to ”skip” the middle term when squared both sides of the equal sign and then expanded by the help of the binominal theorm? Step 5-6 if counted from the top. $\endgroup$ – IGotAQuestion Apr 16 '18 at 20:00
  • $\begingroup$ @IGotAQuestion - To be honest, I'm not sure. I was always taught you could do this if the numbers weren't in brackets, but the more I think about it, the less it makes sense (for example $3+9=a \rightarrow 12^2=a^2 \rightarrow a^2=144$. Which is a totally different answer to: $9^2+3^2=a^2 \rightarrow a^2=90$). Given the answer above looks reasonable, I am assuming it is still correct, but I can't be sure now. I think we will need someone else to clarify whether this is correct or not. Sorry. $\endgroup$ – PhysicsGuy123 Apr 17 '18 at 6:14
  • $\begingroup$ @IGotAQuestion. I have worked it out the long way now and the answer remains the same. I'm still not sure if this is a coincidence or whether the original way is a valid way of calculating the result. Anyway, I have edited the answer so you can see how I worked it out. $\endgroup$ – PhysicsGuy123 Apr 17 '18 at 15:40
0
$\begingroup$

Half the isosceles triangle is a right triangle with legs $7$ and $b$, and hypotenuses $c=9+4\sqrt 2-b$. Pythagoras says $$ 7^2+b^2 = (9+4\sqrt 2-b)^2=(9+4\sqrt 2)^2+2(9+4\sqrt 2) b + b^2.$$ As $b^2$ cancels, you obtain $$ b=\frac{7^2-(9+4\sqrt 2)^2}{2(9+4\sqrt 2)}.$$ From there you get the area of the isosceles triangle as $$ A=7b.$$

$\endgroup$

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.