# Area of an isosceles triangle with height $7$ and perimeter $18+8\sqrt{2}$

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.

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

## 2 Answers

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;

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

• 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. – IGotAQuestion Apr 16 '18 at 20:00
• @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. – PhysicsGuy123 Apr 17 '18 at 6:14
• @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. – PhysicsGuy123 Apr 17 '18 at 15:40

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 $$A=7b.$$