My hypotenuse is 1 and the area is .125... can someone find out how to get the base + height when just given the hypotenuse and area of a triangle?
I know finding the area of a triangle is 1/2base(height), but since I have neither of those, I can't use that formula. I saw a formula somewhere that showed that all you had to do to figure out the base + height was solve b+h= square root(hypotenuse squared plus 4 times the area) but that doesn't seem right. Do y'all have any formulas you think can solve this?
... Would the 90 degrees from the right triangle help me solve it?
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$\begingroup$First let us write down the data provided.Lets take hypotenuse of the triangle as H and the base and the height as b and h. Now we have two equations for this problem.$$ Area= \frac{bh}{2} $$$$ H^2=b^2+h^2$$
Now using the first equation,$$ Area=\frac{bh}{2} $$$$ 0.125=\frac{bh}{2} $$$$ 0.25=bh $$$$ \frac{1}{4}=bh $$
$$ h= {1\over 4b}$$
Now substitute h for the second equation.$$ H^2=h^2+b^2 $$$$ 1=h^2+b^2 $$$$ 1=\left({1\over 4b}\right)^2+b^2 $$and thus we get a nice quadratic equation $16b^4-16b^2+1=0$
If we solve for b,we get:$$ b=\pm\sqrt{2\pm\sqrt 3}\over 2 $$ as b is a breadth and physical quantities cannot be negative, $$ b=\sqrt{2\pm\sqrt 3}\over 2 $$and then,b =0.9656 or b=0.2598When values of b is substituted to $ h= {1\over 4b},$
h=0.258 0r h=0.96227
$\endgroup$ 1 $\begingroup$Let make a general formula for this. Let $A$ be area and $H$ be hypotenuse. Let $b$ and $h$ be base and height of right angled triangle.
$ \frac{hb}{2}=A , h^2+b^2=H^2 $
$$ h^2+b^2+2bh=2A+H^2 $$
$$ (h+b)=\sqrt{2A+H^2} $$
We know from the first equation I wrote $ b=\frac{2A}{h} $.
Thus the equation reduces to become $ \frac{2A}{h}+h=\sqrt{2A+H^2} $
Multiplying both sides by h,$$ h^2 - h\sqrt{2A+H^2}+2A=0 $$
Thus the formula is made. Put the values and get the answer
$\endgroup$ 1 $\begingroup$Given the length of the hypotenuse $|AB|=c=1$ and the area$S=\tfrac18$ of the right-angled triangle $ABC$, find $|AC|$ and$|BC|$.
Since $\angle C=90^\circ$, the point $C$ must be located on the circle with diameter $AB$.
Using well-known formula for the area of triangle\begin{align}
S&=\tfrac12\,|AB|\cdot|CD|
=\tfrac12\,c\,h_c
,
\end{align}
we can find that\begin{align}
|CD|=h_c
&=\frac{2S}c=
\tfrac14
.
\end{align}
And we also have three more right-angled triangles:$DOC$, $DCA$ and $DBC$, so we can easily find that
\begin{align} \triangle DOC:\quad |OD|&=\sqrt{\tfrac{c^2}4-h_c^2} ,\\ |AD|&=\tfrac c2-|OD| ,\\ |BD|&=c-|AD| ,\\ \triangle DCA:\quad |AC|&=\sqrt{|AD|^2+h_c^2} ,\\ \triangle DCA:\quad |AB|&=\sqrt{|BD|^2+h_c^2} . \end{align}
Thus the answer can be expressed in terms of given $c,S$ as a pair\begin{align} \tfrac12\sqrt{2c^2\pm2\sqrt{c^4-16S^2}} \end{align}which for $c=1,\ S=\tfrac18$ becomes
\begin{align} \tfrac14\Big(\sqrt6\pm\sqrt2\Big) . \end{align}
$\endgroup$ $\begingroup$Algebraic derivation solution when $ (h^2, A)=\text{ (hypotenuse$^2$, Area) } $ are given is suggested. We derive a formula.
Let the sides be $(a,b)$.( Base and height in your notation in a right triangle).
$$ ab/2= A, 4A=2 ab;\; h^2=a^2+b^2 ;\tag1 $$
Add, subtract separately and find root
$$ a+b=\sqrt{h^2+4A};\quad a-b=\sqrt{h^2-4A} ;\tag 2 $$Half sum and half difference for sides.$$ \{2a=\sqrt{h^2+4A}+\sqrt{h^2-4A},\quad 2b=\sqrt{h^2+4A}-\sqrt{h^2-4A}\}\tag 3$$Plug in the values for specific cases, $ A= \frac18, h=1;$$$ (a,b)= \frac{\sqrt{6}\pm\sqrt{2}}{4}.$$
$\endgroup$ $\begingroup$Supposing the hypotenuse is not the base, let $b$ be the length of the base and $h$ the height.
As the triangle is right, we have $b^2+h^2=1$
We learn the area is $0.125$ so we have $\frac{1}{2}bh = 0.125$
Adding four times this second equation to the first, we have $b^2+2bh+h^2=1.5$
Factoring, $(b+h)^2=1.5$ and so $b+h=\frac{\sqrt{6}}{2}$
Going back to the second, we have $b=\frac{1}{4h}$ so we have $\frac{1}{4h}+h=\frac{\sqrt{6}}{2}$
Rearranging, $h^2-\frac{\sqrt{6}}{2}h+\frac{1}{4}=0$ and from this we can use our familiar quadratic formula to solve for $h$ and then substitute in to solve for $b$.
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