An isosceles trapezoid is given with diagonal $25$ $cm$ and area $300$ $cm^2.$ Find the altitude and the midsegment.
Let $DD_1$ and $CC_1$ be the altitudes of the trapezoid through $D$ and $C,$ respectively. We have $CD=C_1D_1=b$ and $AD_1=BC_1=\dfrac{a-b}{2}.$ From the given area we can derive $(AB+CD)\cdot DD_1=600.$ How to use the diagonal? Thank you in advance! Any help would be appreciated.
I tried to apply Pythagorean theorem for triangle $ACC_1,$ but it didn't seem to help a lot.
If using coordinates, let $ABCD$ be that trapezioid with $AB||CD$ and $AB>CD$.
Let's say $A(-\frac{a}{2},0)$, $B(\frac{a}{2},0)$, $C(\frac{b}{2},h)$, $D(-\frac{b}{2},h)$, then $AC=BD=25=\sqrt{\left(\frac{a+b}{2}\right)^2+h^2}$, $\frac{AB+CD}{2}h=300=(a+b)h$, so we have only $a+b=x$ and $h$. $$\begin{cases} \frac{x^2}{4}+h^2=625,\\ xh=300 \end{cases}$$ $$\begin{cases} \frac{\left(\frac{300}{h}\right)^2}{4}+h^2=625,\\ x=\frac{300}{h} \end{cases}$$ $$\left(\frac{150}{h}\right)^2+h^2=625$$ $$150^2+h^4=625h^2$$ $$D=625^2-4\cdot 150^2=25^2(625-4*6^2)=25^2\cdot 481$$ $$h^2=\frac{625\pm 25\sqrt{481}}{2},\,x=\frac{300}{h}$$