In rectangle $ABCD$ a circle is drawn passing through point $C$ and touching the sides $AB$ and $AD$ at points $M$ and $N$ respectively. The length of altitude from $C$ To $MN$ is $5$ units. Find the area of rectangle .

I tried using all possible methods like coordinate geometry and trigonometry but could not succeed. Please help. Thanks in advance

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Let $O$ be a center of the circle, $AD=a$, $AB=b$ and $ON=r$.

Thus, by the Pythagoras theorem $$OC^2=ND^2+MB^2$$ or $$r^2=(a-r)^2+(b-r)^2,$$ which gives $$r=a+b-\sqrt{2ab},$$ $$ND=a-r=\sqrt{2ab}-b,$$ $$MB=b-r=\sqrt{2ab}-a$$ and $$MN=\sqrt(a+b-\sqrt{2ab}).$$ Thus, $$\frac{5\cdot\sqrt2(a+b-\sqrt{2ab})}{2}=S_{\Delta MNC}=S_{ABCD}-S_{\Delta MNA}-S_{\Delta DNC}-S_{\Delta MBC}=$$ $$=\frac{2ab-(a+b-\sqrt{2ab})^2-(\sqrt{2ab}-b)b-(\sqrt{2ab}-a)a}{2}=$$ $$=\frac{a\sqrt{2ab}+b\sqrt{2ab}-2ab}{2}=\frac{\sqrt{2ab}(a+b-\sqrt{2ab})}{2},$$ which gives $$ab=25$$ and $$S_{ABCD}=25.$$ Done!


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