Given squares $M$ and $N$ are inscribed in a right triangle $ABC.$ If area$(M)=441$ and area$(N)=440,$ then find area of the triangle $ABC.$

Question

Given squares $M$ and $N$ are inscribed in right triangle $ABC$ as shown in fig. If the area of $M$ is $441$ and area of $N$ is $440$, then find the area of the triangle $ABC.$

My attempt: I did some lengthy calculation and I ended up with the equation

$k^3-882k^2-400k+400×441=0$

Where $k=ab, a=BC, b=AC$

Answer 1

Let $x$ and $y$ be a sides-lengths of the squares with areas $M$ and $N$ respectively.

Thus, by the similarity we obtain: $$\frac{x}{b}=\frac{a-x}{a}$$ and $$\frac{\frac{ab}{\sqrt{a^2+b^2}}-y}{\frac{ab}{\sqrt{a^2+b^2}}}=\frac{y}{\sqrt{a^2+b^2}},$$ which gives $$x=\frac{ab}{a+b}$$ and $$y=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2}.$$ Thus, $$a+b=\frac{k}{21}$$ and we obtain: $$\frac{k\sqrt{\frac{k^2}{441}-2k}}{\frac{k^2}{441}-k}=\sqrt{440}$$ or $$\frac{\sqrt{\frac{k^2}{441}-2k}}{\frac{k}{441}-1}=\sqrt{440}$$ or $$441(k^2-2\cdot441k)=440(k-441)^2$$ or $$441(k-441)^2-441^3=440(k-441)^2$$ or $$(k-441)^2=\left(21^3\right)^2,$$ which gives $$k-441=21^3$$ or $$k=9702$$ and the area is equal to $4851.$

Answer 2

We deduce easily (from the left triangle) that $a+b=ab/\sqrt{M}$ hence $$(a+b)^2=\frac{(ab)^2}{M}.$$

Now consider the right triangle; its area is $ab/2$. Let $c^2=a^2+b^2$. It consists of three triangles and a square. The area of the upper left triangle equals due to similarity $$\frac12ab\frac{N}{a^2},$$ analogue for the other two triangles. Let's sum up: $$\frac12ab\frac{N}{a^2}+\frac12ab\frac{N}{b^2}+\frac12ab\frac{N}{c^2}+N=\frac12ab$$ that is $$N\left(\frac{b}{a}+\frac ab+\frac{ab}{a^2+b^2} \right)+2N=ab.$$ Observe that $$\frac{b}{a}+\frac ab=\frac{a^2+b^2}{ab}=\frac{(a+b)^2}{ab}-2=\frac{ab}{M}-2$$ exploiting $(a+b)^2=\frac{(ab)^2}{M}$.

Solving the quadratic $$N\left(\frac{ab}{M}-2+\frac{1}{\frac{ab}{M}-2}\right)+2N=ab$$ for $ab$ gives the positive solution $$ab=M-\frac{M^2}{\sqrt{M(M-N)}}.$$

In the special case $N=M-1$ we arrive in $ab=M(1+\sqrt M)$ so if $M=441$ we get $ab=9702$.