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ABCD is a rectangle and M is the midpoint of CD.The inradii of triangles ADM and ABM are 3 and 4 respectively.Then find the area of the rectangle.

Let $AD=a,DM=b$ so $AM=\sqrt{a^2+b^2}$
In a triangle ,inradius$=r=\frac{abc}{4Rs}$
where $a,b,c$ are the side lengths and R,r,s are circumradius,inradius and semiperimeter respectively.
Using this relation in the triangle $ADM$,$r=\frac{ab\sqrt{a^2+b^2}}{4(\frac{a+b+\sqrt{a^2+b^2}}{2})\frac{\sqrt{a^2+b^2}}{2}}$

So,$3a+3b+3\sqrt{a^2+b^2}=ab$
Squaring both sides and after simplification we get
$6(a+b)=18+ab$
And the area of the rectangle is $2ab$.I need to find $2ab,$I am not able to find it.Please help me.

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continuing from your method: $AB=2b$ and $AM=MB=\sqrt{a^2+b^2}$

Now $$Ar(\Delta ABM)=ab=r_{ABM}\times s_{ABM}=4\frac{(2b+2\sqrt{a^2+b^2})}{2}$$ So

$$ab=4(b+\sqrt{a^2+b^2}) \tag{1}$$ and from your equation we have

$$3a+3(b+\sqrt{a^2+b^2})=ab \tag{2}$$ $\implies$

$$3a+\frac{3ab}{4}=ab \implies 3+\frac{3b}{4}=b \implies b=12$$

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