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Question Number 185077 by emmanuelson123 last updated on 16/Jan/23

Answered by mahdipoor last updated on 16/Jan/23

$$getsideofrectangleareA,B$$$$\mathrm{\Delta }1\equiv \mathrm{\Delta }2\equiv \mathrm{\Delta }3\Rightarrow$$$$sideof\mathrm{\Delta }1:A,B,\sqrt{A^2+B^2}$$$$and{r}_1:\frac{1}{2}[(A+B)-\sqrt{A^2+B^2}]$$$$sideof\mathrm{\Delta }2:\frac{AB}{\sqrt{A^2+B^2}},\frac{B^2}{\sqrt{A^2+B^3}},B$$$$and{r}_2:\frac{1}{2}[(A+B)-\sqrt{A^2+B^2}]\times \frac{B}{\sqrt{A^2+B^2}}$$$$sideof\mathrm{\Delta }3:\frac{A^2}{\sqrt{A^2+B^2}},\frac{AB}{\sqrt{A^2+B^3}},A$$$$and{r}_2:\frac{1}{2}[(A+B)-\sqrt{A^2+B^2}]\times \frac{A}{\sqrt{A^2+B^2}}$$$$weknow:$$$$\mathrm{\Sigma }r=120=\frac{1}{2}[A+B-\sqrt{A^2+B^2}]\times [\frac{A+B}{\sqrt{A^2+B^2}}+1]$$$$=\frac{{(A+B)}^2-(A^2+B^2)}{2\sqrt{A^3+B^2}}=\frac{AB}{\sqrt{A^2+B^2}}$$$$and64=\frac{A^2}{\sqrt{A^2+B^2}}$$$$\Rightarrow \Rightarrow B=15\sqrt{289}A=8\sqrt{289}$$$$\Rightarrow \Rightarrow S=AB=15\times 8\times 289$$

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