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Question Number 189567 by normans last updated on 18/Mar/23

Commented by normans last updated on 18/Mar/23

$$findtheareaoftriangle??$$

Answered by mr W last updated on 18/Mar/23

$$\frac{x}{2}+2x+x+\frac{x}{2}=4x\iff 90°$$$$2x\iff 45°=\mathrm{\angle }B$$$$x\iff 22.5°=\mathrm{\angle }C$$$$\Rightarrow \mathrm{\angle }A=180°-45°-22.5°=112.5°$$$$\frac{b}{\sin \mathrm{\angle }B}=\frac{c}{\sin \mathrm{\angle }C}=2R$$$$\mathrm{\Delta }=\frac{bc\sin \mathrm{\angle }A}{2}$$$$=2R^2\sin \mathrm{\angle }A\sin \mathrm{\angle }B\sin \mathrm{\angle }C$$$$=2R^2\cos 22.5°\sin 45°\sin 22.5°$$$$=R^2{\sin }^{2}45°$$$$=\frac{R^2}{2}=\frac{4^2}{2}=8✓$$

Answered by HeferH last updated on 18/Mar/23

Commented by HeferH last updated on 18/Mar/23

$$\mathrm{A}+\mathrm{B}=\frac{4\times 4}{2}={8\mathrm{u}}^2$$

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