Question-189567

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

Commented by normans last updated on 18/Mar/23

$$\:{find}\:{the}\:{area}\:{of}\:{triangle}??\: \\ $$

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

(x/2)+2x+x+(x/2)=4x ⇔ 90° 2x ⇔ 45°=∠B x ⇔ 22.5°=∠C ⇒∠A=180°−45°−22.5°=112.5° (b/(sin ∠B))=(c/(sin ∠C))=2R Δ=((bc sin ∠A)/2) =2R^2 sin ∠A sin ∠B sin ∠C =2R^2 cos 22.5° sin 45° sin 22.5° =R^2 sin^2 45° =(R^2 /2)=(4^2 /2)=8 ✓

$$\frac{{x}}{\mathrm{2}}+\mathrm{2}{x}+{x}+\frac{{x}}{\mathrm{2}}=\mathrm{4}{x}\:\Leftrightarrow\:\mathrm{90}° \\ $$$$\mathrm{2}{x}\:\Leftrightarrow\:\mathrm{45}°=\angle{B} \\ $$$${x}\:\Leftrightarrow\:\mathrm{22}.\mathrm{5}°=\angle{C} \\ $$$$\Rightarrow\angle{A}=\mathrm{180}°−\mathrm{45}°−\mathrm{22}.\mathrm{5}°=\mathrm{112}.\mathrm{5}° \\ $$$$\frac{{b}}{\mathrm{sin}\:\angle{B}}=\frac{{c}}{\mathrm{sin}\:\angle{C}}=\mathrm{2}{R} \\ $$$$\Delta=\frac{{bc}\:\mathrm{sin}\:\angle{A}}{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{2}{R}^{\mathrm{2}} \:\mathrm{sin}\:\angle{A}\:\mathrm{sin}\:\angle{B}\:\mathrm{sin}\:\angle{C} \\ $$$$\:\:\:\:=\mathrm{2}{R}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{22}.\mathrm{5}°\:\mathrm{sin}\:\mathrm{45}°\:\mathrm{sin}\:\mathrm{22}.\mathrm{5}° \\ $$$$\:\:\:\:={R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\mathrm{45}° \\ $$$$\:\:\:\:=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{8}\:\checkmark \\ $$

Answered by HeferH last updated on 18/Mar/23