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Question Number 187819 by normans last updated on 22/Feb/23

Answered by mr W last updated on 22/Feb/23

Commented by mr W last updated on 22/Feb/23

$$a=10$$$$b=8$$$$\cos \alpha =\frac{2c}{a+b}$$$${ED}^2=a^2+c^2-2ac\times \frac{2c}{a+b}$$$${BD}^2=a^2+4c^2-4ac\times \frac{2c}{a+b}$$$${ED}^2+{BD}^2=c^2$$$$a^2+c^2-2ac\times \frac{2c}{a+b}+a^2+4c^2-4ac\times \frac{2c}{a+b}=c^2$$$$c^2=\frac{a^2(a+b)}{2(2a-b)}$$$$\Rightarrow c=a\sqrt{\frac{a+b}{2(2a-b)}}$$$$BC=\sqrt{{(a+b)}^2-{\left(2c\right)}^2}=\sqrt{\frac{b(a^2-b^2)}{2a-b}}$$$$areaoftriangleABC:$$$$\mathrm{\Delta }=\frac{\left(2c\right)\times BC}{2}=a\sqrt{\frac{a+b}{2(2a-b)}}\sqrt{\frac{b(a^2-b^2)}{2a-b}}$$$$\mathrm{\Delta }=\frac{a(a+b)}{2a-b}\sqrt{\frac{b(a-b)}{2}}$$$$\mathrm{\Delta }=\frac{10\times 18}{12}\sqrt{\frac{8\times 2}{2}}=30\sqrt{2}✓$$

Commented by normans last updated on 22/Feb/23

$$nicesolution,thankyousir$$

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