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Question Number 176437 by cherokeesay last updated on 19/Sep/22

Answered by HeferH last updated on 19/Sep/22

$$bysimilartriangles:$$$$b=AD=\frac{8\sqrt{5}}{2+2\sqrt{5}}=\frac{\left(4\sqrt{5}\right)}{(1+\sqrt{5})}$$$$h=4$$$$A_{△ABD}=\frac{bh}{2}=\frac{8\sqrt{5}}{\sqrt{5}+1}=10-2\sqrt{5}$$$$$$

Commented by cherokeesay last updated on 19/Sep/22

$$nice,thankyou!$$

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