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Question Number 198477 by a.lgnaoui last updated on 20/Oct/23

$$\mathrm{find}\mathrm{the}\mathrm{Area}\left(\mathrm{ABCD}\right)$$

Commented by a.lgnaoui last updated on 20/Oct/23

Commented by mr W last updated on 21/Oct/23

$$Area\left(ABCD\right)=\frac{x^2-y^2}{4}$$

Answered by mr W last updated on 21/Oct/23

Commented by mr W last updated on 21/Oct/23

$$b^2=d^2+f^2...\left(i\right)$$$$a^2=c^2+{(f+c+d)}^2=c^2+f^2+c^2+d^2+2fc+2fd+2cd...\left(ii\right)$$$$\left(ii\right)-\left(i\right):$$$$a^2-b^2=2c^2+2fc+2fd+2cd$$$$\Rightarrow c^2+fc+fd+cd=\frac{a^2-b^2}{2}$$$$\left[ABCD\right]=\frac{df}{2}+\frac{c(f+c+d)}{2}$$$$=\frac{c^2+df+cf+cd}{2}=\frac{a^2-b^2}{4}✓$$

Commented by a.lgnaoui last updated on 21/Oct/23

$$\mathrm{exactly}\mathrm{thanks}$$

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