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Question Number 83268 by M±th+et£s last updated on 29/Feb/20

Answered by mr W last updated on 29/Feb/20

Commented by mr W last updated on 29/Feb/20

$$a(b-d)=2\times 5=10$$$$\Rightarrow ab-10=ad...\left(i\right)$$$$b(a-c)=2\times 3=6$$$$\Rightarrow ab-6=bc...\left(ii\right)$$$$cd=2\times 4=8...\left(iii\right)$$$$\left(i\right)\times \left(ii\right):$$$$(ab-10)(ab-6)=abcd$$$$(ab-10)(ab-6)=8ab$$$${\left(ab\right)}^2-24\left(ab\right)+60=0$$$$\Rightarrow ab=12+2\sqrt{21}=areaofrectangle$$$$A_{\mathrm{\Delta }DXY}=ab-(3+4+5)=2\sqrt{21}$$$$$$$$generally:$$$$(ab-2A)(ab-2C)=2Bab$$$${\left(ab\right)}^2-2(A+B+C)\left(ab\right)+4AC=0$$$$ab=A+B+C+\sqrt{{(A+B+C)}^2-4AC}$$$$A_{\mathrm{\Delta }DXY}=\sqrt{{(A+B+C)}^2-4AC}$$

Commented by M±th+et£s last updated on 29/Feb/20

$$thankyousir$$

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