Answer :
[tex]\left(\frac{d}{2}\right)^2+\left(\frac{d+2}{2}\right)^2=29^2\\\\\frac{d^2}{4}+\frac{d^2+4d+4}{4}=841\ \ \ \ /\cdot4\\\\d^2+d^2+4d+4=3364\\\\2d^2+4d+4-3364=0\\\\2d^2+4d-3360=0\ \ \ \ /:2\\\\d^2+2d-1680=0[/tex]
[tex]a=1;\ b=2;\ c=-1680\\\\\Delta=b^2-4ac\to\Delta=2^2-4\cdot1\cdot(-1680)=4+6720=6724\\\\\sqrt\Delta=\sqrt{6724}=82\\\\d_1=\frac{-b-\sqrt\Delta}{2a}\to d_1=\frac{-2-82}{2\cdot1} < 0\\\\d_2=\frac{-b+\sqrt\Delta}{2a}\to d_2=\frac{-2+82}{2\cdot1}=\frac{80}{2}=40\ (cm)\\\\d=40cm;\ d+2=42cm\\\\A_r=\frac{d(d+2)}{2}\to A_r=\frac{40\cdot42}{2}=840\ (cm^2)[/tex]
[tex]a=1;\ b=2;\ c=-1680\\\\\Delta=b^2-4ac\to\Delta=2^2-4\cdot1\cdot(-1680)=4+6720=6724\\\\\sqrt\Delta=\sqrt{6724}=82\\\\d_1=\frac{-b-\sqrt\Delta}{2a}\to d_1=\frac{-2-82}{2\cdot1} < 0\\\\d_2=\frac{-b+\sqrt\Delta}{2a}\to d_2=\frac{-2+82}{2\cdot1}=\frac{80}{2}=40\ (cm)\\\\d=40cm;\ d+2=42cm\\\\A_r=\frac{d(d+2)}{2}\to A_r=\frac{40\cdot42}{2}=840\ (cm^2)[/tex]
