[tex]side \ length : \ a=29 \ cm \\ one \ diagonal: \ d \\the \ second \ diagonal: \ e=d+2 \\\\ use \ the \ Pythagorean \ Theorem\\\\\left(\frac{d}{2}\right)^2+\left(\frac{e}{2}\right)^2=a^2\\\\\left(\frac{d}{2}\right)^2+\left(\frac{d+2}{2}\right)^2=29^2[/tex]
[tex]\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=\Delta=2^2-4\cdot1\cdot(-1680)=4+6720=6724\\\\\sqrt\Delta=\sqrt{6724}=82\\\\d_1=\frac{-b-\sqrt\Delta}{2a} =\frac{-2-\sqrt{6724}}{2 }=\frac{-2-82}{2}=-\frac{84}{2}=-42 \\ \\d_{1} < 0[/tex]
[tex]d_2=\frac{-b+\sqrt\Delta}{2a} =\frac{-2+\sqrt{6724}}{2}=\frac{-2+82}{2 }=\frac{80}{2}=40\ cm \\\\d=40cm \\ e=d+2=40+2=42 \ cm\ \\\\Area \ a \ rhombus: \\\\ A =\frac{1}{2}d\cdot e \\ \\ A =\frac{1}{2}\cdot 40 \cdot 42 =20\cdot 42=840\ cm^2 Answer: \ Area \ a \ rhombus \ is \ 840 \ cm^2.[/tex]
