Given vertices $A(2,3)A(2,\; 3)$A(2,3), $B(3,0)B(3,\; 0)$B(3,0), and $C(9,2)C(9,\; 2)$C(9,2):

1. Area: Length $AB=\sqrt{(3-2{)}^2+(0-3{)}^2}=\sqrt{10}AB\; =\; \backslash sqrt\{(3\; –\; 2)^2\; +\; (0\; –\; 3)^2\}\; =\; \backslash sqrt\{10\}$AB=(3−2)2+(0−3)2​=10​. Width $AC=\sqrt{(9-2{)}^2+(2-3{)}^2}=\sqrt{50}AC\; =\; \backslash sqrt\{(9\; –\; 2)^2\; +\; (2\; –\; 3)^2\}\; =\; \backslash sqrt\{50\}$AC=(9−2)2+(2−3)2​=50​. Area $=Length\times Width=\sqrt{10}\times \sqrt{50}=10\sqrt{5}=\; \backslash text\{Length\}\; \backslash times\; \backslash text\{Width\}\; =\; \backslash sqrt\{10\}\; \backslash times\; \backslash sqrt\{50\}\; =\; 10\backslash sqrt\{5\}$=Length×Width=10​×50​=105​.
2. Diagonal $BDBD$BD: Assuming $D(9,3)D(9,\; 3)$D(9,3): $BD=\sqrt{(9-3{)}^2+(3-0{)}^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}BD\; =\; \backslash sqrt\{(9\; –\; 3)^2\; +\; (3\; –\; 0)^2\}\; =\; \backslash sqrt\{36\; +\; 9\}\; =\; \backslash sqrt\{45\}\; =\; 3\backslash sqrt\{5\}$BD=(9−3)2+(3−0)2​=36+9​=45​=35​.

Thus, the area is $10\sqrt{5}10\backslash sqrt\{5\}$105​ and diagonal $BDBD$BD is $3\sqrt{5}3\backslash sqrt\{5\}$35​.