D. 20 cm^{2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n}

Answer :\u00a0<\/b>D. 20 cm2<\/sup><\/h3>\n

\n

Explanation:<\/h2>\n<\/div>\n

\n

\n

Perimeter of rectangle = 18 cm<\/b>
\n2 ( l + b ) = 18 cm
\nl + b = 18 \/ 2
\n= 9 cm
\nl + b = 9 cm<\/p>\n

Diagonal of rectangle = \u221a41 cm<\/b>
\nWe know that, Diagonal of rectangle = \u221a [ ( length )^{2<\/sup>\u00a0+ ( breadth )2<\/sup>\u00a0]\n\u221a( l2<\/sup>\u00a0+ b2<\/sup>) = \u221a41
\nl2<\/sup>\u00a0+ b2<\/sup>\u00a0= ( \u221a41 )2<\/sup>
\nl2<\/sup>\u00a0+ b2<\/sup>\u00a0= 41
\nl2<\/sup>\u00a0+ b2<\/sup>\u00a0+ 2 lb – 2 lb = 41
\n( l + b )2<\/sup>\u00a0– 2 lb = 41<\/p>\n}

Since , we have\u00a0l + b = 9\u00a0<\/b>, so now
\n( 9 )^{2<\/sup>\u00a0– 2 l b = 41
\n81 – 2 lb = 41
\n2 lb = 81 – 41
\n2 lb = 40
\nlb = 40 \/ 2 = 20 cm\u00a02<\/sup>
\nl \u00d7 b = 20 cm2<\/sup>
\nTherefore,\u00a0area of rectangular field =\u00a020 cm2<\/sup><\/b><\/p>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"}

Answer :\u00a0D. 20 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