Question

the length of a diagonal of a rectangle of length 16 centimetre is 20 cm find its area

Answer 1

using Pythagoras theorem
${16}^{2} + {x}^{2} = {20}^{2}$
here x = other side soo
$400 - 256 = {x}^{2}$
${x}^{2} = 144$

${12}^{2} = 144$
hence
$x = 12$
hey bro I was taught Pythagoras theorem in 6th what hell you talking bout. now area
$16 \times 12$
which is 192 sq.cm

Answer 2

$\bf{ \pmb{ \underline{ \gray{GIVEN}} }}\\ \cal \: Length \: of \: R ectangle \: is \: 16 cm \\ \cal \: Length \: of \: its \: Diagonal \: is \: 20 \: cm \\ \\ \bf { \pmb { \underline{ \gray{ To \: F ind }}}}\\ \large \: \blue{ \tt{ Area \: of \: Rectangle}} \\ \\ \huge \bf { \pmb{ \overline{\underline{ \color{black}{ \mid \: SOLUTION \: \mid}}}}} \\ \\ \large\tt{ \pmb{ \red{FORMULA \: \: \: USED}}} \\ \small \orange{\bf[\:{{(Diagonal)}^{2} = {(Length)}^{2} + {(Breadth)}^{2}}\:]} \\ \\ \sf{ \pmb{ CALCULATION }}\\ \rm \: (20)² = (16)² + (b)² \\ 400 = 257 + (b)² \\ 400 - 256 = (b)² \\ 144 = b² \\ \rm \: \sqrt{144 }= b \\ \rm \: 12 = b \\ \cal \blue{Thus, \: The \: Breadth \: of \: Rectangle \: is \: 12 \: cm . }\\ \\ \large\tt{ \pmb{ \red{ FORMULA \: \: \: USED}}} \\ \small \orange{ \bf[\:{Area \: of \: Rectangle =Length\times{Breadth}}\:] }\\ \\ \sf{ \pmb{CALCULATION }}\\ \rm Length × Breadth \\ 16 × 12 \\ 192 \\ \cal \blue{ Thus, \: The \: Area \: of \: Rectangle \: is \: 192 \: cm² }$