Question

find the digonal of rectangle whose lenth is 35cm and breadth is 18 cm​

Answer 1

√(35²+18²)

√(1225+324)

√1549

39.35cm

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Answer 2

$\sf \underline{\large{Given - }}$

$\sf \dashrightarrow \: Length \: of \: rectangle = 35 \: cm$

$\sf \dashrightarrow \: Breadth \: of \: rectangle = 18 \: cm$

$\sf \underline{\large{To \: find - }}$

$\sf \dashrightarrow \: Diagonal \: of \: rectangle = \: ?$

$\sf \underline{\large{Solution - }}$

$\sf \dashrightarrow \: {\boxed{ \sf Diagonal \: of \: rectangle = \: \sqrt{(Length )^2 + (Breadth)^2} }}$

$\sf Substitute \: the \: values,$

$\sf \dashrightarrow \: \sf Diagonal = \: \sqrt{(Length )^2 + (Breadth)^2}$

$\sf \dashrightarrow \: \sf Diagonal = \: \sqrt{(35 )^2 + (18)^2}$

$\sf \dashrightarrow \: \sf Diagonal = \: \sqrt{1225 + 324}$

$\sf \dashrightarrow \: \sf Diagonal = \: \sqrt{1549}$

$\sf \dashrightarrow \: \sf Diagonal = \: 39.35 \: cm$

$\bf{Hence},$

$\sf \dashrightarrow \: \boxed{ \sf Diagonal \:of \: rectangle = \: 39.35 \: cm}$

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$\bf Some \: other \: formulae :$

$\sf \dashrightarrow \: Area \: of \: rectangle = length \times Breadth$

$\sf \dashrightarrow \: Length \: of \: rectangle = \dfrac{Area}{Breadth}$

$\sf \dashrightarrow \: Breadth \: of \: rectangle = \dfrac{Area}{length}$

$\sf \dashrightarrow \: Perimeter \: of \: rectangle = 2(Length + Breadth)$

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$\bf Three \: properties \: of \: rectangle :$

$\sf (1). \:All \:the \:angle \:of \:a\: rectangle \: are \:90\degree.$

$\sf(2). \:Opposite \:side \:of\: a\: rectangle\: are \:equal\: and \: parallel.$

$\sf(3). \:Diagonal \:of\: a \:rectangle\: bisect \:each\:other.$