Question

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Diagonal of rectangle is 29cm and breadth is 20cm find its length area and perimeter

Answer 1

Given,

$\begin{gathered}\blue{ \boxed{\boxed{\begin{array}{cc} \rm \to \: diagonal \: of \: rectangle \: \: d = 29 \: cm \\ \\ \rm \to \: breadth \: of \: rectangle \: \: b = 20 \: cm \\ \\ \underline{ \blue{ \sf \: we \: have \: to \: find \: : }} \\ \\ \rm \to \: area \: of \: \: rectangle \: = A \\ \\ \rm \to \: perimeter \: of \: rectangle \: = S\end{array}}}} \end{gathered}$

$\begin{gathered} \small{\green{ \boxed{\boxed{\begin{array}{cc} \underline{\downarrow \bf \: formula \: \downarrow} \\ \\ \sf \: we \: know \: that \: for \: recrangle \\ \\ \odot \: \rm \: area \: A = length \times breadth \\ \\ \rm \implies \:A = l \times b \\ \\ \\ \odot \: \rm \: perimeter \: S = 2(length + breadth) \\ \\ \rm \implies \: S =2(l + b) \end{array}}}}}\end{gathered}$

At first we have to find length (l)

we know that, for rectangle.

$\rm \: diagonal \: \: d = \sqrt{ {l}^{2} + {b}^{2} }$

So,

According to the question,

$\begin{gathered}29 = \sqrt{ {l}^{2} + {(20)}^{2} } \\ \\ \rm \implies\: {(29)}^{2} = {l}^{2} + 400 \\ \\ \rm \implies\:l = 21 \: cm \\ \\ \rm \: area \: A = 21 \times 20 = 420 \: {cm}^{2} \\ \\ \rm \: perimeter \: S = 2(21 + 20) = 82 \: cm\end{gathered}$

Answer 2

Answer:

At first we have to find length (l)

we know that, for rectangle.

\rm \: diagonal \: \: d = \sqrt{ {l}^{2} + {b}^{2} }diagonald=

l

2

+b

2

So,

According to the question,

\begin{gathered}\begin{gathered}29 = \sqrt{ {l}^{2} + {(20)}^{2} } \\ \\ \rm \implies\: {(29)}^{2} = {l}^{2} + 400 \\ \\ \rm \implies\:l = 21 \: cm \\ \\ \rm \: area \: A = 21 \times 20 = 420 \: {cm}^{2} \\ \\ \rm \: perimeter \: S = 2(21 + 20) = 82 \: cm\end{gathered} \end{gathered}

29=

l

2

+(20)

2

⟹(29)

2

=l

2

+400

⟹l=21cm

areaA=21×20=420cm

2

perimeterS=2(21+20)=82cm