Question

please answer it fastly​

Answer 1

Step-by-step explanation:

Diagonal = 13

then the length breadth will be

H²= l²+b²

13²= l²+b²

169= 144+ 25

length = 12

and breadth= 5

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Answer 2

Given that , The Perimeter of a rectangular room is 34 m and the Length of Diagonal is 13 m .

❍ Let's say that Length and Breadth of Rectangular room be l & b , respectively .

⠀⠀⠀⠀⠀⠀CASE I : The Perimeter of a rectangular room is 34 m .

As , We know that ,

$\qquad \star \:\underline {\boxed{ \pink {\bf{ Perimeter _{(Rectangle )} =\: 2\times \bigg( Length \:+\:Breadth\bigg)}}}}$

$\qquad \dashrightarrow \sf Perimeter_{( Rectangular \:room)}\:=\: 2\times \big( Length \:+\:Breadth\big)$

$\qquad \dashrightarrow \sf 34\:=\: 2\times ( l \:+\:b)$

$\qquad \dashrightarrow \sf \dfrac{34}{2}\:=\: ( l \:+\:b)$

$\qquad \dashrightarrow \sf \cancel {\dfrac{34}{2}}\:=\: ( l \:+\:b)$

$\qquad \dashrightarrow \sf 17\:=\: l \:+\:b$

$\qquad \dashrightarrow \sf l \:=\:\:17 - b$

$\qquad \dashrightarrow \sf l \:=\:\:17 - b \qquad \bigg\lgroup \sf{ \:Equation \:1 \:}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀CASE II : The Diagonal of a rectangular room is 13 m .

As , We know that ,

• By Pythagoras Theorem :

$\qquad\star\: \underline {\boxed{ \pink {\frak { \:(Diagonal)^2 \:\:= \: ( \:Length \:)^2 \:+\:(Breadth)^2 }}}}$

$\qquad \dashrightarrow \sf \:(Diagonal)^2 \:\:= \: ( \:Length \:)^2 \:+\:(Breadth)^2$

$\qquad \dashrightarrow \sf \:(13)^2 \:\:= \: ( \:l \:)^2 \:+\:(b)^2$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: ( \:l \:)^2$

$\qquad \dashrightarrow \sf \sqrt{\:(13)^2 - (b)^2 \:}\:= \: \:l \:$

• Now by Squaring both sides :

$\qquad \dashrightarrow \sf \sqrt{\:(13)^2 - (b)^2 \:}\:= \: \:l \:$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: (\:l )^2\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Equation \: 1 \:[\: as \:,\: value \:of \:l \;]\::}}$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: (\:l )^2\:$

$\qquad \dashrightarrow \sf l \:=\:\:17 - b \qquad \bigg\lgroup \sf{ \:Equation \:1 \:}\bigg\rgroup$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: (\:l )^2\:$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: (\:17 - b )^2\:$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: 17^2 + b^2 - 2 (17)(b)\:$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: 17^2 + b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:(13)^2 - (b)^2 \:\:= \: 17^2 + b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:169 - (b)^2 \:\:= \: 289 + b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:169 - (b)^2 \:\:= \: 289 + b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:169 \:\:= \: 289 + b^2 + b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:0 \:\:= \: 289 - 169 + 2b^2 - 34b\:$

$\qquad \dashrightarrow \sf \:0 \:\:= \: 120 + 2b^2 - 34b\:$

$\qquad \dashrightarrow \sf \: \: 2b^2 - 34b\: + 120 \:= 0$

$\qquad \dashrightarrow \sf \: \: b^2 - 17b\: + 60 \:= 0$

$\qquad \dashrightarrow \sf \: \: b^2 - 12b\: - 5b + 60 \:= 0$

$\qquad \dashrightarrow \sf \: \: b( b - 12)\: - 5( b - 12 ) \:= 0$

$\qquad \dashrightarrow \sf \: \: ( b - 5)\: ( b - 12 ) \:= 0$

$\qquad \therefore \: \underline {\boxed {\pmb{\frak{\purple { b \: \:(\:or \:Breadth \:)\:=\: 12 \:\:or \: 5 \: }}}}}$

⠀⠀⠀⠀⠀⠀If b is equal to 12 , then l will be :

$\qquad \dashrightarrow \sf l \:=\:\:17 - b \qquad \bigg\lgroup \sf{ \:Equation \:1 \:}\bigg\rgroup$

$\qquad \dashrightarrow \sf l \:=\:\:17 - b$

$\qquad \dashrightarrow \sf l \:=\:\:17 - 12$

$\qquad \dashrightarrow \sf l \:=\:\:5$

$\qquad \therefore \: \underline {\boxed {\pmb{\frak{\purple { l \:\:(\:or \:Length \:)\:\:=\: \: 5 \:m }}}}}$

⠀⠀⠀⠀⠀⠀If b is equal to 5 , then l will be :

$\qquad \dashrightarrow \sf l \:=\:\:17 - b$

$\qquad \dashrightarrow \sf l \:=\:\:17 - 5$

$\qquad \dashrightarrow \sf l \:=\:\:17 - 5$

$\qquad \therefore \: \underline {\boxed {\pmb{\frak{\purple { l \:\:(\:or \:Length \:)\:\:=\: \: 12 \:m }}}}}$

⠀⠀⠀⠀⠀⠀Therefore , If Length is equal 12 m then Breadth is equal to 5 m Or, If Length is equal to 5 m then Breadth is equal to 12 m

⠀⠀⠀⠀⠀⠀AND ,

⠀⠀⠀⠀⠀⠀If Breadth is equal 12 m then Length is equal to 5 m Or, If Breadth is equal to 5 m then Length is equal to 12 m .

$\qquad \therefore \underline {\sf Hence, \: The \: Dimensions \:of \:\:Rectangular \:room \: are \:\bf 5 \: m \:\& \:12 \:m \:\sf , respectively.}$