Question

refer to attachment

no spam​

Answer 1

$\huge \sf \underline \red{Answer \: }$

$\sf \underline \purple{ \therefore \: length = 10m \: or \: 4m}$

$\sf \underline \purple { \therefore \: breadth = 14 - 10 = 4m \: or \: 14 - 4 = 10m}$

$\huge \sf \underline \pink{Solution}$

$\sf{Let \: the \: length \: of \: rectangle \: be \: x}$

$\sf \underline{we \: know \: that}$

$\sf \underline{Given \: perimeter = 28}$

$\: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \tt{perimeter = 2(l + b) = 28\: }}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{l + b = \dfrac{28}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{ = 14}$

$\sf \underline{ \therefore \: breadth \: of \: rectangle = 14 - x}$

• Now given area is 40 square meter

$\: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \tt{area = length \times breadth }}}}}}$

$\: \: \: \: \: \: \: \sf \implies{x(14 - x) = 14x - {x}^{2}}$

$\sf \underline{By \: problem}$

$\sf \implies{ 14x - {x}^{2} = 40}$

$\sf \implies{ {x}^{2} - 14x + 40 = 0}$

$\sf \implies{ {x}^{2} - 10x - 4x + 40 = 0 }$

$\sf \implies{ x(x - 10) - 4(x - 10) = 0}$

$\sf \implies{(x - 10)(x - 4) = 0}$

$\sf \implies{ x - 10 = 0 \: or \: x - 4 = 0}$

$\sf \implies{x = 10 \: or \: 4}$

$\sf \underline{ \therefore \: length = 10m \: or \: 4m}$

$\sf \underline { \therefore \: breadth = 14 - 10 = 4m \: or \: 14 - 4 = 10m}$

Answer 2

Answer:

Answer

[ tex ]\sf \underline \purple{ \therefore \: length = 10m \: or \: 4m}

∴length=10mor4m

\sf \underline \purple { \therefore \: breadth = 14 - 10 = 4m \: or \: 14 - 4 = 10m}

∴breadth=14−10=4mor14−4=10m

\huge \sf \underline \pink{Solution}

Solution

\sf{Let \: the \: length \: of \: rectangle \: be \: x}Letthelengthofrectanglebex

\sf \underline{we \: know \: that}

weknowthat

\sf \underline{Given \: perimeter = 28}

Givenperimeter=28

\: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \tt{perimeter = 2(l + b) = 28\: }}}}}}

perimeter=2(l+b)=28

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{l + b = \dfrac{28}{2}}⟹l+b=

2

28

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies{ = 14}⟹=14

\sf \underline{ \therefore \: breadth \: of \: rectangle = 14 - x}

∴breadthofrectangle=14−x

Now given area is 40 square meter

\: \: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \red{ \tt{area = length \times breadth }}}}}}

area=length×breadth

\: \: \: \: \: \: \: \sf \implies{x(14 - x) = 14x - {x}^{2}}⟹x(14−x)=14x−x

2

\sf \underline{By \: problem}

Byproblem

\sf \implies{ 14x - {x}^{2} = 40}⟹14x−x

2

=40

\sf \implies{ {x}^{2} - 14x + 40 = 0}⟹x

2

−14x+40=0

\sf \implies{ {x}^{2} - 10x - 4x + 40 = 0 }⟹x

2

−10x−4x+40=0

\sf \implies{ x(x - 10) - 4(x - 10) = 0}⟹x(x−10)−4(x−10)=0

\sf \implies{(x - 10)(x - 4) = 0}⟹(x−10)(x−4)=0

\sf \implies{ x - 10 = 0 \: or \: x - 4 = 0}⟹x−10=0orx−4=0

\sf \implies{x = 10 \: or \: 4}⟹x=10or4

\sf \underline{ \therefore \: length = 10m \: or \: 4m}

∴length=10mor4m

\sf \underline { \therefore \: breadth = 14 - 10 = 4m \: or \: 14 - 4 = 10m}

∴breadth=14−10=4mor14−4=10m[/tex]

Step-by-step explanation:

Hope this answer will help you.✌️