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Consider a rectangle that has a perimeter of [tex]80\ \text{cm}[/tex]. Write a function [tex]A(l)[/tex] that represents the area o

By Melody

Consider a rectangle that has a perimeter of [tex]80\ \text{cm}[/tex]. Write a function [tex]A(l)[/tex] that represents the area of the rectangle with length [tex]l[/tex].​

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2 thoughts on “Consider a rectangle that has a perimeter of [tex]80\ \text{cm}[/tex]. Write a function [tex]A(l)[/tex] that represents the area o”

  1. [tex]\large\underline{\sf{Given- }}[/tex]

    [tex]\rm :\longmapsto\:Length_{(rectangle)} \: = \: L \: cm[/tex]

    [tex]\rm :\longmapsto\:Perimeter_{(rectangle)} = \: 80 \: cm[/tex]

    [tex]\large\underline{\sf{To\:Find – }}[/tex]

    [tex]\rm :\longmapsto\:Area_{(rectangle)}[/tex]

    [tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]

    [tex] \boxed{ \sf \: Area_{(rectangle)} = Length_{(rectangle)} \times Breadth_{(rectangle)}}[/tex]

    [tex] \boxed{ \sf \: Perimeter_{(rectangle)} = 2(Length_{(rectangle)} + Breadth_{(rectangle)})}[/tex]

    [tex]\large\underline{\sf{Solution-}}[/tex]

    Given that ,

    • Length of rectangle = L cm

    Let

    • Breadth of rectangle = ‘b’ cm

    Since,

    • Perimeter of rectangle = 80 cm

    We know,

    [tex] \sf \: Perimeter_{(rectangle)} = 2(Length_{(rectangle)} + Breadth_{(rectangle)})[/tex]

    [tex]\rm :\longmapsto\:80 = 2(L + b)[/tex]

    [tex]\rm :\longmapsto\:L \: + \: b \: = \: 40[/tex]

    [tex]\bf\implies \:b \: = \: 40 – L \: – – – (1)[/tex]

    Now,

    We have,

    • Length of rectangle = L cm
    • Breadth of rectangle, b = (40 – L) cm

    So,

    Area of rectangle is

    [tex] \sf \: Area_{(rectangle)} = Length_{(rectangle)} \times Breadth_{(rectangle)}[/tex]

    [tex]\rm :\longmapsto\:Area_{(rectangle)} = L \: \times \: (40 – L)[/tex]

    [tex]\rm :\longmapsto\:Area_{(rectangle)} = 40L\: – \: {L}^{2} [/tex]

    Additional Information :-

    [tex] \boxed{ \sf \: Area_{(square)} = {(side)}^{2}} [/tex]

    [tex] \boxed{ \sf \: Area_{(circle)} = \pi \: {r}^{2} }[/tex]

    [tex] \boxed{ \sf \: Perimeter_{(square)} = 4 \times side}[/tex]

    [tex] \boxed{ \sf \: Perimeter_{(circle)} = 2\pi \: r}[/tex]

    [tex] \boxed{ \sf \: Area_{(right \: \triangle)} = \dfrac{1}{2} \times base \times height}[/tex]

    [tex] \boxed{ \sf \: Area_{(parallelogram)} = base \times height}[/tex]

    [tex]\boxed{ \sf \: Area_{(rhombus)}=base \times height=\dfrac{1}{2}(product \: of \: diagonals)}[/tex]

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