Question

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

Answer 1

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

$\rm :\longmapsto\:Length_{(rectangle)} \: = \: L \: cm$

$\rm :\longmapsto\:Perimeter_{(rectangle)} = \: 80 \: cm$

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

$\rm :\longmapsto\:Area_{(rectangle)}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: Area_{(rectangle)} = Length_{(rectangle)} \times Breadth_{(rectangle)}}$

$\boxed{ \sf \: Perimeter_{(rectangle)} = 2(Length_{(rectangle)} + Breadth_{(rectangle)})}$

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

Given that ,

• Length of rectangle = L cm

Let

• Breadth of rectangle = 'b' cm

Since,

• Perimeter of rectangle = 80 cm

We know,

$\sf \: Perimeter_{(rectangle)} = 2(Length_{(rectangle)} + Breadth_{(rectangle)})$

$\rm :\longmapsto\:80 = 2(L + b)$

$\rm :\longmapsto\:L \: + \: b \: = \: 40$

$\bf\implies \:b \: = \: 40 - L \: - - - (1)$

Now,

We have,

• Length of rectangle = L cm

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

So,

Area of rectangle is

$\sf \: Area_{(rectangle)} = Length_{(rectangle)} \times Breadth_{(rectangle)}$

$\rm :\longmapsto\:Area_{(rectangle)} = L \: \times \: (40 - L)$

$\rm :\longmapsto\:Area_{(rectangle)} = 40L\: - \: {L}^{2}$

Additional Information :-

$\boxed{ \sf \: Area_{(square)} = {(side)}^{2}}$

$\boxed{ \sf \: Area_{(circle)} = \pi \: {r}^{2} }$

$\boxed{ \sf \: Perimeter_{(square)} = 4 \times side}$

$\boxed{ \sf \: Perimeter_{(circle)} = 2\pi \: r}$

$\boxed{ \sf \: Area_{(right \: \triangle)} = \dfrac{1}{2} \times base \times height}$

$\boxed{ \sf \: Area_{(parallelogram)} = base \times height}$

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

Answer 2

It is the correct answer.

Step-by-step explanation:

Hope this attachment helps you.