Question

A rhombus is inscribed in a rectangle that is w meters wide with a perimeter of 24 m. Each vertex of the rhombus is a midpoint of a side of the rectangle. Express the area of the rhombus as a function of the​ rectangle's width.

Answer 1

P=2(L+W)
P=24
24=2(L+W)
12=L+W
L=12-W
area of the rhombus
see diagram
as you can see, the area of the rhombus is 1/2LW
L=12-W
area=(1/2)(12-W)(W)
$A(W)= \frac{12W-W^2}{2}$

Answer 2

The area of the rhombus as a function of the​ rectangle's width is $\frac{12w-w^2}{2}$.

Let the width of rectangle be $w$.

Perimeter of a rectangle $= 2(l+w)$, where $l$ is the length and $w$ is the width.

Perimeter of the rectangle $= 24$ m.

$2(l+w)=24$

$l+w=12$

$l=12-w$

Area of a rhombus $= \frac{1}{2} \times d_1 \times d_2$,where $d_1$ and $d_2$ are the diagonals.

Here, it can be seen that the diagonals of the rhombus are equal to the length and width of the rectangle.

So, Area of a rhombus $= \frac{1}{2} \times (12-w) \times w$

$=\frac{12w-w^2}{2}$

So, the area of the rhombus as a function of the​ rectangle's width is $\frac{12w-w^2}{2}$.

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