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.

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)
\(A(W)= \frac{12W-W^2}{2}\)

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kobenhavn kobenhavn · Answer 2 · 2021-10-02T04:26:58-04:00

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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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.

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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.