Question 24 the area of a rectangle is 33 yd2 , and the length of the rectangle is 5 yd less than twice the width. find the dimensions of the rectangle.

Respuesta :

Let w and 2w-5 be the width and length. Then:
w(2w-5)=33
2w²-5w-33=0
(2w-11 )(w+3 )=0
w=11/2 or -3
If the width is 11/2 yd, then the length is 6 yds
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The dimensions of the rectangle are length is 6 yd and width is 11/2 yd.

What is a rectangle?

A rectangle is one of the types of quadrilaterals in which all four angles are right angles or equal to 90 degrees. It is four-sided polygon in which the opposite sides are parallel and equal to each other. The dimensions of rectangle are length and width.

For the given situation,

Let l be the length of the rectangle and

Let w be the width of the rectangle

The length of the rectangle is 5 yd less than twice the width,

[tex]l=2w-5[/tex]

The area of a rectangle = 33 yd^2

The formula for area of rectangle is

[tex]A=lw[/tex]

⇒ [tex]33=(2w-5)(w)[/tex]

⇒ [tex]33=2w^{2} -5w[/tex]

⇒ [tex]2w^{2} -5w-33=0[/tex]

This equation can be factored as

⇒ [tex]2w^{2} +6w-11w-33=0[/tex]

⇒ [tex]2w(w+3)-11(w+3)=0[/tex]

⇒ [tex](2w-11)(w+3)=0[/tex]

⇒ [tex]2w-11=0[/tex] or [tex]w+3=0[/tex]

⇒ [tex]w=\frac{11}{2}[/tex] or [tex]w=-3[/tex]

Dimension cannot be negative. So we take width as w = 11/2.

Now, length is

⇒ [tex]l=2(\frac{11}{2} )-5[/tex]

⇒ [tex]l=11-5[/tex]

⇒ [tex]l=6[/tex]

Hence we can conclude that the dimensions of the rectangle are length is 6 yd and width is 11/2 yd.

Learn more about the rectangles here

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