the length of a rectangle is 4 meters less than twice its width. The area of the rectangle is 70. Find the dimensions of the rectangle

Answer:  The width is 7 meters. The length is 10 meters.

Step-by-step explanation:   Area = Length × width

Length is 2w - 4  

Substitute that value for length, then solve for w.

70 = w(2w - 4)

70 = 2w² - 4w  reorganize to quadratic equation form

2w² - 4w -70 = 0   These are All even numbers; factor out 2 (divide all by 2)

w² - 2w - 35 = 0  factor this

(w - 7)(w + 5)    set each factor = 0 and calculate the value of w.

w+5=0  w= -5  (disregard this one because dimensions of real rectangles can't be negative)

w - 7 = 0  w = 7   The width is 7

Substitute into the original expressions to get the value of Length.

2(7) -4 = L    14 - 4 = L  The Length is 10.

Check by substituting into the original equation:

w × 2w -4 = 70

7 × 2(7) -4 = 70

7 × 10 = 70    TRUE!

w

Dimensions of the rectangle are length=10 meter and width= 7 meter

A rectangle is a quadrilateral with four right angles.

Given,

The length of a rectangle is 4 meter less than twice its width

Consider,

l is the length of the rectangle

w is the width of the rectangle

l=2w-4

Area of the rectangle = Length × Width

70=l × w

Substitute the value of l is the second equation

70=(2w-4)×w

[tex]2w^{2} -4w=70\\2w^{2} -4w-70=0[/tex]

Take 2 outside

[tex]2(w^{2}-2w-35)=0\\ w^{2} -2w-35=0\\(w+5)(w-7)=0[/tex]

w=7,-5

Hence w wont be negative value

Therefore w=7 meter

Then,

l×w=70

[tex]l=\frac{70}{7}\\[/tex]

l=10 meter

Hence, the dimensions of the rectangle are length=10 meter and width= 7 meter

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