An architect wants to draw a rectangle with a diagonal of 15 centimeters. The length of the rectangle is to be 6 centimeters more than twice the width. What dimensions should she make the rectangle

Answer :

Answer:

She should make a rectangle with dimensions 14.4 cm by 4.2 cm.

Step-by-step explanation:

  • The diagonal of a rectangle is represented by [tex]\sqrt{length^2+width^2}[/tex].
  • Area of rectangle is = length×width
  • Perimeter of a  rectangle = 2(length+width).

Assume x be the width of the rectangle.

The length of the rectangle is to be 6 cm more than twice the width.

The length of the rectangle is= (2x+6) cm

Then the diagonal of the rectangle is [tex]\sqrt{length^2+width^2}[/tex]

                                                             [tex]=\sqrt{(2x+6)^2+x^2}[/tex]

                                                            [tex]=\sqrt{4x^2+24x+36+x^2}[/tex]

                                                             [tex]=\sqrt{5x^2+24x+36}[/tex] cm

According to the problem,

[tex]\sqrt{5x^2+24x+36}=15[/tex]

Squaring both sides

[tex]\Rightarrow{5x^2+24x+36}=15^2[/tex]

[tex]\Rightarrow{5x^2+24x+36}=225[/tex]

[tex]\Rightarrow{5x^2+24x+36-225=0[/tex]

[tex]\Rightarrow{5x^2+24x-189=0[/tex]

⇒5x²+45x-21x-189=0

⇒5x(x+9)-21(x+9)=0

⇒(x+9)(5x-21)=0

⇒x+9=0  or,   5x-21=0

[tex]\Rightarrow x=-9, \frac{21}5[/tex]

⇒x= -9, 4.2

Since the width of a rectangle can not negative.

So, x=4.2 cm

The width of rectangle is = 4.2 cm

The length of the rectangle is =(2×4.2+6)

                                                  =14.4 cm

She should make a rectangle with dimensions 14.4 cm by 4.2 cm.

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