Question

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 1

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 $\sqrt{length^2+width^2}$.
• 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 $\sqrt{length^2+width^2}$

                                                             $=\sqrt{(2x+6)^2+x^2}$

                                                            $=\sqrt{4x^2+24x+36+x^2}$

                                                             $=\sqrt{5x^2+24x+36}$ cm

According to the problem,

$\sqrt{5x^2+24x+36}=15$

Squaring both sides

$\Rightarrow{5x^2+24x+36}=15^2$

$\Rightarrow{5x^2+24x+36}=225$

$\Rightarrow{5x^2+24x+36-225=0$

$\Rightarrow{5x^2+24x-189=0$

⇒5x²+45x-21x-189=0

⇒5x(x+9)-21(x+9)=0

⇒(x+9)(5x-21)=0

⇒x+9=0  or,   5x-21=0

$\Rightarrow x=-9, \frac{21}5$

⇒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.