Question

An architect wants to draw a rectangle with a diagonal of 17 millimeters. The length of the rectangle is to be 9 millimeters
more than triple the width. What dimensions should she make the retangle?​

Answer 1

Let x represent width of rectangle.

We have been given that the length of the rectangle is to be 9 millimeters  more than triple the width. So length of rectangle would be $3x+9$.

We are also told that an architect wants to draw a rectangle with a diagonal of 17 millimeters.

We know that sides of rectangle are perpendicular to each other. So it will form a right triangle.

Now we will use Pythagoras theorem to solve for x as:

$x^2+\left(3x+9\right)^2=17^2$

$x^2+9x^2+54x+81=289$

$10x^2+54x+81-289=289-289$

$10x^2+54x-208=0$

Now we will use quadratic formula to solve for x as:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-54\pm\sqrt{54^2-4\cdot \:10\left(-208\right)}}{2\cdot \:10}$

$x=\frac{-54\pm\sqrt{2916+8320}}{20}$

$x=\frac{-54\pm\sqrt{11236}}{20}$

$x=\frac{-54\pm106}{20}$

$x=\frac{-54-106}{20},x=\frac{-54+106}{20}$

$x=\frac{-160}{20},x=\frac{52}{20}$

$x=-8,x=2.6$

Since width cannot be negative, therefore, the width of the rectangle would be 2.6 millimeters.

The length of rectangle would be $3x+9\Rightarrow 3(2.6)+9=7.8+9=16.8$.

Therefore, the length of the rectangle would be 16.8 millimeters.