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=(-b\pm √(b^2-4ac))/(2a)`

`x=(-54\pm√(54^2-4\cdot \:10\left(-208\right)))/(2\cdot \:10)`

`x=(-54\pm√(2916+8320))/(20)`

`x=(-54\pm√(11236))/(20)`

`x=(-54\pm106)/(20)`

`x=(-54-106)/(20),x=(-54+106)/(20)`

`x=(-160)/(20),x=(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.