Question

an architect wants to draw a rectangle with a diagonal of 25 millimeters. the length of the rectangle is to be 10 millimeters more than twice the width. what dimensions should she make the rectangle? answer length = millimeters width = millimeters

Explanation:

Step1: Define variables

Let the width of the rectangle be $x$ millimeters. Then the length is $2x + 10$ millimeters.

Step2: Apply Pythagorean theorem

In a rectangle, $width^{2}+length^{2}=diagonal^{2}$. So $x^{2}+(2x + 10)^{2}=25^{2}$.
Expand $(2x + 10)^{2}$ using $(a + b)^{2}=a^{2}+2ab + b^{2}$, we get $x^{2}+4x^{2}+40x + 100 = 625$.
Combine like - terms: $5x^{2}+40x+100 - 625=0$, which simplifies to $5x^{2}+40x - 525 = 0$.
Divide through by 5: $x^{2}+8x - 105 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+8x - 105=(x + 15)(x - 7)=0$.
Set each factor equal to zero: $x+15 = 0$ gives $x=-15$ (rejected since length cannot be negative), $x - 7=0$ gives $x = 7$.

Step4: Find the length

If $x = 7$, then the length $l=2x + 10=2\times7+10=24$.

Answer:

Length = 24 millimeters
Width = 7 millimeters