Question

an architect wants to draw a rectangle with a diagonal of 13 millimeters. the length of the rectangle is to be 3 millimeters more than triple the width. what dimensions should she use for the rectangle?

Explanation:

Step1: Let the width be $x$ mm.

The length is $2x + 3$ mm.

Step2: Apply Pythagorean theorem.

In a rectangle, $l^{2}+w^{2}=d^{2}$, so $(2x + 3)^{2}+x^{2}=13^{2}$.
Expand: $4x^{2}+12x + 9+x^{2}=169$.
Combine like - terms: $5x^{2}+12x+9 - 169 = 0$, i.e., $5x^{2}+12x - 160 = 0$.

Step3: Solve the quadratic equation $ax^{2}+bx + c = 0$ ($a = 5$, $b = 12$, $c=-160$) using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

First, calculate the discriminant $\Delta=b^{2}-4ac=(12)^{2}-4\times5\times(-160)=144 + 3200=3344$.
Then $x=\frac{-12\pm\sqrt{3344}}{10}=\frac{-12\pm57.83}{10}$.
We take the positive root since length cannot be negative. $x=\frac{-12 + 57.83}{10}=4.583\approx4$ (rounded to the nearest whole number).

Step4: Find the length.

Length $l=2x + 3=2\times4+3=11$ mm.
Width $w = 4$ mm.

Answer:

Length = 11 millimeters
Width = 4 millimeters