Question

An architect wants to draw a rectangle with a diagonal of 13 centimeters. The length of the rectangle is to be 3 centimeters more than triple the width. What dimensions should she make the rectangle?​

Answer 1

Answer:

The rectangle is 3.2 cm by 12.6 cm

Step-by-step explanation:

See attached image for a diagram.

Choose w to represent the width because the length is described by referring to the width:  it's 3 more than (add 3) triple (multiplied by 3) the width.

Length = 3w + 3

The diagonal forms two right triangles, each with leg = w, other leg = 3w + 3, hypotenuse  =  13.

The Pythagorean Theorem says $(\text{leg})^2+(\text{leg})^2=(\text{hypotenuse})^2$ so

$w^2+(3w+3)^2=13^2\\\\w^2+9w^2+18w+9=169\\\\10w^2+18w-160=0\\\\5w^2+9w-80=0$

Now solve using the Quadratic Formula with $a=5,\,b=9,\,c=-80$.

$w=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\w=\frac{-9\pm\sqrt{81+1600}}{10}\\\\w=\frac{-9\pm\sqrt{1681}}{10}\\\\w=\frac{-9\pm41}{10}\\\\w=-5\text{ or }w=3.2$

The negative root makes no sense as a distance, so the width of the rectangle is 3.2 cm. The length is 3(3.2) + 3 = 12.6 cm.