How do you solve this?

Mr. Hamilton is placing a support plank along the diagonal of a gate. The height of the gate is 5 feet, and the diagonal is 1 foot longer than the width of the gate, as shown below. What is the width, in feet, of the gate?

Feb 13, 2017

$12$ feet

Explanation:

By Pythagoras:

${\left(w + 1\right)}^{2} = {w}^{2} + {5}^{2}$

That is:

${w}^{2} + 2 w + 1 = {w}^{2} + 25$

Subtracting ${w}^{2} + 1$ from both sides, we find:

$2 w = 24$

Divide both sides by $2$ to get:

$w = 12$

So the gate is $12$ feet wide.

$\textcolor{w h i t e}{}$
Footnote

The first few right angled triangles with sides $a$, $b$ and $b + 1$ have sides of lengths:

$3 , 4 , 5$

$5 , 12 , 13$

$7 , 24 , 25$

$9 , 40 , 41$

In general, they take the form:

$a , \frac{{a}^{2} - 1}{2} , \frac{{a}^{2} + 1}{2}$

for any odd value of $a \ge 3$

Feb 13, 2017

$\text{width "=12" feet}$

Explanation:

Using $\textcolor{b l u e}{\text{Pythagoras' theorem}}$ on the right triangle.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.

$\Rightarrow {\left(w + 1\right)}^{2} = {w}^{2} + {5}^{2}$

$\Rightarrow {w}^{2} + 2 w + 1 = {w}^{2} + 25$

subtract ${w}^{2}$ from both sides.

$\cancel{{w}^{2}} \cancel{- {w}^{2}} + 2 w + 1 = \cancel{{w}^{2}} \cancel{- {w}^{2}} + 25$

$\Rightarrow 2 w + 1 = 25$

subtract 1 from both sides.

$2 w \cancel{+ 1} \cancel{- 1} = 25 - 1$

$\Rightarrow 2 w = 24$

divide both sides by 2

$\frac{\cancel{2} w}{\cancel{2}} = \frac{24}{2}$

$\Rightarrow w = 12 \text{ is the solution}$

That is, the width of the gate is 12 feet.