# How do you solve this?

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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?

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?

##### 2 Answers

#### Explanation:

By Pythagoras:

#(w+1)^2 = w^2+5^2#

That is:

#w^2+2w+1 = w^2+25#

Subtracting

#2w = 24#

Divide both sides by

#w = 12#

So the gate is

**Footnote**

The first few right angled triangles with sides

#3, 4, 5#

#5, 12, 13#

#7, 24, 25#

#9, 40, 41#

In general, they take the form:

#a, (a^2-1)/2, (a^2+1)/2#

for any odd value of

#### Explanation:

Using

#color(blue)"Pythagoras' theorem"# on the right triangle.

#color(orange)"Reminder" # The square on the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides.

#rArr(w+1)^2=w^2+5^2#

#rArrw^2+2w+1=w^2+25# subtract

#w^2# from both sides.

#cancel(w^2)cancel(-w^2)+2w+1=cancel(w^2)cancel(-w^2)+25#

#rArr2w+1=25# subtract 1 from both sides.

#2wcancel(+1)cancel(-1)=25-1#

#rArr2w=24# divide both sides by 2

#(cancel(2) w)/cancel(2)=24/2#

#rArrw=12" is the solution"# That is, the width of the gate is 12 feet.