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?
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.
