We want to construct a sign in the shape of the right triangle.If the longer leg of the sign is two inches more than the shorter leg and the hypotenuse is two inches less than twice the shorter leg...
We want to construct a sign in the shape of the right triangle.
If the longer leg of the sign is two inches more than the shorter leg and the hypotenuse is two inches less than twice the shorter leg , determine all the sides of the sign .
We'll note the shorter leg, namely one cathetus, as x.
The longer cathetus is x + 2.
The hypotenuse is 2x - 2.
Now, we'll apply the relation that puts together all the sides of a right angled triangle, the Pythagorean theorem:
hypotenuse^2 = shorter cathetus^2 + longer cathetus^2
We'll substitute the hypotenuse and the cathetus by the relations above:
(2x-2)^2 = x^2 + (x+2)^2
We'll expand the squares from both sides and we'll get:
4x^2 - 8x + 4 = x^2 + x^2 + 4x + 4
We'll eliminate and combine like terms:
4x^2 - 8x = 2x^2 + 4x
We'll subtract 2x^2 + 4x both sides:
4x^2 - 8x - 2x^2 - 4x = 0
We'll combine like terms:
2x^2 - 12x = 0
We'll divide by 2:
x^2 - 6x = 0
We'll factorize by x:
x(x-6) = 0
x1 = 0
x-6 = 0
x2 = 6
Since the measure of a cathetus cannot be zero, we'll reject the first solution and we'll keep the second one.
So, the shorter leg of the sign has the measure of 6 inches, the longer leg has the measure of 8 inches and the hypotenuse has the measure of 10 inches.
given longer leg of the right angled triangle is 2 inch more than the shorter leg.
So let the measurement of longer leg l and shoter legs s be (x+2) and x respectively
The hypotenuse = 2 inch less than twice shorter leg. So,
hypotenuse h = -2 +2x.
Therefore by Pythagoras theorem,
l ^2+ s^2 = h^2,
(x+2)^2 +x^2 = (-2+2x)^2, is the equation to be solved for x.
x^2+4x+4 +x^2 = 4-8x+4x^2
4x^2-8x+4 - 2x^2 -4x - 4 = 0
2x^2-12x = 0
2x(x -6) = 0
x= 0. Or x = 6.
x = 6 .
sorter leg x = 6 inch
longer leg l = x+2 = 8 inch
hypotenuse h = -2+2x = -2+2*6 = 10 inch
Tally by Pythagoras:
l^2+s^2 = 8^2+6^2 = 100.
h^2 = 10^2 = 100.