A triangle has a perimeter of 165 cm.The first side is 65 cm less
than twice the second side.The third side is 10 cm less than the second side.
Write and solve an equation to find the length of each side of the triangle.
2006-10-05 12:02:26 · 5 answers · asked by techline210 2 in Science & Mathematics ➔ Mathematics
Let x = first side, y = second side and z = third side, then
P = x + y + z = 165
x = 2y - 65
z = y - 10
Plug (2y-65) in for x and (y-10) in for z in the first equation.
2y - 65 + y + y - 10 = 165
Combine like terms
4y - 75 = 165
Add 75 to each side
4y = 240
Divide both sides by 4
y = 60 = length of second side
x = 2(60) - 65 = 120 - 65 = 55 = length of first side
z = 60 - 10 = 50 = length of third side
2006-10-05 12:10:38 · answer #1 · answered by MsMath 7 · 0⤊ 0⤋
have x = length of first side
y = length of the second side
z = length of the third side
x = 2y - 65 (the first side is 65 cm less than twice the second side)
z = y - 10 (the third side is 10 cm less than the second side)
to find the perimeter of a triangle, you add up all the lengths of the sides:
x + y + z = 165
now plug in the other equations into this equation:
x + y + z = 165
(2y - 65) + y + (y - 10) = 165
now solve for y
2y - 65 + y + y - 10 = 165
4y - 75 = 165
add 75 to both sides:
4y = 240
divide both sides by 4
y = 60
now we know that the length of the second side is 60, so plug that into the other equations:
x = 2y - 65
x = 2(60) -65
x = 120 - 65
x = 55
z = y - 10
z = 60 - 10
z = 50
the three sides are 55, 50 and 60
2006-10-05 19:20:04 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Let x,y and z be the lengths of the first, second and third sides.
Then x=2y-65, and z=y-10. Also, x+y+z=165.
In the equation x+y+z=165, replace z with y-10, and get x+2y-10=165, or x+2y=175.
Then, you have the system of two equations and two unknowns,
x+2y=175
x-2y=-65
Add the two equations together, and you get 2x=110, so x=55. That means y=60, and z=50.
2006-10-05 19:08:37 · answer #3 · answered by James L 5 · 1⤊ 0⤋
Let the sides be a, b, c.
a+b+c = 165
a = 2*b - 65
c = b - 10
A classic case of three equations in three unknowns. I would solve by substitution: Put in a from the second equation into the first. Put c from the third into the first. Now the first eq has only b as a variable. solve for b. Then get a and c from the next two equations.
2006-10-05 19:09:05 · answer #4 · answered by gp4rts 7 · 1⤊ 0⤋
2x-65+x+x-10=165
55=1st side
60= 2nd side
50=3rs side
i hope u are younger that 14 becaue that is how old i am
2006-10-05 19:16:15 · answer #5 · answered by zj 2 · 0⤊ 0⤋