Let the sides of the triangle be x, y, and z.
x + y + z = 12
xy/2 = 6
x + y = 12 - z
xy/2 = 6
(12 - z)z/2 = 6
144 - 12z + z^2/2 = 6
z^2 - 12z + 12 = 0
z = 6 ± √36
z = 6 ± 6
z = 12 or 0
If z = 12, then x + y = 0, which is not possible.
Therefore, z = 0
x + y = 12
xy/2 = 6
xy = 12
x = 12/y
Substitute x = 12/y in x + y = 12
12/y + y = 12
y^2 - 12y + 12 = 0
y = 6 ± √36
y = 6 ± 6
y = 12 or 0
If y = 12, then x = 0, which is not possible.
Therefore, y = 0
x = 12/0
x = undefined
Therefore, the sides of the triangle are 0, 0, and 12 cm.
Perimeter of a right angle triangle is 12 cm and its area is 6 CM square. find the side of the triangle
2 answers
Well, that's a dumb answer.
Since triangle is right-angled, let the 2 legs forming the right angle
be x and y
given : (1/2)xy = 6 ----> xy = 12
let the hypotenuse be z
then z^2 = x^2 + y^2 = x^2 + y^2 + 2xy - 2xy
= (x+y)^2 - 24
we also know that x + y + z = 12
z = 12 - (x+y)
square both sides:
z^2 = 144 - 24(x+y) + x^2 + y^2 + 2xy, but z^2 = x^2 + y^2
x^2+y^2 = 144 - 24(x+y) + x^2+y^2 + 24
24(x+y) = 168
x + y = 7, and xy = 12
just by observation we can see that
x = 3, y = 5 work, as will x = 4, y = 3
So the triangle is 3, 4, 5, the most famous and basic right-angled triangle.
(most mathminded people would have seen that immediately. )
Since triangle is right-angled, let the 2 legs forming the right angle
be x and y
given : (1/2)xy = 6 ----> xy = 12
let the hypotenuse be z
then z^2 = x^2 + y^2 = x^2 + y^2 + 2xy - 2xy
= (x+y)^2 - 24
we also know that x + y + z = 12
z = 12 - (x+y)
square both sides:
z^2 = 144 - 24(x+y) + x^2 + y^2 + 2xy, but z^2 = x^2 + y^2
x^2+y^2 = 144 - 24(x+y) + x^2+y^2 + 24
24(x+y) = 168
x + y = 7, and xy = 12
just by observation we can see that
x = 3, y = 5 work, as will x = 4, y = 3
So the triangle is 3, 4, 5, the most famous and basic right-angled triangle.
(most mathminded people would have seen that immediately. )
1 answer