Find the measure of each side of isosceles triangle ABC with AB = BC if AB = 4y , BC = 3y + 2 , and AC = 3y.
Well, since triangle ABC is isosceles, that means that AB is equal to BC. So we can set up an equation:
4y = 3y + 2
To solve this equation, we need to isolate y:
4y - 3y = 2
y = 2
Now that we've found the value of y, we can substitute it back into the expressions for AB, BC, and AC:
AB = 4(2) = 8
BC = 3(2) + 2 = 8
AC = 3(2) = 6
So, the measure of each side of triangle ABC is AB = 8, BC = 8, and AC = 6.
To find the measure of each side of the isosceles triangle ABC, we need to set up an equation using the given lengths of AB, BC, and AC.
Since AB = BC, we can set up an equation as follows:
4y = 3y + 2
To solve for y, we can subtract 3y from both sides of the equation:
4y - 3y = 3y + 2 - 3y
Simplifying:
y = 2
Now that we have found the value of y, we can substitute it back into the equation to find the lengths of the sides of the triangle.
AB = 4y = 4(2) = 8
BC = 3y + 2 = 3(2) + 2 = 8
AC = 3y = 3(2) = 6
Therefore, the measures of each side of the isosceles triangle ABC are AB = 8, BC = 8, and AC = 6.
To find the measure of each side of the isosceles triangle ABC, we need to equate the lengths of the two equal sides.
Given:
AB = BC = 4y (let's call this equation 1)
AC = 3y
Since AB = BC, we can equate the expressions:
4y = 3y + 2
To solve this equation, subtract 3y from both sides:
4y - 3y = 3y + 2 - 3y
y = 2
Now we can substitute the value of y back into equation 1 to find the measure of each side:
AB = BC = 4y = 4(2) = 8
Therefore, each side of isosceles triangle ABC measures 8 units.
If AB=BC, 4y = 3y + 2
y = 2
You should be able to take it from here.