the perimeter of isosceles triangle is 70 d. of the length of a leg of the triangle is 5 dm less than twice the lenth of the base, find the lengths of the sides of the triangle.
Let x = length of base, then 2x-5 = length of legs (sides).
2(2x-5) + x = 70
Solve for x.
To find the lengths of the sides of an isosceles triangle, we can start by setting up equations based on the given information.
Let's represent the length of the base of the triangle as "b" and the length of each leg as "l."
We know that the perimeter of the isosceles triangle is 70d, so we can write the equation:
Perimeter = 2l + b = 70d.
We are also given that the length of each leg is 5dm less than twice the length of the base, so we can write another equation:
l = 2b - 5.
Now we have a system of equations:
2l + b = 70d,
l = 2b - 5.
To solve this system of equations, we can substitute the second equation into the first equation:
2(2b - 5) + b = 70d.
Simplifying this equation gives us:
4b - 10 + b = 70d,
5b - 10 = 70d,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d + 10,
5b = 70d ,
b = (70d + 10)/5,
b = 14d + 2.
Now that we know the value of b, we can substitute it back into the equation l = 2b - 5 to find the length of each leg:
l = 2(14d + 2) - 5,
l = 28d + 4 - 5,
l = 28d - 1.
Therefore, the lengths of the sides of the isosceles triangle are:
Base (b) = 14d + 2,
Legs (l) = 28d - 1.