Answer:
Given: Consider an isosceles triangle ABC as shown below in the attachment.
Perimeter = 20 in. and the mid segment through the legs of an isosceles triangle is 4 in.
An isosceles triangle is a triangle that has two sides of equal length.
Therefore, in triangle ABC, AB = AC = a (let) andf BC = b (let)
Construct a line DE parallel to BC.
The Triangle Mid segment Theorem states that in a triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length.
As we know the mid segment is 4 in. i.e, DE = 4 in.
therefore, by mid segment theorem the parallel side i.e, BC = 2 DE
then; BC = [tex]2 \cdot 4 = 8[/tex] in.
or BC = b = 8 in.
Perimeter of an isosceles triangle is the sum of the three lengths.
I,e P = a+a+b = 2a +b ; where P is the perimeter , a and b are the sides of the triangle.
Now, substituting the values of b = 8 in in the given perimeter formula to solve for a;
[tex]2a+8 =20[/tex]
Subtract 8 from both sides of an equation;
2a+8-8 = 20 -8
Simplify:
2a = 12
Divide both sides by 2 we get;
[tex]\frac{2a}{2} = \frac{12}{2}[/tex]
Simplify:
a =6 in.
Therefore, the sides of the given triangle is; a= 6 in and b = 8 in
