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Special Isosceles Triangle Properties - Problem 1

Teacher/Instructor Brian McCall
Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

Given the perimeter of an isosceles triangle and the length of a leg of the triangle, it is possible to find the measure of the base side. Recall that the two legs of an isosceles triangle are the same. This means that two sides of the isosceles triangle have the same length. The base side length of the triangle is twice the length of the leg subtracted from the given perimeter.

Then, also recall that the altitude of an isosceles triangle is the median of the base. So, the altitude divides the base into two equal segments. Therefore, the distance from the altitude to the base vertex of the isosceles triangle is half of the length of the base length.

Let’s look at a problem where we can apply what we know about the special segment in an isosceles triangle. The problem says if the perimeter of ABC, our triangle, is 36cm and if AB is equal to 10cm, find the segment DC.

So let’s start by writing in what we know. Well we’re given that AB is equal 10cm, since we have an isosceles triangle which I know from these markings, I can say that BC must also be 10 centimetres. So if I add up these three sides including the base, I get 36.

So what I’m going to do is I’m going to split this up into 2 pieces called x, but why can I do that? Because this altitude in my isosceles triangle from the vertex angle, is also the median, so what this point does it bisects this line segment AC. So let’s write an equation.

We know that 36 is the sum of our total perimeter, so that’s 10 plus 10 which in my head I’m going to do is 20, plus x and x which is 2x. So if I solve this equation, I’m going to subtract 20 from both sides and I get 16 equals 2x and if I divide by 2, I see that x must equal 8. X also happens to be DC, so find line segment DC, that’s just going to be 8cm.

The key to this problem is remembering that this altitude is also the median of this base.

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