01.008 In an isosceles triangle, a median is drawn to a side equal to 4

In an isosceles triangle, a median is drawn to a side equal to 4. Find the base of the triangle if the median is 3 long

Solution.

Let’s break down this problem step by step. We have an isosceles triangle. Let the two equal sides be denoted as ‘a’, and the base as ‘b’. We are given that one of the equal sides is 4 cm. Let’s say we have triangle ABC, where AC = BC = 4 cm.

A median is drawn to one of the equal sides. Let’s say the median CD is drawn to side AB. No, that’s not right. The problem states “a median is drawn to a side equal to 4 cm”. This means the median is drawn to one of the *equal* sides, let’s say AC. Let D be the midpoint of AC. So, AD = DC = 2 cm. The median BD is given as 3 cm.

Our goal is to find the length of the base, AB. Let’s call the base ‘x’.

To solve this, we can use the Law of Cosines. Consider triangle ABC. Let the angle at vertex C be $\alpha$ .

Using the Law of Cosines in triangle ABC to find the base ‘x’:

x^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos \alpha

Substituting the given values, AC = 4 and BC = 4:

x^2 = 4^2 + 4^2 - 2 \cdot 4 \cdot 4 \cdot \cos \alpha

x^2 = 16 + 16 - 32 \cos \alpha

x^2 = 32 - 32 \cos \alpha \quad (Equation \ 1)

Now, let’s consider the triangle formed by the median. We have triangle BDC, where BC = 4 cm, DC = 2 cm (since D is the midpoint of AC), and the median BD = 3 cm. The angle at C is still $\alpha$ .

Using the Law of Cosines in triangle BDC to find the length of the median ‘m’ (which is 3 cm):

m^2 = BC^2 + DC^2 - 2 \cdot BC \cdot DC \cdot \cos \alpha

Substituting the given values, BC = 4, DC = 2, and m = 3:

3^2 = 4^2 + 2^2 - 2 \cdot 4 \cdot 2 \cdot \cos \alpha

9 = 16 + 4 - 16 \cos \alpha

9 = 20 - 16 \cos \alpha

Now, we can solve this equation for

\cos \alpha

16 \cos \alpha = 20 - 9

16 \cos \alpha = 11

\cos \alpha = \frac{11}{16} \quad (Value \ of \ \cos \alpha)

Now we have the value for $\cos \alpha$ . We can substitute this back into Equation 1 to find $x^2$ :

x^2 = 32 - 32 \left( \frac{11}{16} \right)

x^2 = 32 - 2 \cdot 11

x^2 = 32 - 22

x^2 = 10

To find ‘x’, we take the square root:

x = \sqrt{10}

The length of the base is

\sqrt{10}

cm.

Answer:

\sqrt{10}

cm.

Solution.

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