Altitude of Isosceles triangle

In this chapter, we will derive the formula to calculate the altitude length of isosceles triangle.

Deriving Isosceles altitude formula

The triangle in which two sides are of equal length are called isosceles triangle.

Here length of sides are given as;
AB = a cm
AC = a cm
BC = b cm

Calculating altitude of isosceles triangle

Note that in Isosceles triangle, the altitude divides the base into two equal parts.

So, BM = MC = b/2

Now applying Pythagoras theorem in triangle ABM.

\mathtt{AB^{2} =\ BM^{2} +AM^{2}}

Putting the values, we get;

\mathtt{a^{2} =\ \left(\frac{b}{2}\right)^{2} +h^{2}}\\\ \\ \mathtt{h^{2} =\ a^{2} -\frac{b^{2}}{4}}\\\ \\ \mathtt{h^{2} =\ \frac{4a^{2} -b^{2}}{4}}\\\ \\ \mathtt{h=\sqrt{\frac{4a^{2} -b^{2}}{4}}}\\\ \\ \mathtt{h\ =\ \frac{1}{2}\sqrt{4a^{2} -b^{2}}}

Using the above formula, we can calculate the length of the altitude of isosceles triangle.

You have to remember this formula for examination purpose.

Calculating Isosceles triangle altitude – Solved Problems

Example 01
Given below is the Isosceles triangle ABC with sides AB = AC = 4 cm and side BC = 5 cm. Find the length of altitude h.

Problems on calculating isosceles triangle

The formula for altitude of isosceles triangle is given as;

\mathtt{h\ =\ \frac{1}{2}\sqrt{4a^{2} -b^{2}}}

Putting the values;

\mathtt{h\ =\ \frac{1}{2}\sqrt{4( 4)^{2} -5^{2}}}\\\ \\ \mathtt{h\ =\ \frac{1}{2}\sqrt{64-25}}\\\ \\ \mathtt{h\ =\ \frac{1}{2}\sqrt{39}}

Hence, the length of altitude is \mathtt{h\ =\ \frac{1}{2}\sqrt{39}} cm.

Example 02
Given below is the isosceles triangle. Find the length of altitude BQ.

Altitude of isosceles triangle solved examples

Since altitude BQ intersect side AC, we will consider AC as a base of triangle.

Formula for altitude of isosceles triangle is given as;

\mathtt{h\ =\ \frac{1}{2}\sqrt{4a^{2} -b^{2}}}

Putting the values;

\mathtt{h\ =\ \frac{1}{2}\sqrt{4( 6)^{2} -4^{2}}}\\\ \\ \mathtt{h\ =\ \frac{1}{2}\sqrt{144-16}}\\\ \\ \mathtt{h\ =\ \frac{1}{2}\sqrt{128}}\\\ \\ \mathtt{h\ =\ \frac{8\sqrt{2}}{2}}\\\ \\ \mathtt{h\ =\ 4\sqrt{2} \ cm\ }

Hence, \mathtt{4\sqrt{2} \ cm} is the length of required altitude.

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