Question

In isosceles triangle ABC, AC=BC,AB=6 in, line CD is perpendicular to line AB, and CD= square root of 3.
Find the perimeter of the isosceles triangle.

Answer 1

Let's denote the lengths of the sides of isosceles triangle \( ABC \):

- Since \( AC = BC \), let's denote these equal sides as \( x \).
- The base \( AB \) is given as \( 6 \) inches.

We have a perpendicular \( CD \) from point \( C \) to line \( AB \), and it is given that \( CD = \sqrt{3} \) inches.

To find \( x \) (the lengths of sides \( AC \) and \( BC \)), we can position the triangle in a coordinate plane for easier calculations:

- Let point \( A \) be at \( (-3, 0) \), point \( B \) be at \( (3, 0) \), and point \( C \) be at \( (0, h) \) where \( h \) is the height from point \( C \) to line \( AB \).
- Hence, \( CD = h = \sqrt{3} \), so \( h = \sqrt{3} \).

Now we apply the Pythagorean theorem in triangle \( ACD \):

\[
AC^2 = AD^2 + CD^2
\]

Now, we find \( AD \):

- \( AD \) can be found as half the base \( AB \), which is:

\[
AD = \frac{AB}{2} = \frac{6}{2} = 3
\]

Now substituting our known lengths into the equation:

\[
AC^2 = AD^2 + CD^2
\]

This gives:

\[
x^2 = 3^2 + (\sqrt{3})^2
\]

Calculating, we have:

\[
x^2 = 9 + 3 = 12
\]

Taking the square root of both sides, we find:

\[
x = \sqrt{12} = 2\sqrt{3}
\]

Now the lengths of the sides of triangle \( ABC \) are:

- \( AC = BC = 2\sqrt{3} \)
- \( AB = 6 \)

The perimeter \( P \) of triangle \( ABC \) is:

\[
P = AB + AC + BC = 6 + 2\sqrt{3} + 2\sqrt{3}
\]

Simplifying further:

\[
P = 6 + 4\sqrt{3}
\]

Thus, the perimeter of isosceles triangle \( ABC \) is given by:

\[
\boxed{6 + 4\sqrt{3}}
\]