Question

An altitude is drawn from the vertex of an isosceles triangle, forming a right angle and two congruent triangles. As a result, the altitude cuts the base into two equal segments. The length of the altitude is 18 inches, and the length of the base is 17 inches. Find the triangle’s perimeter. Round to the nearest tenth of an inch.

Answer 1

To find the perimeter of the isosceles triangle with altitude, we work through the problem step by step.

Given:
- The length of the altitude (height) \( h = 18 \) inches.
- The length of the base \( b = 17 \) inches.

Since the altitude divides the base into two equal segments, each segment will be:
\[
\frac{b}{2} = \frac{17}{2} = 8.5 \text{ inches.}
\]

Now, our altitude, base segment, and the side of the triangle can be understood as forming a right triangle. The altitude forms a right triangle with half of the base, where one leg is the height \( h \), the other leg is half of the base \( \frac{b}{2} = 8.5 \) inches, and the hypotenuse is the length of one of the equal sides \( s \) of the isosceles triangle.

Using the Pythagorean theorem, we have:
\[
s^2 = h^2 + \left(\frac{b}{2}\right)^2.
\]
Substituting the known values:
\[
s^2 = 18^2 + 8.5^2.
\]

Calculating \( 18^2 \) and \( 8.5^2 \):
\[
18^2 = 324,
\]
\[
8.5^2 = 72.25.
\]

Thus, we can find \( s^2 \):
\[
s^2 = 324 + 72.25 = 396.25.
\]

Now we take the square root to find \( s \):
\[
s = \sqrt{396.25} \approx 19.9 \text{ inches.}
\]

Now we can find the perimeter \( P \) of the triangle. The perimeter of the isosceles triangle is given by:
\[
P = 2s + b.
\]
Substituting \( s \) and \( b \) into the formula:
\[
P = 2(19.9) + 17,
\]
\[
P = 39.8 + 17 = 56.8 \text{ inches.}
\]

Therefore, the perimeter of the triangle, rounded to the nearest tenth of an inch, is:
\[
\boxed{56.8} \text{ inches.}
\]