1. **Problem Statement:** We have an isosceles triangle with an altitude drawn from the vertex to the base, forming two congruent right triangles. The altitude length is 18 inches, and the base length is 17 inches. We need to find the perimeter of the triangle. 2. **Key Properties:** - The altitude divides the base into two equal segments, so each segment is $\frac{17}{2} = 8.5$ inches. - The altitude is perpendicular to the base, forming right triangles. - The two legs of each right triangle are the altitude (18 inches) and half the base (8.5 inches). 3. **Find the length of the equal sides:** Use the Pythagorean theorem for one of the right triangles: $$ \text{side} = \sqrt{(18)^2 + (8.5)^2} $$ Calculate inside the square root: $$ 18^2 = 324, \quad 8.5^2 = 72.25 $$ So, $$ \text{side} = \sqrt{324 + 72.25} = \sqrt{396.25} $$ 4. **Simplify the square root:** $$ \sqrt{396.25} \approx 19.9 $$ 5. **Calculate the perimeter:** The perimeter $P$ is the sum of the base and the two equal sides: $$ P = 17 + 2 \times 19.9 = 17 + 39.8 = 56.8 $$ **Final answer:** The perimeter of the triangle is approximately **56.8 inches**.