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 29 inches, and the length of the base is 18 inches. find the triangles perimeter. round to the nearest tenth of an inch.

Explanation:

Step1: Find the length of the equal sides

Since the altitude of an isosceles triangle bisects the base, each segment of the base is $\frac{18}{2}=9$ inches.
Let the length of the equal side of the isosceles triangle be $l$. Using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$ (where $a = 9$, $b=29$ and $c = l$), we have $l=\sqrt{9^{2}+29^{2}}$.

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Step2: Calculate the perimeter

The perimeter $P$ of the isosceles triangle is $P = 2l+18$. Substitute $l=\sqrt{922}\approx30.4$ into the formula.

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Answer:

The perimeter of the isosceles triangle is $78.8$ inches.