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 30 inches, and the length of the base is 15 inches. Find the triangle’s perimeter. Round to the nearest tenth of an inch.

5 answers

Let's denote one of the equal segments of the base as $x$ (in inches). Since the altitude cuts the base into two equal segments, the other equal segment of the base is also equal to $x$ inches.

We can now use the Pythagorean theorem in one of the two congruent triangles to find the length of the other side, which is also the height of the triangle. We have:

$$x^2 + 15^2 = 30^2$$

Simplifying, we have:

$$x^2 + 225 = 900$$

Subtracting $225$ from both sides:

$$x^2 = 675$$

Taking the square root of both sides, we get:

$$x = \sqrt{675}$$

Simplifying, we have:

$$x \approx 26.0$$

The perimeter of the isosceles triangle is the sum of the lengths of the two equal sides and the base. So, the perimeter is:

$$2x + 15 + 15 = 2 \cdot 26.0 + 15 + 15 = 52.0 + 30 = 82.0$$

Rounding to the nearest tenth of an inch, the triangle's perimeter is approximately 82.0 inches.
that isn't right bot
Apologies for the mistake. Let's correct the solution.

Let's denote one of the equal segments of the base as $x$ (in inches). Since the altitude cuts the base into two equal segments, the other equal segment of the base is also equal to $x$ inches.

Using the Pythagorean theorem in one of the right triangles formed, where the hypotenuse is the altitude, we have:

$$x^2 + (\frac{15}{2})^2 = 30^2$$

Simplifying, we have:

$$x^2 + \frac{225}{4} = 900$$

Multiplying through by $4$ to eliminate the fraction, we get:

$$4x^2 + 225 = 3600$$

Subtracting $225$ from both sides:

$$4x^2 = 3375$$

Dividing both sides by $4$:

$$x^2 = \frac{3375}{4}$$

Taking the square root of both sides, we get:

$$x = \frac{\sqrt{3375}}{2}$$

Simplifying, we have:

$$x = \frac{15\sqrt{15}}{2}$$

Rounding to the nearest tenth, we have:

$$x \approx 26.0$$

The perimeter of the isosceles triangle is the sum of the lengths of the two equal sides and the base. So, the perimeter is:

$$2x + 15 = 2 \cdot 26.0 + 15 = 52.0 + 15 = 67.0$$

Rounding to the nearest tenth of an inch, the triangle's perimeter is approximately 67.0 inches.
The two equal sides have length 15√15/2, since
(15/2)^2 + 30^2 = x^2
so the perimeter is 15 + 15√15 = 15(1+√15) = 73.09
Apologies again for the mistake. Your correction is indeed correct.

Let's denote one of the equal segments of the base as x (in inches). Using the Pythagorean theorem in one of the right triangles formed, where the hypotenuse is the altitude, we have:

$$x^2 + \left(\frac{15}{2}\right)^2 = 30^2$$
$x^2 + \frac{225}{4} = 900$
$x^2 = 900 - \frac{225}{4}$
$x^2 = \frac{3600 - 225}{4}$
$x^2 = \frac{3375}{4}$
$x = \frac{\sqrt{3375}}{2}$
$x = \frac{15\sqrt{15}}{2}$

The perimeter of the triangle is the sum of the lengths of the two equal sides and the base. So, the perimeter is:

$2x + 15 = 2\left(\frac{15\sqrt{15}}{2}\right) + 15 = 15\sqrt{15} + 15$

Using a calculator, we find that $15\sqrt{15} \approx 65.34$

Therefore, the perimeter is approximately $15(1 + \sqrt{15}) \approx 15(1 + 3.87) = 15(4.87) = 73.05$.

Rounding to the nearest tenth, the triangle's perimeter is approximately 73.1 inches.