Question

two triangles have equal perimeters of 14 cm. one is equilateral, the other is isosceles. explain how their altitudes will compare (greater, smaller, or equal) and justify using reasoning or formulas.

Explanation:

Step1: Find side lengths

For the equilateral triangle with perimeter \(P = 16\mathrm{cm}\), since \(P=3s\) (where \(s\) is the side - length), then \(s=\frac{16}{3}\mathrm{cm}\). The altitude \(h_{1}\) of an equilateral triangle is given by \(h_{1}=\frac{\sqrt{3}}{2}s\). Substituting \(s = \frac{16}{3}\), we get \(h_{1}=\frac{\sqrt{3}}{2}\times\frac{16}{3}=\frac{8\sqrt{3}}{3}\approx4.62\mathrm{cm}\).

Step2: Consider a scalene triangle

Let's assume a scalene triangle with perimeter \(P = 16\mathrm{cm}\). For example, consider a triangle with side - lengths \(a = 5\mathrm{cm}\), \(b=6\mathrm{cm}\), \(c = 5\mathrm{cm}\). Using Heron's formula \(s=\frac{a + b + c}{2}=\frac{16}{2}=8\mathrm{cm}\). The area \(A=\sqrt{s(s - a)(s - b)(s - c)}=\sqrt{8(8 - 5)(8 - 6)(8 - 5)}=\sqrt{8\times3\times2\times3}=\sqrt{144}=12\mathrm{cm}^{2}\). If we assume the base \(b = 6\mathrm{cm}\), then using the formula \(A=\frac{1}{2}bh\), we get \(h_{2}=\frac{2A}{b}\). Substituting \(A = 12\) and \(b = 6\), we have \(h_{2}=4\mathrm{cm}\)

Answer:

The altitude of the equilateral triangle is greater than the altitude of the scalene triangle. In general, for a given perimeter, the equilateral triangle (a regular polygon) encloses the maximum area among all triangles. Using the formula \(A=\frac{1}{2}bh\), for a fixed perimeter, when the area is maximized (in the case of an equilateral triangle) and if we assume a non - degenerate base (non - zero length), the altitude of the equilateral triangle will be greater than the altitude of a non - equilateral (scalene) triangle.