Question

the legs of a right isosceles triangle are congruent to the legs of an isosceles triangle with a 36° vertex angle. which triangle has the greater perimeter

choose the correct answer below.

a. the non - right triangle has a greater perimeter, because the base angles are smaller than those in the right triangle.
b. the non - right triangle has a greater perimeter, because the base will be shorter in the right triangle.
c. the right triangle has a greater perimeter, because its legs are longer than the non - right triangle.
d. the right triangle has a greater perimeter, because the base will be shorter in the non - right triangle.

Explanation:

Let the length of the legs of both triangles be \(l\).

• For the right isosceles triangle:
• The base angles are \(45^{\circ}\) each. Using the Pythagorean theorem, the length of the hypotenuse \(h=\sqrt{l^{2}+l^{2}}=\sqrt{2}l\approx1.414l\). The perimeter \(P_{1}=2l + \sqrt{2}l=(2 + \sqrt{2})l\approx3.414l\).
• For the isosceles triangle with a \(36^{\circ}\) vertex angle:
• The base angles are \(\frac{180^{\circ}- 36^{\circ}}{2}=72^{\circ}\) each. Using the law of cosines, if the sides are \(a = b=l\) (the legs) and \(c\) is the base, then \(c^{2}=l^{2}+l^{2}-2l\times l\times\cos36^{\circ}\). Since \(\cos36^{\circ}\approx0.809\), \(c^{2}=2l^{2}(1 - 0.809)=2l^{2}\times0.191\), so \(c=\sqrt{2\times0.191}l\approx0.618l\). The perimeter \(P_{2}=2l + c\approx2l+0.618l = 2.618l\) (This is incorrect. Let's use the law of sines. Let \(a = b=l\). By the law of sines \(\frac{c}{\sin36^{\circ}}=\frac{l}{\sin72^{\circ}}\), so \(c=\frac{l\sin36^{\circ}}{\sin72^{\circ}}\approx\frac{l\times0.588}{0.951}\approx0.618l\). But actually, if we consider the fact that in a triangle, when the vertex angle is smaller (while the equal sides are of the same length), the base is shorter. Wait, no. Wait, another approach:
• In a triangle, given two sides \(a\) and \(b\) (here \(a = b=l\)), the length of the third side \(c\) is given by \(c^{2}=a^{2}+b^{2}-2ab\cos\theta\) (\(\theta\) is the included angle). When \(\theta = 90^{\circ}\), \(c=\sqrt{2}l\), when \(\theta=36^{\circ}\), \(c=\sqrt{2l^{2}(1 - \cos36^{\circ})}\). Since \(\cos\theta\) decreases as \(\theta\) increases from \(0^{\circ}\) to \(180^{\circ}\). The base of the non - right triangle is longer.
• Another way: In a triangle, the larger the angle opposite a side, the longer the side. In the right isosceles triangle, the angles are \(45^{\circ},45^{\circ},90^{\circ}\). In the isosceles triangle with a \(36^{\circ}\) vertex angle, the base angles (\(72^{\circ}\)) are larger than \(45^{\circ}\). The side opposite the \(72^{\circ}\) angle (the base of the non - right triangle) is longer than the side opposite the \(45^{\circ}\) angle (the base of the right - isosceles triangle). Since the legs are equal in length (\(l\)), the non - right triangle has a greater perimeter. The base angles of the non - right triangle (\(72^{\circ}\)) are larger than the base angles of the right - isosceles triangle (\(45^{\circ}\)), and using the law of sines \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), the base of the non - right triangle is longer. So the non - right triangle has a greater perimeter.

Answer:

A. The non - right triangle has a greater perimeter, because the base angles are smaller than those in the right triangle.