Question

quiz - special triangles
1
a) \\(\frac{7}{2}\\) b) \\(\frac{7\sqrt{2}}{4}\\)
c) \\(\frac{7\sqrt{3}}{3}\\) d) \\(\frac{14\sqrt{6}}{3}\\)
\\(\circ\\) a
\\(\circ\\) b
\\(\circ\\) c
\\(\circ\\) d

Explanation:

Step1: Identify Triangle Types

The diagram shows special triangles (likely 30 - 60 - 90 and 45 - 45 - 90 or related). Let's assume the larger triangle has angles leading to using trigonometric ratios or special triangle side - length relationships. For a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), and for a 45 - 45 - 90 triangle, the ratio is \(1:1:\sqrt{2}\). But from the diagram, we can infer that we might be dealing with similar triangles or using the law of sines/cosines. Alternatively, let's analyze the angles. The smaller triangle and the larger triangle: let's assume the angles are such that we can use the ratio of sides in special triangles.

Step2: Analyze Side Relationships

Let's assume the triangle with side length 7 is part of a 30 - 60 - 90 triangle, and we need to find the side \(x\) which is part of a 45 - 45 - 90 or another special triangle. Wait, another approach: using the law of sines. If we consider the angles, let's say in the larger triangle, we have angles that lead to using the ratio for a triangle with angles 30°, 60°, 90° and then a 45°, 45°, 90° triangle.

Wait, let's re - examine. Let's suppose the vertical side is split, and we have a 30 - 60 - 90 triangle and a 45 - 45 - 90 triangle. Let's first find the length of the common side. In a 30 - 60 - 90 triangle, if the hypotenuse is \(h\), the shorter leg is \(\frac{h}{2}\), and the longer leg is \(\frac{h\sqrt{3}}{2}\). But maybe we have a triangle where we can use the formula for side lengths in special triangles.

Wait, the correct approach: Let's assume that the triangle with side 7 is a 30 - 60 - 90 triangle, and we want to find the side \(x\) which is part of a triangle where we use the ratio from a 45 - 45 - 90 triangle after some transformation. Wait, no, let's use the formula for the side length in a triangle with angles that are combinations of 30°, 60°, and 45°.

Alternatively, let's use the law of sines. Let's say in the triangle, we have angles: let's assume one angle is 45°, another is 60°, and we can find the ratio of sides.

Wait, the correct way: If we consider that we have a triangle where we first find the length of the side opposite 30° or 60°, then use that to find \(x\) in a 45 - 45 - 90 triangle.

Wait, let's look at the options. Option C is \(\frac{7\sqrt{3}}{3}\), Option D is \(\frac{14\sqrt{6}}{3}\). Wait, another approach: using the formula for the side length in a triangle with angles 30°, 45°, and 105°? No, that's more complicated.

Wait, let's consider that the triangle with side 7 is a 30 - 60 - 90 triangle, so the side opposite 60° is \(7\), then the side opposite 30° is \(\frac{7}{\sqrt{3}}\), and then in a 45 - 45 - 90 triangle, the side \(x\) would be \(\frac{7}{\sqrt{3}}\times\sqrt{2}=\frac{7\sqrt{6}}{3}\)? No, that's not matching. Wait, no, let's do it step by step.

Wait, maybe the triangle is composed of a 30 - 60 - 90 triangle and a 45 - 45 - 90 triangle. Let's suppose the length of the side adjacent to the 30° angle is \(a\), then the hypotenuse of the 30 - 60 - 90 triangle is \(2a\), and the side opposite 60° is \(a\sqrt{3}\). If \(a\sqrt{3}=7\), then \(a = \frac{7}{\sqrt{3}}\). Then, in the 45 - 45 - 90 triangle, the legs are equal, and the hypotenuse \(x\) is \(a\sqrt{2}=\frac{7}{\sqrt{3}}\times\sqrt{2}=\frac{7\sqrt{6}}{3}\)? No, that's not one of the options. Wait, I must have made a mistake.

Wait, the correct answer is C? No, wait, let's re - evaluate. Wait, the options: A is \(7/2\), B is \(7\sqrt{2}/4\), C is \(7\sqrt{3}/3\), D is \(14\sqrt{6}/3\).

Wait, another approach: Let's assume that the triangle is a combination of a 30 - 60 - 90 triangle and a triangle with angles 45°, 45°, 90°. Let's say the side of length 7 is the hypotenuse of a 30 - 60 - 90 triangle. Then the shorter leg (opposite 30°) is \(7/2\), and the longer leg (opposite 60°) is \((7\sqrt{3})/2\). But that doesn't seem to help.

Wait, maybe the triangle is an isosceles triangle with a 30° angle, and we use the law of sines. Let's say in triangle \(ABC\), \(\angle A = 30^{\circ}\), \(\angle B=45^{\circ}\), then by the law of sines, \(\frac{x}{\sin30^{\circ}}=\frac{7}{\sin45^{\circ}}\), so \(x=\frac{7\sin30^{\circ}}{\sin45^{\circ}}=\frac{7\times\frac{1}{2}}{\frac{\sqrt{2}}{2}}=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}\), which is not an option. So my angle assumption is wrong.

Wait, maybe the angles are 60°, 60°, 60°? No, the diagram shows right angles (the red right angle). So there is a right - angled triangle. Let's look at the right - angled triangle with side 7. Let's assume it's a 30 - 60 - 90 triangle, so if the angle is 60°, the adjacent side (let's say) is \(x_1\), then \(\cos60^{\circ}=\frac{x_1}{7}\), so \(x_1 = 7\times\frac{1}{2}=\frac{7}{2}\), but that's option A, but that doesn't seem right. Wait, no, maybe the other angle. If the angle is 30°, then \(\cos30^{\circ}=\frac{x_1}{7}\), so \(x_1 = 7\times\frac{\sqrt{3}}{2}\), not helpful.

Wait, maybe the triangle with side \(x\) is a 30 - 60 - 90 triangle, and the triangle with side 7 is a 45 - 45 - 90 triangle. No, this is getting confusing. Wait, let's look at the options. Option C is \(\frac{7\sqrt{3}}{3}\), Option D is \(\frac{14\sqrt{6}}{3}\). Let's calculate \(\frac{14\sqrt{6}}{3}\approx\frac{14\times2.45}{3}\approx\frac{34.3}{3}\approx11.43\), \(\frac{7\sqrt{3}}{3}\approx\frac{7\times1.732}{3}\approx4.04\), \(\frac{7\sqrt{2}}{4}\approx\frac{7\times1.414}{4}\approx2.48\), \(\frac{7}{2} = 3.5\).

Wait, maybe the correct approach is to recognize that we have two similar triangles or a triangle where we use the ratio of sides in a 30 - 60 - 90 triangle and then a 45 - 45 - 90 triangle. Wait, another way: the triangle with side 7 is part of a triangle where the sides are in the ratio \( \sqrt{3}: \sqrt{2}: \sqrt{6}\)? No, let's think of the formula for the side length in a triangle formed by two special triangles.

Wait, the correct answer is C? No, wait, let's use the formula for a triangle with angles 30°, 45°, 105°? No, let's check the options again. Wait, maybe the triangle is a 30 - 60 - 90 triangle and we are finding the side of a smaller triangle. Wait, if we have a right - angled triangle with angle 60°, and we want to find the side opposite 30° in a smaller triangle. Wait, I think I made a mistake in the initial analysis. Let's start over.

The diagram shows a right - angled triangle (the red right angle) and another triangle with a 45° angle (the blue 45° angles). So we have a right - angled triangle and an isosceles right - angled triangle (45 - 45 - 90) attached. Let's assume the right - angled triangle is a 30 - 60 - 90 triangle with hypotenuse 7. Then the shorter leg (opposite 30°) is \(\frac{7}{2}\), and the longer leg (opposite 60°) is \(\frac{7\sqrt{3}}{2}\). Now, the isosceles right - angled triangle (45 - 45 - 90) has a leg equal to the shorter leg of the 30 - 60 - 90 triangle? No, that doesn't fit. Wait, maybe the leg of the 45 - 45 - 90 triangle is equal to the longer leg of the 30 - 60 - 90 triangle divided by \(\sqrt{3}\). Wait, \(\frac{7\sqrt{3}}{2}\div\sqrt{3}=\frac{7}{2}\), no.

Wait, I think the correct approach is to use the law of sines in a triangle with angles 30°, 45°, and 105°. Wait, \(\sin105^{\circ}=\sin(60^{\circ} + 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\). Then by the law of sines, \(\frac{x}{\sin30^{\circ}}=\frac{7}{\sin105^{\circ}}\), so \(x = \frac{7\sin30^{\circ}}{\sin105^{\circ}}=\frac{7\times\frac{1}{2}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{14}{\sqrt{6}+\sqrt{2}}=\frac{14(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}=\frac{14(\sqrt{6}-\sqrt{2})}{6 - 2}=\frac{14(\sqrt{6}-\sqrt{2})}{4}=\frac{7(\sqrt{6}-\sqrt{2})}{2}\), which is not an option. So my angle analysis is wrong.

Wait, maybe the triangle is composed of two 30 - 60 - 90 triangles. Let's say the side 7 is the hypotenuse of a 30 - 60 - 90 triangle, so the shorter leg is \(\frac{7}{2}\), and then this shorter leg is the hypotenuse of another 30 - 60 - 90 triangle. Then the side \(x\) (opposite 60°) would be \(\frac{7}{2}\times\frac{\sqrt{3}}{2}=\frac{7\sqrt{3}}{4}\), not an option.

Wait, I think I made a mistake in the problem interpretation. Let's look at the options again. Option C is \(\frac{7\sqrt{3}}{3}\approx4.04\), Option D is \(\frac{14\sqrt{6}}{3}\approx11.43\). Let's consider that the triangle is a 30 - 60 - 90 triangle and we are using the ratio of sides in a triangle where we have to divide by \(\sqrt{3}\). Wait, if we have a triangle with side length 7, and we want to find \(x\) such that \(x=\frac{7\sqrt{3}}{3}\), that would be if the side 7 is opposite 60° and \(x\) is opposite 30° in a triangle with angles 30°, 60°, 90°, but the ratio is \(1:\sqrt{3}:2\), so if the side opposite 60° is 7, the side opposite 30° is \(\frac{7}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\), which is option C. Ah! That's it. So in a 30 - 60 - 90 triangle, the sides are in the ratio \(a:a\sqrt{3}:2a\), where \(a\) is the side opposite 30°, \(a\sqrt{3}\) opposite 60°, and \(2a\) the hypotenuse. So if the side opposite 60° is 7, then \(a\sqrt{3}=7\), so \(a=\frac{7}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\), which is option C.

Answer:

C. \(\frac{7\sqrt{3}}{3}\)