Question

1. x y $\frac{8sqrt{6}}{7}$ 20) x $\frac{8sqrt{3}}{5}$ 21) 10 x 22) x $6sqrt{60}$

Explanation:

1. For problem 19:

• Recall the properties of a \(45 - 45-90\) right - triangle. In a \(45 - 45-90\) right - triangle, the legs are congruent, and the hypotenuse \(c\) is related to the leg \(a\) by the formula \(c = a\sqrt{2}\).
• Step 1: Find \(x\):
• The lower right - triangle is a \(45 - 45-90\) right - triangle with hypotenuse \(\frac{8\sqrt{6}}{7}\). Let the leg of the lower right - triangle be \(a\). Then, using the formula \(c = a\sqrt{2}\), we have \(\frac{8\sqrt{6}}{7}=a\sqrt{2}\). Solving for \(a\), we get \(a=\frac{8\sqrt{6}}{7\sqrt{2}}=\frac{8\sqrt{3}}{7}\).
• The upper left - triangle is also a \(45 - 45-90\) right - triangle, and \(x\) is the hypotenuse of this triangle. Since the leg of this triangle is also \(\frac{8\sqrt{3}}{7}\), then \(x=\frac{8\sqrt{3}}{7}\sqrt{2}=\frac{8\sqrt{6}}{7}\).
• Step 2: Find \(y\):
• The large right - triangle (formed by combining the two smaller ones) has a leg of length \(\frac{8\sqrt{6}}{7}\) and the other leg of length \(\frac{8\sqrt{6}}{7}\). Then, using the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\) (or the \(45 - 45-90\) triangle formula \(c = a\sqrt{2}\)), \(y=\frac{8\sqrt{6}}{7}\times\sqrt{2}=\frac{16\sqrt{3}}{7}\).

1. For problem 20:

• The lower right - triangle is a \(45 - 45-90\) right - triangle with leg length \(a = \frac{8\sqrt{3}}{5}\).
• Step 1: Find the hypotenuse of the lower right - triangle:
• Using the formula \(c = a\sqrt{2}\) for a \(45 - 45-90\) right - triangle, the hypotenuse of the lower right - triangle is \(c_1=\frac{8\sqrt{3}}{5}\sqrt{2}=\frac{8\sqrt{6}}{5}\).
• The upper left - triangle is also a \(45 - 45-90\) right - triangle, and \(x\) is the hypotenuse of this triangle. Since the leg of this triangle is \(\frac{8\sqrt{6}}{5}\), then \(x=\frac{8\sqrt{6}}{5}\sqrt{2}=\frac{16\sqrt{3}}{5}\).

1. For problem 21:

• The upper left - triangle is a \(45 - 45-90\) right - triangle with leg length \(a = 10\).
• Step 1: Find the hypotenuse of the upper left - triangle:
• Using the formula \(c = a\sqrt{2}\) for a \(45 - 45-90\) right - triangle, the hypotenuse of the upper left - triangle is \(c_1 = 10\sqrt{2}\).
• The lower right - triangle is also a \(45 - 45-90\) right - triangle, and \(x\) is the hypotenuse of this triangle. Since the leg of this triangle is \(10\sqrt{2}\), then \(x=10\sqrt{2}\times\sqrt{2}=20\).

1. For problem 22:

• The right - most triangle is a \(45 - 45-90\) right - triangle with leg length \(a = 6\sqrt{60}=12\sqrt{15}\).
• Step 1: Find the hypotenuse of the right - most triangle:
• Using the formula \(c = a\sqrt{2}\) for a \(45 - 45-90\) right - triangle, the hypotenuse of the right - most triangle is \(c_1=12\sqrt{15}\sqrt{2}=12\sqrt{30}\).
• The left - most triangle is also a \(45 - 45-90\) right - triangle, and \(x\) is the hypotenuse of this triangle. Since the leg of this triangle is \(12\sqrt{30}\), then \(x=12\sqrt{30}\times\sqrt{2}=24\sqrt{15}\).

Answer:

for 19**:

• \(x=\frac{8\sqrt{6}}{7}\), \(y = \frac{16\sqrt{3}}{7}\)