Question

in exercises 23 and 24, (a) find the length of each side of the right triangle, and (b) show that these lengths satisfy the pythagorean theorem. 23. y (4, 5) 5 4 3 2 (0, 2) (4, 2) 1 1 2 3 4 5 x 24. y (13, 5) 8 4 (1, 0) (13, 0) x in exercises 25 - 34, (a) plot the points, (b) find the distance between the points, and (c) find the midpoint of the line

Explanation:

1. For Exercise 23:

• (a) Find the length of each - side of the right - triangle:
• Step 1: Find the length of the horizontal side:
• The two endpoints of the horizontal side are \((0,2)\) and \((4,2)\). Using the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), when \(x_1 = 0,y_1 = 2,x_2 = 4,y_2 = 2\), we have \(d_1=\sqrt{(4 - 0)^2+(2 - 2)^2}=\sqrt{4^2+0^2}=4\).
• Step 2: Find the length of the vertical side:
• The two endpoints of the vertical side are \((4,2)\) and \((4,5)\). Using the distance formula, when \(x_1 = 4,y_1 = 2,x_2 = 4,y_2 = 5\), we have \(d_2=\sqrt{(4 - 4)^2+(5 - 2)^2}=\sqrt{0^2+3^2}=3\).
• Step 3: Find the length of the hypotenuse:
• The two endpoints of the hypotenuse are \((0,2)\) and \((4,5)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), when \(x_1 = 0,y_1 = 2,x_2 = 4,y_2 = 5\), we have \(d_3=\sqrt{(4 - 0)^2+(5 - 2)^2}=\sqrt{4^2 + 3^2}=\sqrt{16+9}=\sqrt{25}=5\).
• (b) Show that these lengths satisfy the Pythagorean Theorem:
• The Pythagorean Theorem states that for a right - triangle with side lengths \(a\), \(b\), and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).
• Here, \(a = 3\), \(b = 4\), and \(c = 5\). Then \(a^{2}+b^{2}=3^{2}+4^{2}=9 + 16=25\) and \(c^{2}=5^{2}=25\). So \(a^{2}+b^{2}=c^{2}\).

1. For Exercise 24:

• (a) Find the length of each side of the right - triangle:
• Step 1: Find the length of the horizontal side:
• The two endpoints of the horizontal side are \((1,0)\) and \((13,0)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), when \(x_1 = 1,y_1 = 0,x_2 = 13,y_2 = 0\), we have \(d_1=\sqrt{(13 - 1)^2+(0 - 0)^2}=\sqrt{12^2+0^2}=12\).
• Step 2: Find the length of the vertical side:
• The two endpoints of the vertical side are \((13,0)\) and \((13,5)\). Using the distance formula, when \(x_1 = 13,y_1 = 0,x_2 = 13,y_2 = 5\), we have \(d_2=\sqrt{(13 - 13)^2+(5 - 0)^2}=\sqrt{0^2+5^2}=5\).
• Step 3: Find the length of the hypotenuse:
• The two endpoints of the hypotenuse are \((1,0)\) and \((13,5)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), when \(x_1 = 1,y_1 = 0,x_2 = 13,y_2 = 5\), we have \(d_3=\sqrt{(13 - 1)^2+(5 - 0)^2}=\sqrt{12^2+5^2}=\sqrt{144 + 25}=\sqrt{169}=13\).
• (b) Show that these lengths satisfy the Pythagorean Theorem:
• The Pythagorean Theorem states that \(a^{2}+b^{2}=c^{2}\). Here, \(a = 5\), \(b = 12\), and \(c = 13\). Then \(a^{2}+b^{2}=5^{2}+12^{2}=25+144 = 169\) and \(c^{2}=13^{2}=169\). So \(a^{2}+b^{2}=c^{2}\).

Answer:

• Exercise 23:
• (a) Horizontal side: 4, Vertical side: 3, Hypotenuse: 5.
• (b) Satisfies the Pythagorean Theorem since \(3^{2}+4^{2}=5^{2}\).
• Exercise 24:
• (a) Horizontal side: 12, Vertical side: 5, Hypotenuse: 13.
• (b) Satisfies the Pythagorean Theorem since \(5^{2}+12^{2}=13^{2}\).