Question

independent practice
find the measures of the sides of $\triangle abc$ and classify each triangle by its sides.

1. $a(5, 4)$, $b(3, -1)$, $c(7, -1)$\t\t\t33. $a(-4, 1)$, $b(5, 6)$, $c(-3, -7)$
2. $a(-7, 9)$, $b(-7, -1)$, $c(4, -1)$

Explanation:

Step1: Recall Distance Formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). We'll apply this to find the lengths of \(AB\), \(BC\), and \(AC\) for each triangle.

For Problem 32: \(A(5, 4)\), \(B(3, -1)\), \(C(7, -1)\)

Step2: Calculate \(AB\)

\(x_1 = 5, y_1 = 4\); \(x_2 = 3, y_2 = -1\)
\(AB = \sqrt{(3 - 5)^2 + (-1 - 4)^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\)

Step3: Calculate \(BC\)

\(x_1 = 3, y_1 = -1\); \(x_2 = 7, y_2 = -1\)
\(BC = \sqrt{(7 - 3)^2 + (-1 - (-1))^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4\)

Step4: Calculate \(AC\)

\(x_1 = 5, y_1 = 4\); \(x_2 = 7, y_2 = -1\)
\(AC = \sqrt{(7 - 5)^2 + (-1 - 4)^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}\)

Step5: Classify Triangle 32

Since \(AB = AC = \sqrt{29}\) and \(BC = 4\), it's an isosceles triangle (two sides equal).

For Problem 33: \(A(-4, 1)\), \(B(5, 6)\), \(C(-3, -7)\)

Step2: Calculate \(AB\)

\(x_1 = -4, y_1 = 1\); \(x_2 = 5, y_2 = 6\)
\(AB = \sqrt{(5 - (-4))^2 + (6 - 1)^2} = \sqrt{9^2 + 5^2} = \sqrt{81 + 25} = \sqrt{106}\)

Step3: Calculate \(BC\)

\(x_1 = 5, y_1 = 6\); \(x_2 = -3, y_2 = -7\)
\(BC = \sqrt{(-3 - 5)^2 + (-7 - 6)^2} = \sqrt{(-8)^2 + (-13)^2} = \sqrt{64 + 169} = \sqrt{233}\)

Step4: Calculate \(AC\)

\(x_1 = -4, y_1 = 1\); \(x_2 = -3, y_2 = -7\)
\(AC = \sqrt{(-3 - (-4))^2 + (-7 - 1)^2} = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}\)

Step5: Classify Triangle 33

Since \(AB \approx 10.3\), \(BC \approx 15.26\), \(AC \approx 8.06\), all sides are different. So it's a scalene triangle.

For Problem 34: \(A(-7, 9)\), \(B(-7, -1)\), \(C(4, -1)\)

Step2: Calculate \(AB\)

\(x_1 = -7, y_1 = 9\); \(x_2 = -7, y_2 = -1\)
\(AB = \sqrt{(-7 - (-7))^2 + (-1 - 9)^2} = \sqrt{0^2 + (-10)^2} = \sqrt{100} = 10\)

Step3: Calculate \(BC\)

\(x_1 = -7, y_1 = -1\); \(x_2 = 4, y_2 = -1\)
\(BC = \sqrt{(4 - (-7))^2 + (-1 - (-1))^2} = \sqrt{11^2 + 0^2} = \sqrt{121} = 11\)

Step4: Calculate \(AC\)

\(x_1 = -7, y_1 = 9\); \(x_2 = 4, y_2 = -1\)
\(AC = \sqrt{(4 - (-7))^2 + (-1 - 9)^2} = \sqrt{11^2 + (-10)^2} = \sqrt{121 + 100} = \sqrt{221} \approx 14.87\)

Step5: Classify Triangle 34

Since \(AB = 10\), \(BC = 11\), \(AC \approx 14.87\), all sides are different? Wait, no—wait, \(AB\) is vertical (same \(x\)-coordinate), \(BC\) is horizontal (same \(y\)-coordinate), so angle at \(B\) is right angle? Wait, \(AB\) length 10, \(BC\) length 11, \(AC\) length \(\sqrt{11^2 + 10^2} = \sqrt{221}\), so by Pythagoras, \(AB^2 + BC^2 = 100 + 121 = 221 = AC^2\). So it's a right - scalene triangle (right angle and all sides different in length except the right - angled sides which are 10 and 11).

Answer:

Problem 32:

• \(AB=\sqrt{29}\), \(BC = 4\), \(AC=\sqrt{29}\); Triangle Type: Isosceles.

Problem 33:

• \(AB=\sqrt{106}\), \(BC=\sqrt{233}\), \(AC=\sqrt{65}\); Triangle Type: Scalene.

Problem 34:

• \(AB = 10\), \(BC = 11\), \(AC=\sqrt{221}\); Triangle Type: Right - Scalene (since \(AB^2+BC^2 = AC^2\)).