Question

drag the tiles to the boxes to form correct pairs. match each set of vertices to the triangle they form. right scalene obtuse scalene right isosceles acute isosceles acute equilateral a(2,4), b(4,5), c(3,6) a(3,5), b(3,4), c(5,4) a(2,4), b(3,5), c(2,6) a(3,5), b(5,6), c(3,0)

Explanation:

To solve this, we use the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the side lengths, then classify the triangle by side (isosceles, scalene, equilateral) and angle (acute, right, obtuse) using the Pythagorean theorem or angle properties.

1. Triangle \(A(2,4), B(4,5), C(3,6)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(4 - 2)^2 + (5 - 4)^2} = \sqrt{4 + 1} = \sqrt{5}\)
• \(BC = \sqrt{(3 - 4)^2 + (6 - 5)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(AC = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1 + 4} = \sqrt{5}\)

Step 2: Classify by sides

Two sides (\(AB, AC\)) are equal → isosceles.

Step 3: Classify by angles

Check the largest side (\(AB = AC = \sqrt{5}\), \(BC = \sqrt{2}\); largest side is \(\sqrt{5}\)).
Using the Law of Cosines for angle at \(B\) (opposite \(AC\)):
\(\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{5 + 2 - 5}{2 \cdot \sqrt{5} \cdot \sqrt{2}} = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}} > 0\) → angle \(B\) is acute. All angles are acute (since it’s isosceles and no right/obtuse signs) → acute isosceles.

2. Triangle \(A(3,5), B(3,4), C(5,4)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(3 - 3)^2 + (4 - 5)^2} = \sqrt{0 + 1} = 1\)
• \(BC = \sqrt{(5 - 3)^2 + (4 - 4)^2} = \sqrt{4 + 0} = 2\)
• \(AC = \sqrt{(5 - 3)^2 + (4 - 5)^2} = \sqrt{4 + 1} = \sqrt{5}\)

Step 2: Classify by sides

All sides are different → scalene.

Step 3: Classify by angles

Check \(AB^2 + BC^2 = 1^2 + 2^2 = 5\), and \(AC^2 = (\sqrt{5})^2 = 5\) → \(AB^2 + BC^2 = AC^2\) → right triangle. Thus, right scalene.

3. Triangle \(A(2,4), B(3,5), C(2,6)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(3 - 2)^2 + (5 - 4)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(BC = \sqrt{(2 - 3)^2 + (6 - 5)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(AC = \sqrt{(2 - 2)^2 + (6 - 4)^2} = \sqrt{0 + 4} = 2\)

Step 2: Classify by sides

Two sides (\(AB, BC\)) are equal → isosceles.

Step 3: Classify by angles

Check \(AB^2 + BC^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 4\), and \(AC^2 = 2^2 = 4\) → \(AB^2 + BC^2 = AC^2\) → right triangle. Thus, right isosceles.

4. Triangle \(A(3,5), B(5,6), C(3,0)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(5 - 3)^2 + (6 - 5)^2} = \sqrt{4 + 1} = \sqrt{5}\)
• \(BC = \sqrt{(3 - 5)^2 + (0 - 6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\)
• \(AC = \sqrt{(3 - 3)^2 + (0 - 5)^2} = \sqrt{0 + 25} = 5\)

Step 2: Classify by sides

All sides are different → scalene.

Step 3: Classify by angles

Check the largest side (\(BC = 2\sqrt{10} \approx 6.32\)).
\(AB^2 + AC^2 = 5 + 25 = 30\), \(BC^2 = 40\). Since \(AB^2 + AC^2 < BC^2\) → angle at \(A\) is obtuse (Law of Cosines: \(\cos A < 0\)) → obtuse scalene.

Final Matches:

• \(A(2,4), B(4,5), C(3,6)\) → acute isosceles
• \(A(3,5), B(3,4), C(5,4)\) → right scalene
• \(A(2,4), B(3,5), C(2,6)\) → right isosceles
• \(A(3,5), B(5,6), C(3,0)\) → obtuse scalene

(Note: The “acute equilateral” tile is not used here, as no triangle has all sides equal.)

Answer:

To solve this, we use the distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) to find the side lengths, then classify the triangle by side (isosceles, scalene, equilateral) and angle (acute, right, obtuse) using the Pythagorean theorem or angle properties.

1. Triangle \(A(2,4), B(4,5), C(3,6)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(4 - 2)^2 + (5 - 4)^2} = \sqrt{4 + 1} = \sqrt{5}\)
• \(BC = \sqrt{(3 - 4)^2 + (6 - 5)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(AC = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1 + 4} = \sqrt{5}\)

Step 2: Classify by sides

Two sides (\(AB, AC\)) are equal → isosceles.

Step 3: Classify by angles

Check the largest side (\(AB = AC = \sqrt{5}\), \(BC = \sqrt{2}\); largest side is \(\sqrt{5}\)).
Using the Law of Cosines for angle at \(B\) (opposite \(AC\)):
\(\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{5 + 2 - 5}{2 \cdot \sqrt{5} \cdot \sqrt{2}} = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}} > 0\) → angle \(B\) is acute. All angles are acute (since it’s isosceles and no right/obtuse signs) → acute isosceles.

2. Triangle \(A(3,5), B(3,4), C(5,4)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(3 - 3)^2 + (4 - 5)^2} = \sqrt{0 + 1} = 1\)
• \(BC = \sqrt{(5 - 3)^2 + (4 - 4)^2} = \sqrt{4 + 0} = 2\)
• \(AC = \sqrt{(5 - 3)^2 + (4 - 5)^2} = \sqrt{4 + 1} = \sqrt{5}\)

Step 2: Classify by sides

All sides are different → scalene.

Step 3: Classify by angles

Check \(AB^2 + BC^2 = 1^2 + 2^2 = 5\), and \(AC^2 = (\sqrt{5})^2 = 5\) → \(AB^2 + BC^2 = AC^2\) → right triangle. Thus, right scalene.

3. Triangle \(A(2,4), B(3,5), C(2,6)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(3 - 2)^2 + (5 - 4)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(BC = \sqrt{(2 - 3)^2 + (6 - 5)^2} = \sqrt{1 + 1} = \sqrt{2}\)
• \(AC = \sqrt{(2 - 2)^2 + (6 - 4)^2} = \sqrt{0 + 4} = 2\)

Step 2: Classify by sides

Two sides (\(AB, BC\)) are equal → isosceles.

Step 3: Classify by angles

Check \(AB^2 + BC^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 4\), and \(AC^2 = 2^2 = 4\) → \(AB^2 + BC^2 = AC^2\) → right triangle. Thus, right isosceles.

4. Triangle \(A(3,5), B(5,6), C(3,0)\)

Step 1: Calculate side lengths

• \(AB = \sqrt{(5 - 3)^2 + (6 - 5)^2} = \sqrt{4 + 1} = \sqrt{5}\)
• \(BC = \sqrt{(3 - 5)^2 + (0 - 6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\)
• \(AC = \sqrt{(3 - 3)^2 + (0 - 5)^2} = \sqrt{0 + 25} = 5\)

Step 2: Classify by sides

All sides are different → scalene.

Step 3: Classify by angles

Check the largest side (\(BC = 2\sqrt{10} \approx 6.32\)).
\(AB^2 + AC^2 = 5 + 25 = 30\), \(BC^2 = 40\). Since \(AB^2 + AC^2 < BC^2\) → angle at \(A\) is obtuse (Law of Cosines: \(\cos A < 0\)) → obtuse scalene.

Final Matches:

• \(A(2,4), B(4,5), C(3,6)\) → acute isosceles
• \(A(3,5), B(3,4), C(5,4)\) → right scalene
• \(A(2,4), B(3,5), C(2,6)\) → right isosceles
• \(A(3,5), B(5,6), C(3,0)\) → obtuse scalene

(Note: The “acute equilateral” tile is not used here, as no triangle has all sides equal.)