QUESTION IMAGE
Question
in exercises 7–10, classify △abc by its sides. then determine whether it is a right triangle. (see example 2.)
- a(2, 3), b(6, 3), c(2, 7)
- a(3, 3), b(6, 9), c(6, −3)
- a(1, 9), b(4, 8), c(2, 5)
- a(−2, 3), b(0, −3), c(3, −2)
Let's solve problem 7 first (we can solve others similarly).
Problem 7:
We have points \( A(2, 3) \), \( B(6, 3) \), \( C(2, 7) \).
Step 1: Calculate the lengths of the sides using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
- Length of \( AB \):
\( x_1 = 2, y_1 = 3 \); \( x_2 = 6, y_2 = 3 \)
\( AB=\sqrt{(6 - 2)^2+(3 - 3)^2}=\sqrt{4^2 + 0^2}=\sqrt{16}=4 \)
- Length of \( AC \):
\( x_1 = 2, y_1 = 3 \); \( x_2 = 2, y_2 = 7 \)
\( AC=\sqrt{(2 - 2)^2+(7 - 3)^2}=\sqrt{0^2+4^2}=\sqrt{16}=4 \)
- Length of \( BC \):
\( x_1 = 6, y_1 = 3 \); \( x_2 = 2, y_2 = 7 \)
\( BC=\sqrt{(2 - 6)^2+(7 - 3)^2}=\sqrt{(- 4)^2+4^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2} \)
Step 2: Classify the triangle by sides and check for right triangle
- By sides: Since \( AB = AC = 4 \), it is an isosceles triangle.
- Check for right triangle: Use Pythagorean theorem. \( AB^2+AC^2=4^2 + 4^2=16 + 16 = 32 \) and \( BC^2=(4\sqrt{2})^2=32 \). Since \( AB^2 + AC^2=BC^2 \), it is a right - isosceles triangle (right triangle with \( \angle A = 90^{\circ} \))
Problem 8:
Points \( A(3, 3) \), \( B(6, 9) \), \( C(6,-3) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1 = 3,y_1 = 3;x_2 = 6,y_2 = 9 \)
\( AB=\sqrt{(6 - 3)^2+(9 - 3)^2}=\sqrt{3^2+6^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5} \)
- \( AC \): \( x_1 = 3,y_1 = 3;x_2 = 6,y_2=-3 \)
\( AC=\sqrt{(6 - 3)^2+(-3 - 3)^2}=\sqrt{3^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5} \)
- \( BC \): \( x_1 = 6,y_1 = 9;x_2 = 6,y_2=-3 \)
\( BC=\sqrt{(6 - 6)^2+(-3 - 9)^2}=\sqrt{0^2+(-12)^2}=\sqrt{144}=12 \)
Step 2: Classify
- By sides: \( AB = AC=3\sqrt{5} \), isosceles triangle.
- Check for right triangle: \( AB^2+AC^2=(3\sqrt{5})^2+(3\sqrt{5})^2 = 45+45 = 90 \), \( BC^2 = 144 \). \( AB^2+AC^2
eq BC^2 \). Check \( AB^2+BC^2=(3\sqrt{5})^2+12^2=45 + 144 = 189
eq AC^2 \), \( AC^2+BC^2=(3\sqrt{5})^2+12^2=45 + 144 = 189
eq AB^2 \). Not a right triangle.
Problem 9:
Points \( A(1, 9) \), \( B(4, 8) \), \( C(2, 5) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1 = 1,y_1 = 9;x_2 = 4,y_2 = 8 \)
\( AB=\sqrt{(4 - 1)^2+(8 - 9)^2}=\sqrt{3^2+(-1)^2}=\sqrt{9 + 1}=\sqrt{10} \)
- \( AC \): \( x_1 = 1,y_1 = 9;x_2 = 2,y_2 = 5 \)
\( AC=\sqrt{(2 - 1)^2+(5 - 9)^2}=\sqrt{1^2+(-4)^2}=\sqrt{1 + 16}=\sqrt{17} \)
- \( BC \): \( x_1 = 4,y_1 = 8;x_2 = 2,y_2 = 5 \)
\( BC=\sqrt{(2 - 4)^2+(5 - 8)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4 + 9}=\sqrt{13} \)
Step 2: Classify
- By sides: All sides \( \sqrt{10},\sqrt{13},\sqrt{17} \) are of different lengths, scalene triangle.
- Check for right triangle: \( AB^2 = 10,AC^2 = 17,BC^2 = 13 \). \( AB^2+BC^2=10 + 13=23
eq AC^2 \), \( AB^2+AC^2=10 + 17 = 27
eq BC^2 \), \( AC^2+BC^2=17 + 13=30
eq AB^2 \). Not a right triangle.
Problem 10:
Points \( A(-2, 3) \), \( B(0,-3) \), \( C(3,-2) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1=-2,y_1 = 3;x_2 = 0,y_2=-3 \)
\( AB=\sqrt{(0 + 2)^2+(-3 - 3)^2}=\sqrt{2^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10} \)
- \( AC \): \( x_1=-2,y_1 = 3;x_2 = 3,y_2=-2 \)
\( AC=\sqrt{(3 + 2)^2+(-2 - 3)^2}=\sqrt{5^2+(-5)^2}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2} \)
- \( BC \): \( x_1 = 0,y_1=-3;x_2 = 3,y_2=-2 \)
\( BC=\sqrt{(3 - 0)^2+(-2 + 3)^2}=\sqrt{3^2+1^2}=\sqrt{9 + 1}=\sqrt{10} \)
Step 2: Classify
- By sides: \( AB = 2\sqrt{10},BC=\sqrt{10},AC = 5\sqrt{2} \). Check \( AB^2=(2\sqrt{10})^2 = 40 \), \( BC^2 = 10 \), \( AC^2=50 \). Since \( AB^2+BC^2=40 + 10=50=AC^2 \), it is a right triangle (right - angled at \( B \)) and also a scalene triangle (since \( AB
eq BC
eq AC \))
Final Answers (for each problem):
- Problem 7:…
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
Let's solve problem 7 first (we can solve others similarly).
Problem 7:
We have points \( A(2, 3) \), \( B(6, 3) \), \( C(2, 7) \).
Step 1: Calculate the lengths of the sides using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
- Length of \( AB \):
\( x_1 = 2, y_1 = 3 \); \( x_2 = 6, y_2 = 3 \)
\( AB=\sqrt{(6 - 2)^2+(3 - 3)^2}=\sqrt{4^2 + 0^2}=\sqrt{16}=4 \)
- Length of \( AC \):
\( x_1 = 2, y_1 = 3 \); \( x_2 = 2, y_2 = 7 \)
\( AC=\sqrt{(2 - 2)^2+(7 - 3)^2}=\sqrt{0^2+4^2}=\sqrt{16}=4 \)
- Length of \( BC \):
\( x_1 = 6, y_1 = 3 \); \( x_2 = 2, y_2 = 7 \)
\( BC=\sqrt{(2 - 6)^2+(7 - 3)^2}=\sqrt{(- 4)^2+4^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2} \)
Step 2: Classify the triangle by sides and check for right triangle
- By sides: Since \( AB = AC = 4 \), it is an isosceles triangle.
- Check for right triangle: Use Pythagorean theorem. \( AB^2+AC^2=4^2 + 4^2=16 + 16 = 32 \) and \( BC^2=(4\sqrt{2})^2=32 \). Since \( AB^2 + AC^2=BC^2 \), it is a right - isosceles triangle (right triangle with \( \angle A = 90^{\circ} \))
Problem 8:
Points \( A(3, 3) \), \( B(6, 9) \), \( C(6,-3) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1 = 3,y_1 = 3;x_2 = 6,y_2 = 9 \)
\( AB=\sqrt{(6 - 3)^2+(9 - 3)^2}=\sqrt{3^2+6^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5} \)
- \( AC \): \( x_1 = 3,y_1 = 3;x_2 = 6,y_2=-3 \)
\( AC=\sqrt{(6 - 3)^2+(-3 - 3)^2}=\sqrt{3^2+(-6)^2}=\sqrt{9 + 36}=\sqrt{45}=3\sqrt{5} \)
- \( BC \): \( x_1 = 6,y_1 = 9;x_2 = 6,y_2=-3 \)
\( BC=\sqrt{(6 - 6)^2+(-3 - 9)^2}=\sqrt{0^2+(-12)^2}=\sqrt{144}=12 \)
Step 2: Classify
- By sides: \( AB = AC=3\sqrt{5} \), isosceles triangle.
- Check for right triangle: \( AB^2+AC^2=(3\sqrt{5})^2+(3\sqrt{5})^2 = 45+45 = 90 \), \( BC^2 = 144 \). \( AB^2+AC^2
eq BC^2 \). Check \( AB^2+BC^2=(3\sqrt{5})^2+12^2=45 + 144 = 189
eq AC^2 \), \( AC^2+BC^2=(3\sqrt{5})^2+12^2=45 + 144 = 189
eq AB^2 \). Not a right triangle.
Problem 9:
Points \( A(1, 9) \), \( B(4, 8) \), \( C(2, 5) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1 = 1,y_1 = 9;x_2 = 4,y_2 = 8 \)
\( AB=\sqrt{(4 - 1)^2+(8 - 9)^2}=\sqrt{3^2+(-1)^2}=\sqrt{9 + 1}=\sqrt{10} \)
- \( AC \): \( x_1 = 1,y_1 = 9;x_2 = 2,y_2 = 5 \)
\( AC=\sqrt{(2 - 1)^2+(5 - 9)^2}=\sqrt{1^2+(-4)^2}=\sqrt{1 + 16}=\sqrt{17} \)
- \( BC \): \( x_1 = 4,y_1 = 8;x_2 = 2,y_2 = 5 \)
\( BC=\sqrt{(2 - 4)^2+(5 - 8)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4 + 9}=\sqrt{13} \)
Step 2: Classify
- By sides: All sides \( \sqrt{10},\sqrt{13},\sqrt{17} \) are of different lengths, scalene triangle.
- Check for right triangle: \( AB^2 = 10,AC^2 = 17,BC^2 = 13 \). \( AB^2+BC^2=10 + 13=23
eq AC^2 \), \( AB^2+AC^2=10 + 17 = 27
eq BC^2 \), \( AC^2+BC^2=17 + 13=30
eq AB^2 \). Not a right triangle.
Problem 10:
Points \( A(-2, 3) \), \( B(0,-3) \), \( C(3,-2) \)
Step 1: Calculate side lengths
- \( AB \): \( x_1=-2,y_1 = 3;x_2 = 0,y_2=-3 \)
\( AB=\sqrt{(0 + 2)^2+(-3 - 3)^2}=\sqrt{2^2+(-6)^2}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10} \)
- \( AC \): \( x_1=-2,y_1 = 3;x_2 = 3,y_2=-2 \)
\( AC=\sqrt{(3 + 2)^2+(-2 - 3)^2}=\sqrt{5^2+(-5)^2}=\sqrt{25 + 25}=\sqrt{50}=5\sqrt{2} \)
- \( BC \): \( x_1 = 0,y_1=-3;x_2 = 3,y_2=-2 \)
\( BC=\sqrt{(3 - 0)^2+(-2 + 3)^2}=\sqrt{3^2+1^2}=\sqrt{9 + 1}=\sqrt{10} \)
Step 2: Classify
- By sides: \( AB = 2\sqrt{10},BC=\sqrt{10},AC = 5\sqrt{2} \). Check \( AB^2=(2\sqrt{10})^2 = 40 \), \( BC^2 = 10 \), \( AC^2=50 \). Since \( AB^2+BC^2=40 + 10=50=AC^2 \), it is a right triangle (right - angled at \( B \)) and also a scalene triangle (since \( AB
eq BC
eq AC \))
Final Answers (for each problem):
- Problem 7: Isosceles right triangle.
- Problem 8: Isosceles non - right triangle.
- Problem 9: Scalene non - right triangle.
- Problem 10: Scalene right triangle.