Answer:
The triangle with vertices A (2 , 0) , B (3 , 2) , C (5 , 1) is isosceles right Δ
The triangle with vertices A (-3 , 1) , B (-3 , 4) , C (-1 , 1) is right Δ
The triangle with vertices A (-5 , 2) , B (-4 , 4) , C (-2 , 2) is acute scalene Δ
The triangle with vertices A (-4 , 2) , B (-2 , 4) , C (-1 , 4) is obtuse scalene Δ
Step-by-step explanation:
* Lets explain the relation between the sides and the angles in
a triangle
- The types of the triangles according the length of its sides:
# Equilateral triangle; all its sides are equal in length and all the angles
have measures 60°
# Isosceles triangle; tow sides equal in lengths and the 2 angles not
included between them are equal in measures
# Scalene triangles; all sides are different in lengths and all angles
are different in measures
- The types of the triangles according the measure of its angles:
# Acute triangle; its three angles are acute and the relation between
its sides is the sum of the squares of the two shortest sides is
greater than the square of the longest side
# Obtuse triangle; one angle is obtuse and the other 2 angles are
acute and the relation between its sides is the sum of the squares
of the two shortest sides is smaller than the square of the longest
side
# Right triangle; one angle is right and he other 2 angles are
acute and the relation between its sides is the sum of the squares
of the two shortest sides is equal to the square of the longest side
- The distance between the points 9x1 , y1) and (x2 , y2) is
[tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
* Lets solve the problem
# The triangle with vertices A (2 , 0) , B (3 , 2) , C (5 , 1)
∵ [tex]AB=\sqrt{(3-2)^{2}+(2-0)^{2}}=\sqrt{1+4}=\sqrt{5}[/tex]
∵ [tex]BC=\sqrt{(5-3)^{2}+(1-2)^{2}}=\sqrt{4+1}=\sqrt{5}[/tex]
∵ [tex]AC=\sqrt{(5-2)^{2}+(1-0)^{2}}=\sqrt{9+1}=\sqrt{10}[/tex]
- Lets check the relation between the sides
∵ AB = BC = √5 ⇒ shortest sides
∵ AC = √10
∵ (AB)² + (BC)² = (√5)² + (√5)² = 5 + 5 = 10
∵ (AC)² = (√10)² = 10
∴ The sum of the squares of the shortest sides is equal to the square
of the longest side
∴ Δ ABC is right triangle
∵ Δ ABC has two equal sides
∴ Δ ABC is isosceles right triangle
# The triangle with vertices A (-3 , 1) , B (-3 , 4) , C (-1 , 1)
∵ [tex]AB=\sqrt{(-3--3)^{2}+(4-1)^{2}}=\sqrt{0+9}=3[/tex]
∵ [tex]BC=\sqrt{(-1--3)^{2}+(1-4)^{2}}=\sqrt{4+9}=\sqrt{13}[/tex]
∵ [tex]AC=\sqrt{(-1--3)^{2}+(1-1)^{2}}=\sqrt{4+0}=2[/tex]
- Lets check the relation between the sides
∵ AB = 3
∵ BC = √13 ⇒ longest sides
∵ AC = 2
∵ (AB)² + (AC)² = (3)² + (2)² = 9 + 4 = 13
∵ (BC)² = (√13)² = 13
∴ The sum of the squares of the shortest sides is equal to the square
of the longest side
∴ Δ ABC is right triangle
∴ Δ ABC is right triangle
# The triangle with vertices A (-5 , 2) , B (-4 , 4) , C (-2 , 2)
∵ [tex]AB=\sqrt{(-4--5)^{2}+(4-2)^{2}}=\sqrt{1+4}=\sqrt{5}[/tex]
∵ [tex]BC=\sqrt{(-2--4)^{2}+(2-4)^{2} }=\sqrt{4+4}=\sqrt{8}[/tex]
∵ [tex]AC=\sqrt{(-2--5)^{2}+(2-2)^{2}}=\sqrt{9+0}=3[/tex]
- Lets check the relation between the sides
∵ AB = √5
∵ BC = √8
∵ AC = 3 ⇒ longest sides
∵ (AB)² + (BC)² = (√5)² + (√8)² = 5 + 8 = 13
∵ (AC)² = (3)² = 9
∴ The sum of the squares of the shortest sides is greater than the
square of the longest side
∴ Δ ABC is acute triangle
∵ Δ ABC has three different sides in lengths
∴ Δ ABC is acute scalene triangle
# The triangle with vertices A (-4 , 2) , B (-2 , 4) , C (-1 , 4)
∵ [tex]AB=\sqrt{(-2--4)^{2}+(4-2)^{2}}=\sqrt{4+4}=\sqrt{8}[/tex]
∵ [tex]BC=\sqrt{(-1--2)^{2}+(4-4)^{2} }=\sqrt{1+0}=1[/tex]
∵ [tex]AC=\sqrt{(-1--4)^{2}+(4-2)^{2}}=\sqrt{9+4}=\sqrt{13}[/tex]
- Lets check the relation between the sides
∵ AB = √8
∵ BC = 1
∵ AC = √13 ⇒ longest sides
∵ (AB)² + (BC)² = (√8)² + (1)² = 8 + 1 = 9
∵ (AC)² = (√13)² = 13
∴ The sum of the squares of the shortest sides is smaller than the
square of the longest side
∴ Δ ABC is obtuse triangle
∵ Δ ABC has three different sides in lengths
∴ Δ ABC is obtuse scalene triangle
