Question

Match each set of vertices with the type of triangle they form.

Answer 1

Answer:  The calculations are done below.

Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

$AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.$

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

$AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.$

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

$AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.$

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

$AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.$

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

$AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.$

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

Answer 2

Answer:

Step-by-step explanation:

Given  the following pair of vertices we have to find the nature

first pair of vertices of triangle A(2, 0), B(3, 2), C(5, 1)  

By distance formula

$AB=\sqrt{(3-2)^2+(2-0)^2}=\sqrt5$

$BC=\sqrt{(5-3)^2+(1-2)^2}=\sqrt5$

$AC=\sqrt{(5-2)^2+(1-0)^2}=\sqrt10$

gives $AC^2=AB^2+BC^2$

Hence, right angled triangle

A(4, 2), B(6, 2), C(5, 3.73)

$AB=\sqrt{(6-4)^2+(2-2)^2}=2$

$BC=\sqrt{(5-6)^2+(3.73-2)^2}=\sqrt3.993$

$AC=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt3.992$

Isosceles triangle

A(-5, 2), B(-4, 4), C(-2, 2)

$AB=\sqrt{(-4+5)^2+(4-2)^2}=\sqrt5$

$BC=\sqrt{(-2+4)^2+(2-4)^2}=\sqrt8$

$AC=\sqrt{(5-2)^2+(1-0)^2}=3$

Scalene triangle

A(-3, 1), B(-3, 4), C(-1, 1)

$AB=\sqrt{(-3+3)^2+(4-1)^2}=3$

$BC=\sqrt{(-1+3)^2+(1-1)^2}=2$

$AC=\sqrt{(-1+3)^2+(1-1)^2}=2$

Isosceles triangle

A(-4, 2), B(-2, 4), C(-1, 4)

$AB=\sqrt{(-2+4)^2+(4-2)^2}=\sqrt8$

$BC=\sqrt{(-1+2)^2+(4-4)^2}=1$

$AC=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt13$

Scalene triangle