Question

6) find the perimeter of a triangle ABC which vertices are A(2,1) , B(5,1) and C(5,5).

7) prove that , A(2,1) , B(2,3) and C(4,2) are vertices of an isosceles triangle.

Answer 1

Step-by-step explanation:

6)

vertices are A(2,1) , B(5,1) and C(5,5)

Lets find the length of the sides

$AB = \sqrt{ {(5 - 2)}^{2} + {(1 - 1}^{2} } = \sqrt{ {3}^{2} + {0}^{2} } = \sqrt{9} = 3$

$BC = \sqrt{ {((5 - 5)}^{2} + {(5 - 1)}^{2} } = \sqrt{ {0}^{2} + {4}^{2} } = \sqrt{16} = 4$

$CA = \sqrt{ {(5 - 2)}^{2} + {(5 - 1)}^{2} } = \sqrt{ {3}^{2} + {4}^{2} } = \sqrt{25} = 5$

Perimeter of triangle = Sum of its side = 3+4+5 = 12

7)

Vertices are A(2,1) , B(2,3) and C(4,2)

Lets find out the length of the sides

$AB = \sqrt{ {(2 - 2)}^{2} + {(3 - 1)}^{2} } = \sqrt{ {0}^{2} + {2}^{2} } = \sqrt{4} = 2$

$BC = \sqrt{ {(4 - 2)}^{2} + {(2 - 3)}^{2} } = \sqrt{ {2}^{2} + {( - 1)}^{2} } = \sqrt{5}$

$CA = \sqrt{ {(4 - 2)}^{2} + {(2 - 1)}^{2} } \\ = \sqrt{ {2}^{2} + {1}^{2} } = \sqrt{5}$

As sides BC = CA hence the vertices are of isosceles triangle.