calculate the perimeter of isosceles triangle ABC with vertices A(2;3), B(5;7) and C(-2;6)
Answers
Answer:
The perimeter of isosceles triangle will be(9;13) am I right?
Answer:
Step-by-step explanation:
6)
vertices are A(2,1) , B(5,1) and C(5,5)
Lets find the length of the sides
\begin{gathered}AB = \sqrt{ {(5 - 2)}^{2} + {(1 - 1}^{2} } = \sqrt{ {3}^{2} + {0}^{2} } = \sqrt{9} = 3\\ \end{gathered}AB=(5−2)2+(1−12=32+02=9=3
BC = \sqrt{ {((5 - 5)}^{2} + {(5 - 1)}^{2} } = \sqrt{ {0}^{2} + {4}^{2} } = \sqrt{16} = 4BC=((5−5)2+(5−1)2=02+42=16=4
CA = \sqrt{ {(5 - 2)}^{2} + {(5 - 1)}^{2} } = \sqrt{ {3}^{2} + {4}^{2} } = \sqrt{25} = 5CA=(5−2)2+(5−1)2=32+42=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} = 2AB=(2−2)2+(3−1)2=02+22=4=2
BC = \sqrt{ {(4 - 2)}^{2} + {(2 - 3)}^{2} } = \sqrt{ {2}^{2} + {( - 1)}^{2} } = \sqrt{5}BC=(4−2)2+(2−3)2=22+(−1)2=5
\begin{gathered}CA = \sqrt{ {(4 - 2)}^{2} + {(2 - 1)}^{2} } \\ = \sqrt{ {2}^{2} + {1}^{2} } = \sqrt{5} \end{gathered}CA=(4−2)2+(2−1)2=22+12=5
As sides BC = CA hence the vertices are of isosceles triangle.