ANSWER TO QUESTION 1
We use the distance formula,
[tex]d = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} } [/tex]
to find the length of each side and add them to find the total distance around the triangle.
A(−11, 4) , B(−7, 8) , and C (−4, 4) are the vertices of the triangle.
[tex]|AB|= \sqrt{ {( - 7 - - 11)}^{2} + {(8 - 4)}^{2} } [/tex]
[tex]|AB|= \sqrt{ {( - 7 + 11)}^{2} + {(8 - 4)}^{2} } [/tex]
[tex]|AB|= \sqrt{ {( 4)}^{2} + {(4)}^{2} } [/tex]
[tex]|AB|= \sqrt{ 16 + 16} [/tex]
[tex]|AB|= \sqrt{32} [/tex]
[tex]|AB|= 4\sqrt{2} \: units[/tex]
[tex]|BC|= \sqrt{ {( - 7- -4)}^{2} + {(8- 4)}^{2} } [/tex]
[tex]|BC|= \sqrt{ {( - 7+ 4)}^{2} + {(8- 4)}^{2} } [/tex]
[tex]|BC|= \sqrt{ {( -3)}^{2} + {(4)}^{2} } [/tex]
[tex]|BC|= \sqrt{ 9 + 16} [/tex]
[tex]|BC|= \sqrt{ 25} [/tex]
[tex]|BC|=5 \: units[/tex]
[tex]|AC|= \sqrt{ {( - 11- -4)}^{2} + {(4 - 4)}^{2} } [/tex]
.
[tex]|AC|= \sqrt{ {( - 11 + 4)}^{2} + {(4 - 4)}^{2} } [/tex]
[tex]|AC|= \sqrt{ {( - 7)}^{2} + {(0)}^{2} } [/tex]
[tex]|AC|= \sqrt{ 49} [/tex]
[tex]|AC|= 7 \: units[/tex]
[tex]Perimeter = 7 + 5 + 4 \sqrt{2} = 12 + 4 \sqrt{2} = 17.7 \: units[/tex]
ANSWER TO QUESTION 2
For the rectangle the perimeter is given by the formula,
[tex]p = 2w + 2l[/tex]
we use the distance formula again to find the length and width of the rectangle.
W(−6, 1) , X(−1, 6) , Y(2, 3) , and Z(−3, −2) are the vertices of the rectangle.
[tex]|WX|= \sqrt{ {( - 1 - - 6)}^{2} + (6 - 1) ^{2} } [/tex]
[tex]|WX|= \sqrt{ {( - 1 + 6)}^{2} + (6 - 1) ^{2} } [/tex]
[tex]|WX|= \sqrt{ {( 5)}^{2} + (5) ^{2} } [/tex]
[tex]|WX|= \sqrt{ 25 + 25 } [/tex]
[tex]|WX|= \sqrt{ 50} [/tex]
[tex]|WX|= 5\sqrt{ 2 } [/tex]
This the length of the rectangle.
We now find the width.
[tex]|XY| = \sqrt{ {(2 - - 1)}^{2} + {(3 - 6)}^{2} } [/tex]
[tex]|XY| = \sqrt{ {(2 + 1)}^{2} + {( 3 - 6)}^{2} } [/tex]
[tex]|XY| = \sqrt{ {( 3)}^{2} + {( - 3)}^{2} } [/tex]
[tex]|XY| = 3\sqrt{ 2} [/tex]
[tex]Perimeter = 2(5 \sqrt{2} ) + 2(3 \sqrt{2)} [/tex]
[tex]Perimeter = 2(8\sqrt{2)} = 16 \sqrt{2} \: \: = 22.6 \: units[/tex]
