Answer with Step-by-step explanation:
We are given that
COAL is a rhombus .
CA and LO are the diagonals of rhombus COAL.
CA=27 m
OL=19 m
We know that
Diagonals of rhombus are bisect to each other.
Let P(0,0) be the intersecting point of diagonals
Therefore, CP=PA
LP=PO
Length PO=[tex]\frac{1}{2}LO=\frac{19}{2}=9.5 m[/tex]
Length of CP=[tex]\frac{1}{2}PA=\farc{1}{2}(27)=13.5 m[/tex]
Let C([tex]0,y_2),O(x_1,0),A(0,y_1) \;and\;L(x_2,0)[/tex].
Distance formula:[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the formula
[tex]CP=\sqrt{(0-0)^2+(0-y_2)^2}=13.5[/tex]
[tex]\sqrt{y^2_2}=13.5[/tex]
[tex]y_2=\pm 13.5[/tex]
C(0,13.5) or (0,-13.5)
[tex]\sqrt{(y_2-y_1)^2}=27[/tex]
Substitute the value [tex]y_2=13.5[/tex]
[tex]\sqrt{13.5-y_1}=27[/tex]
[tex]13.5-y_1=\pm 27[/tex]
[tex]y_1=13.5-27=-13.5[/tex]
A(0,-13.5)
[tex]13.5-y_1=-27[/tex]
[tex]y_1=13.5+27=40.5[/tex]
A(0,40.5)
Substitute [tex]y_2=-13.5[/tex]
[tex]\sqrt{(-13.5-y_1)^2}=27[/tex]
[tex]-13.5-y_1=\pm 27[/tex]
[tex]-13.5-y_1=27[/tex]
[tex]y_1=-13.5-27=-40.5[/tex]
A(0,-40.5)
[tex]-13.5-y_1=-27[/tex]
[tex]y_1=-13.5+27=13.5[/tex]
A(0,13.5)
[tex]LP=\sqrt{(x_2-0)^2}=9.5[/tex]
[tex]\sqrt{x^2_2}=9.5[/tex]
[tex]x_2=\pm 9.5[/tex]
L(9.5,0) or L(-9.5,0)
[tex]\sqrt{x_2-x_1)^2}=19[/tex]
[tex](9.5-x_1)^2}=19[/tex]
[tex]9.5-x_1=\pm 19[/tex]
[tex]9.5-x_1=19[/tex]
[tex]x_1=9.5-19=-9.5[/tex]
[tex]9.5-x_1=-19[/tex]
[tex]x_1=9.5+19=28.5[/tex]
[tex](-9.5-x_1)=19[/tex]
[tex]-9.5-x_1=\pm 19[/tex]
[tex]-9.5-x_1=19[/tex]
[tex]x_1=-9.5-19=-28.5[/tex]
[tex]-9.5-x_1=-19[/tex]
[tex]-9.5+19=x_1[/tex]
[tex]x_1=9.5[/tex]
O(-9.5,0) or O(28.5,0) or (-28.5,0) or (9.5,0)
