The area of the polygon JKLMNO is 68.5 square units
The area of triangle MNO
The coordinates of the triangle are:
M = (-1, -5)
N = (-8, -6)
O = (-5, -2)
Calculate the base and the height of the triangle using:
[tex]d = \sqrt{(y_2 -y_1)^2 +(x_2 -x_1)^2[/tex]
So, we have:
[tex]ON = \sqrt{(-5+ 8)^2 +(-2 + 6)^2[/tex]
ON = 5
[tex]OM = \sqrt{(-5+ 1)^2 +(-2 + 5)^2[/tex]
OM = 5
The area is:
Area = 0.5 * ON * OM
This gives
Area = 0.5 * 5 * 5
Evaluate
Area = 12.5
Hence, the area of the triangle MNO is 12.5 square units
The perimeter of rectangle JLMO
The coordinates of the rectangle are:
J = (1, 6)
L = (5, 3)
M = (-1, -5)
O = (-5, -2)
Calculate the length and the width of the rectangle using:
[tex]d = \sqrt{(y_2 -y_1)^2 +(x_2 -x_1)^2[/tex]
So, we have:
[tex]OM = \sqrt{(-5+ 1)^2 +(-2 + 5)^2[/tex]
OM = 5
[tex]ML = \sqrt{(-1 - 5)^2 + (-5 - 3)^2[/tex]
ML = 10
The perimeter is:
Perimeter = 2 * (OM + ML)
This gives
Perimeter = 2 * (5 + 10)
Evaluate
Perimeter = 30
Hence, the perimeter of the rectangle JLMO is 30 units
The area of the polygon
In (a), we have:
The area of the triangle MNO is 12.5 square units
The area of the rectangle JKMO is:
Area = OM * ML
This gives
Area = 5 * 10 = 50
The coordinates of triangle JKL are:
J = (1, 6)
K = (5, 6)
L = (5, 3)
Calculate the base and the height of the triangle using:
[tex]d = \sqrt{(y_2 -y_1)^2 +(x_2 -x_1)^2[/tex]
So, we have:
[tex]KL = \sqrt{(5- 5)^2 +(3 - 6)^2[/tex]
KL = 3
[tex]JK = \sqrt{(1- 5)^2 +(6 - 6)^2[/tex]
JK = 4
The area of JKL is
Area = 0.5 * KL * JK
This gives
Area = 0.5 * 3 * 4
Area = 6
Add the three areas
Total area = 6 + 12.5 + 50
Evaluate
Total area = 68.5
Hence, the area of the polygon JKLMNO is 68.5 square units
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