we know that
The area of a rectangle is equal to
[tex] A=L*W [/tex]
where
L is the length side of the rectangle
w is the width side of the rectangle
To determine the solution we will proceed to graph and calculate the area of each rectangle
case 1) [tex] A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16) [/tex]
using a graph tool
see the attached figure N [tex] 1 [/tex]
Find the distance BC (Length side)
[tex]d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex]d=\sqrt{(13-5)^{2} +(1+5)^{2}}[/tex]
[tex] d=\sqrt{(8)^{2} +(6)^{2}}[/tex]
[tex] d=10\ units[/tex]
Find the distance AB (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(5-8)^{2} +(-5+9)^{2}}[/tex]
[tex] d=\sqrt{(-3)^{2} +(4)^{2}}[/tex]
[tex] d=5\ units[/tex]
Find the area of rectangle ABCD
[tex] A=L*W [/tex]
[tex] A=10*5 [/tex]
[tex] A=50\ units^{2}[/tex]
The area of the rectangle ABCD is [tex] 50\ units^{2}[/tex]
case 2) [tex] E(30, 20), F(39, 29), G(49, 19), H(40, 10) [/tex]
using a graph tool
see the attached figure N [tex] 2 [/tex]
Find the distance EH (Length side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(10-20)^{2} +(40-30)^{2}}[/tex]
[tex] d=\sqrt{(-10)^{2} +(10)^{2}}[/tex]
[tex] d= \sqrt{200}\ units [/tex]
Find the distance EF (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(29-20)^{2} +(39-30)^{2}}[/tex]
[tex] d=\sqrt{(9)^{2} +(9)^{2}}[/tex]
[tex] d= \sqrt{162}\ units [/tex]
Find the area of rectangle EFGH
[tex] A=L*W [/tex]
[tex] A=\sqrt{200} *\sqrt{162}[/tex]
[tex] A=180\ units^{2}[/tex]
The area of the rectangle EFGH is [tex] 180\ units^{2}[/tex]
case 3) [tex] I(-6, 2), J(2, 2), K(2, -8), L(-6, -8) [/tex]
using a graph tool
see the attached figure N [tex] 3 [/tex]
Find the distance JK (Length side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-8-2)^{2} +(2-2)^{2}}[/tex]
[tex] d=\sqrt{(-10)^{2} +(0)^{2}}[/tex]
[tex] d=10\ units[/tex]
Find the distance KL (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-8+8)^{2} +(-6-2)^{2}}[/tex]
[tex] d=\sqrt{(0)^{2} +(-8)^{2}}[/tex]
[tex] d=8\ units[/tex]
Find the area of rectangle IJKL
[tex] A=L*W [/tex]
[tex] A=10*8 [/tex]
[tex] A=80\ units^{2}[/tex]
The area of the rectangle IJKL is [tex] 80\ units^{2}[/tex]
case 4) [tex] M(5, 5), N(11, 5), O(11, -5), P(5, -5) [/tex]
using a graph tool
see the attached figure N [tex] 4 [/tex]
Find the distance NO (Length side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-5-5)^{2} +(11-11)^{2}}[/tex]
[tex] d=\sqrt{(-10)^{2} +(0)^{2}}[/tex]
[tex] d=10\ units[/tex]
Find the distance OP (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-5+5)^{2} +(5-11)^{2}}[/tex]
[tex] d=\sqrt{(0)^{2} +(-6)^{2}}[/tex]
[tex] d=6\ units[/tex]
Find the area of rectangle MNOP
[tex] A=L*W [/tex]
[tex] A=10*6 [/tex]
[tex] A=60\ units^{2}[/tex]
The area of the rectangle MNOP is [tex] 60\ units^{2}[/tex]
case 5) [tex] Q(10, 0), R(15, 5), S(25, -5), T(20, -10) [/tex]
using a graph tool
see the attached figure N [tex] 5 [/tex]
Find the distance RS (Length side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-5-5)^{2} +(25-15)^{2}}[/tex]
[tex] d=\sqrt{(-10)^{2} +(10)^{2}}[/tex]
[tex] d= \sqrt{200}\ units [/tex]
Find the distance ST (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-10+5)^{2} +(20-25)^{2}}[/tex]
[tex] d=\sqrt{(-5)^{2} +(-5)^{2}}[/tex]
[tex] d= \sqrt{50}\ units [/tex]
Find the area of rectangle QRST
[tex] A=L*W [/tex]
[tex] A=\sqrt{200} *\sqrt{50}[/tex]
[tex] A=100\ units^{2}[/tex]
The area of the rectangle QRST is [tex] 100\ units^{2}[/tex]
case 6) [tex] U(0, 5), V(15, 20), W(25, 10), X(10, -5) [/tex]
using a graph tool
see the attached figure N [tex] 6 [/tex]
Find the distance WX (Length side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(-5-10)^{2} +(10-25)^{2}}[/tex]
[tex] d=\sqrt{(-15)^{2} +(-15)^{2}}[/tex]
[tex] d= \sqrt{450}\ units [/tex]
Find the distance VW (Width side)
[tex] d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}[/tex]
[tex] d=\sqrt{(10-20)^{2} +(25-15)^{2}}[/tex]
[tex] d=\sqrt{(-10)^{2} +(10)^{2}}[/tex]
[tex] d= \sqrt{200}\ units [/tex]
Find the area of rectangle UVWX
[tex] A=L*W [/tex]
[tex] A=\sqrt{200} *\sqrt{450}[/tex]
[tex] A=300\ units^{2}[/tex]
The area of the rectangle UVWX is [tex] 300\ units^{2}[/tex]
Using the distance formula, the rectangles and their areas are:
Area of rectangle EFGH = 180 square units
Area of rectangle IJKL = 80 square units
Area of rectangle MNOP= 60 square units
Area of rectangle QRST= 100 square units
Area of rectangle UVWX = 300square units
The distance formula is used to determine the length of a segment on a coordinate or the distance between one vertex and another vertex.
Distance formula = [tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex].
Area of Rectangle ABCD:
A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16)
Area = AB × BC
Using the distance formula, let's find AB and CD:
[tex]AB = \sqrt{(-5 -(-9))^2 + (5 - 8)^2}\\\\AB = \sqrt{16 + 9}\\\\AB = \sqrt{25} \\\\\mathbf{AB = 5 $ units}[/tex]
[tex]BC = \sqrt{(1 - (-5))^2 + (13 - 5)^2}\\\\BC = \sqrt{36 + 64}\\\\BC = \sqrt{100} \\\\\mathbf{BC = 10 $ units}[/tex]
Area of rectangle ABCD = 5 × 10 = 50 square units.
Using the same distance formula and steps as shown above, the area of the rest rectangles would be calculated as follows.
Area of EFGH:
E(30, 20), F(39, 29), G(49, 19), H(40, 10)
Area = EF × FG = √162 × √200
Area of rectangle EFGH = 180 square units
Area of IJKL:
I(-6, 2), J(2, 2), K(2, -8), L(-6, -8)
Area = IJ × JK = 8 × 10
Area of rectangle IJKL = 80 square units
Area of MNOP:
M(5, 5), N(11, 5), O(11, -5), P(5, -5)
Area = MN × NO = 6 × 10
Area of rectangle MNOP= 60 square units
Area of QRST:
Q(10, 0), R(15, 5), S(25, -5), T(20, -10)
Area = QR × RS = √50 × √200
Area of rectangle QRST= 100 square units
Area of UVWX:
U(0, 5), V(15, 20), W(25, 10), X(10, -5)
Area = UV × VW = √450 × √200
Area of rectangle UVWX = 300square units
Learn more about distance formula on:
https://brainly.com/question/661229
