Match the rectangles formed by the sets of points to their corresponding areas.

Tiles

A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16)

50 square units

E(30, 20), F(39, 29), G(49, 19), H(40, 10)

300 square units

I(-6, 2), J(2, 2), K(2, -8), L(-6, -8)

100 square units

M(5, 5), N(11, 5), O(11, -5), P(5, -5)

80 square units

Q(10, 0), R(15, 5), S(25, -5), T(20, -10)

U(0, 5), V(15, 20), W(25, 10), X(10, -5)

Verified Answer :

we know that

The area of a rectangle is equal to

\(A=L*W\)

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) \(A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16)\)

using a graph tool

see the attached figure N \(1\)

Find the distance BC (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(13-5)^{2} +(1+5)^{2}}\)

\(d=\sqrt{(8)^{2} +(6)^{2}}\)

\(d=10\ units\)

Find the distance AB (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(5-8)^{2} +(-5+9)^{2}}\)

\(d=\sqrt{(-3)^{2} +(4)^{2}}\)

\(d=5\ units\)

Find the area of rectangle ABCD

\(A=L*W\)

\(A=10*5\)

\(A=50\ units^{2}\)

The area of the rectangle ABCD is  \(50\ units^{2}\)

case 2) \(E(30, 20), F(39, 29), G(49, 19), H(40, 10)\)

using a graph tool

see the attached figure N \(2\)

Find the distance EH (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(10-20)^{2} +(40-30)^{2}}\)

\(d=\sqrt{(-10)^{2} +(10)^{2}}\)

\(d= \sqrt{200}\ units\)

Find the distance EF (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(29-20)^{2} +(39-30)^{2}}\)

\(d=\sqrt{(9)^{2} +(9)^{2}}\)

\(d= \sqrt{162}\ units\)

Find the area of rectangle EFGH

\(A=L*W\)

\(A=\sqrt{200} *\sqrt{162}\)

\(A=180\ units^{2}\)

The area of the rectangle EFGH is  \(180\ units^{2}\)

case 3) \(I(-6, 2), J(2, 2), K(2, -8), L(-6, -8)\)

using a graph tool

see the attached figure N \(3\)

Find the distance JK (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-8-2)^{2} +(2-2)^{2}}\)

\(d=\sqrt{(-10)^{2} +(0)^{2}}\)

\(d=10\ units\)

Find the distance KL (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-8+8)^{2} +(-6-2)^{2}}\)

\(d=\sqrt{(0)^{2} +(-8)^{2}}\)

\(d=8\ units\)

Find the area of rectangle IJKL

\(A=L*W\)

\(A=10*8\)

\(A=80\ units^{2}\)

The area of the rectangle IJKL is  \(80\ units^{2}\)

case 4)  \(M(5, 5), N(11, 5), O(11, -5), P(5, -5)\)

using a graph tool

see the attached figure N \(4\)

Find the distance NO (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-5-5)^{2} +(11-11)^{2}}\)

\(d=\sqrt{(-10)^{2} +(0)^{2}}\)

\(d=10\ units\)

Find the distance OP (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-5+5)^{2} +(5-11)^{2}}\)

\(d=\sqrt{(0)^{2} +(-6)^{2}}\)

\(d=6\ units\)

Find the area of rectangle MNOP

\(A=L*W\)

\(A=10*6\)

\(A=60\ units^{2}\)

The area of the rectangle MNOP is  \(60\ units^{2}\)

case 5)  \(Q(10, 0), R(15, 5), S(25, -5), T(20, -10)\)

using a graph tool

see the attached figure N \(5\)

Find the distance RS (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-5-5)^{2} +(25-15)^{2}}\)

\(d=\sqrt{(-10)^{2} +(10)^{2}}\)

\(d= \sqrt{200}\ units\)

Find the distance ST (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-10+5)^{2} +(20-25)^{2}}\)

\(d=\sqrt{(-5)^{2} +(-5)^{2}}\)

\(d= \sqrt{50}\ units\)

Find the area of rectangle QRST

\(A=L*W\)

\(A=\sqrt{200} *\sqrt{50}\)

\(A=100\ units^{2}\)

The area of the rectangle QRST is  \(100\ units^{2}\)

case 6)  \(U(0, 5), V(15, 20), W(25, 10), X(10, -5)\)

using a graph tool

see the attached figure N \(6\)

Find the distance WX (Length side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(-5-10)^{2} +(10-25)^{2}}\)

\(d=\sqrt{(-15)^{2} +(-15)^{2}}\)

\(d= \sqrt{450}\ units\)

Find the distance VW (Width side)

\(d=\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}\)

\(d=\sqrt{(10-20)^{2} +(25-15)^{2}}\)

\(d=\sqrt{(-10)^{2} +(10)^{2}}\)

\(d= \sqrt{200}\ units\)

Find the area of rectangle UVWX

\(A=L*W\)

\(A=\sqrt{200} *\sqrt{450}\)

\(A=300\ units^{2}\)

The area of the rectangle UVWX is  \(300\ units^{2}\)

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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

What is the Distance Formula?

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 = \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

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:

\(AB = \sqrt{(-5 -(-9))^2 + (5 - 8)^2}\\\\AB = \sqrt{16 + 9}\\\\AB = \sqrt{25} \\\\\mathbf{AB = 5 $ units}\)

\(BC = \sqrt{(1 - (-5))^2 + (13 - 5)^2}\\\\BC = \sqrt{36 + 64}\\\\BC = \sqrt{100} \\\\\mathbf{BC = 10 $ units}\)

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