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Luden [163]
4 years ago
6

Match the rectangles formed by the sets of points to their corresponding areas. A(-9, 8), B(-5, 5), C(1, 13), D(-3, 16) 50 squar

e 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)
Mathematics
2 answers:
ira [324]4 years ago
5 0

Answer:

Step-by-step explanation:

Area of any rectangle = length x width

Using this:

ABCD: Length = AB=\sqrt{(-5 - -9)² + (5 - 8)²}  = 5

Width =

BC = \sqrt{(1 - -5)² + (13 - 5)²}  = 10

Area =50 sq units

Similarly for EFGH, length =width = 9\sqrt{2}

Area = 162 sq units.

iJKL:

Length = 8 and width = 10

Area = 80 sq units.

MNOP:

Length = width =6

Area = 36 sq units.

QRST: 100 sq units and

TUVW: 300 sq units.

svlad2 [7]4 years ago
4 0

Answer:

Area of ABCD = 50 square units

Area of IJKL = 80 square units

Area of QRST = 100 square units

Area of UVWX = 300 square units

Step-by-step explanation:

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

AB = √(-5 - -9)² + (5 - 8)² = 5

BC = √(1 - -5)² + (13 - 5)² = 10

Area = 5 × 10 = 50 square units

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

IJ = 2 - -6 = 8 ⇒ horizontal line (same y-coordinates)

JK = 2 - -8 = 10 ⇒ vertical line (same x-coordinates)

Area = 8 × 10 = 80 square units

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

QR = √(15 - 10)² + (5 - 0)² = 5√2

RS = √(25 - 15)² + (-5 - 5)² = 10√2

Area = 5√2 × 10√2 = 100 square units

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

UV = √(15 - 0)² + (20 - 5)² = 15√2

VW = √(25 - 15)² + (10 - 20)² = 10√2

A = 15√2 × 10√2 = 300 square units

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When the sum of a number and 3 is subtracted from 10 the result is 5 solve om algebraic equation?
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y = 2

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