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VLD [36.1K]
4 years ago
10

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) arrowBoth arrowBoth arrowBoth arrowBoth

Mathematics
2 answers:
Karolina [17]4 years ago
7 0

Answer:

The area of ABCD is 50 units

The area IJKL is 80 units

The area of QRST is 100 units

The area of UVWX is 300 units

Step-by-step explanation:

spayn [35]4 years ago
6 0

Answer:

The area of ABCD is 50 units²

The area IJKL is 80 units²

The area of QRST is 100 units²

The area of UVWX is 300 units²

Step-by-step explanation:

* Lets revise the area of the rectangle

- The area of any rectangle = its length × its width

- To solve the problem find the lengths of two adjacent sides and

  consider that one of them is the length and the other is the width

- Use the rule of the distance between two points (x1 , y1) and (x2 , y2)

 the distance = √[(x2 - x1)² + (y2 - y1)²]

# In rectangle ABCD

∵ A = (-9 , 8) , B = (-5 , 5) , C = (1 , 13)

∴ AB = √[(-5 - -9)² + (5 - 8)²] = √[(4)² + (-3)²] = √[16 + 9] = √25 = 5 units

∴ BC = √[(1 - -5)² + (13 - 5)²] = √[(6)² + (8)²] = √[36 + 64] = √100 = 10 units

∴ The area of ABCD = 5 × 10 = 50 units²

# In rectangle EFGH

∵ E = (30 , 20) , F = (39 , 29) , G (49 , 19)

∴ EF = √[(39 - 30)² + (29 - 20)²] = √[9² + 9²] = √[81 + 81] = √162 unit

∴ FG = √[(49 - 39)² + (19 - 29)²] = √10² + (-10)²] = √[100 + 100] = √200 units

∴ The area of EFGH = √162 × √200 = 180 units²

# In rectangle IJKL

∵ I = (-6 , 2) , J = (2 , 2) , K = (2 , -8)

∴ IJ = √[(2 - -6)² + (2 - 2)²] = √[8² + 0²] = √8² = 8 units

∴ JK = √[(2 - 2)² + (-8 - 2)²] = √[0² + (-10)²] = √10² = 10 units

∴ The area IJKL = 8 × 10 = 80 units²

# In rectangle MNOP

∵ M = (5 , 5) , N = (11 , 5) , O = (11 , -5)

∴ MN = √[(11 - 5)² + (5 - 5)²] = √[6² + 0²] = √6² = 6 units

∴ NO = √[(11 - 11)² + (-5 - 5)²] = √[0² + (-10)²] = √10² = 10 units

∴ The area of MNOP = 6 × 10 = 60 units²

# In rectangle QRST

∵ Q = (10 , 0) , R = (15 , 5) , S = (25 , -5)

∴ QR = √[(15 - 10)² + (5 - 0)²] = √[5² + 5²] = √[25 + 25] = √50 units

∴ RS = √[(25 - 15)² + (-5 - 5)²] = √[10² + (-10)²] = √[100 + 100] = √200 units

∴ The area of QRST = √50 × √200 = 100 units²

# In rectangle UVWX

∵ U = (0 , 5) , V = (15 , 20) , W = (25 , 10)

∴ UV = √[(15 - 0)² + (20 - 5)²] = √[15² + 15²] = √[225 + 225] = √450 units

∴ VW = √[(25 - 15)² + (10 - 20)²] = √[10² + (-10)²] = √100 + 100 = √200 units

∴ The area of UVWX = √450 × √200 = 300 units²

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Step-by-step explanation:

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2(3x + 6) = 3( x - 9)​
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Step-by-step explanation:

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When the angle of elevation of the sun is 78 degrees, a tree casts a 13 foot shadow. How tall is the tree?
oksian1 [2.3K]

Answer: 61.16 ft

Step-by-step explanation:

We can think in this situation as a triangle rectangle.

where:

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The shadow of the tree is the other cathetus.

We know that the angle of elevation of the sun is 78°, an angle of elevation is measured from the ground, then the adjacent cathetus to this angle is the shadow of the tree. And the opposite cathetus will be the height of the tree.

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