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 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) arrowboth arrowboth arrowboth arrowboth

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

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:

idk

step-by-step explanation:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine