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
