Answer:
38.25 units.
Explanation:
The trapezoid is given with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)
When given vertices ( x1, y1), (x2, y2)
We use the formula:
√(x2 - x1)² + (y2 - y1)² to find the length of the sides of the Trapezoid
Side QR: Q(8, 8), R(14, 16)
= √(x2 - x1)² + (y2 - y1)²
= √(14 - 8)² + (16 - 8)²
= √6² + 8²
= √36 + 64
= √100
= 10 units
Side RS: R(14, 16), S(20, 16)
= √(x2 - x1)² + (y2 - y1)²
= √(20 - 14)² + (16 - 16)²
= √6² + 0²
= √36
= 6 units
Side ST : S(20, 16), T(22, 8)
= √(x2 - x1)² + (y2 - y1)²
= √(22 - 20)² + ( 8 - 16)²
= √2² + (-8)²
= √4 + 64
= √68
= 8.2462112512
≈ nearest hundredth = 8.25 units
Side QT, Q(8, 8), T(22, 8)
= √(x2 - x1)² + (y2 - y1)²
= √(22 - 8)² + (8 - 8)²
= √14² + 0²
= √196
= 14 units
The Perimeter of a Trapezoid is the sum of all it's sides.
P = QR + RS + ST + QT
P = (10 + 6 + 14 + 8.25) units
P = 38.25 units
Therefore, the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8) is 38.25units.