Answer:
Width of Belmont Avenue is 5.2 meters
Step-by-step explanation:
We use the Pythagoras theorem to solve the problem
Pythagoras Theorem states that:
Hypotenuse² = Base² + Perpendicular²
Danville Street, Belmont Avenue and their diagonal forms a right angled triangle
In this case, let the base be Danville Street with width of 8.1 meters
Let the hypotenuse be 9.6 meters
We need to find the perpendicular
Put the values in Pythagoras theorem.
9.6² = 8.1² + perpendicular²
perpendicular² = 9.6²-8.1²
perpendicular² = 92.16 - 65.61
perpendicular² = 26.55
take square root on both sides
perpendicular =5.153
Round of to nearest 10th
Perpendicular = 5.2 meters
Belmont Avenue = 5.2 meters
