Danville Street and Belmont Avenue intersect. If the diagonal distance across the intersection is 9.6 meters and Danville Street is 8.1 meters wide, how wide is Belmont Avenue? If necessary, round to the nearest tenth.

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

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x