Answer: 43.02 miles (Approx)
Step-by-step explanation:
Let the entire situation is shown by a triangle CAB,
In which A shows Almondville, B shows Walnut Grove and C shows Pecan City.
Also, [tex]D\in BC[/tex] such that [tex]AD\perp BC[/tex]
Thus, According to the question, AB = 52.4 miles
AC = 75.3 miles and BC = 91.7 miles.
And, We have to find out AD = ?
Since, CAB and CDA are right angle triangles.
Where, ∠CAB≅∠CDA
∠ACB≅∠ACD
Thus, By AA similarity postulate,
[tex]\triangle CAB\sim \triangle CDA[/tex]
Therefore, by the property of similarity,
[tex]\frac{AB}{DA} = \frac{CB}{AC}[/tex]
⇒ [tex]\frac{52.4}{DA} = \frac{91.7}{75.3}[/tex]
⇒ [tex]DA= \frac{75.3\times 52.4}{91.7}[/tex]
⇒ [tex]DA= \frac{3945}{91.7}[/tex]
⇒ [tex]DA= 43.0285714286 [/tex]≈ 43.02 miles
Thus, the shortest length possible for that road= 43.02 miles (approx)
