Khalil uses the method of similar triangles to find the distance across the river, [tex]\overline{PR}[/tex]
- The distance across the river is [tex]\underline{308.\overline 3 \ feet}[/tex]
Reasons:
The distances between the formed the sight-lines are;
[tex]\overline{RB}[/tex] = 185 feet
[tex]\overline{OC}[/tex] = 275 feet
The distance between the point close to the river and the next point further from the river [tex]\overline{RO}[/tex] = 150 feet
In triangles ΔPRB and ΔPOC, we have;
∠PRE = ∠POC = 90° Given
∠PER ≅ ∠PCO By corresponding angle formed between two parallel lines and a common transversal.
∴ ΔPRE is similar to ΔPOC by Angle-Angle, AA, similarity theorem
Which gives;
[tex]\displaystyle \frac{\overline {PR}}{\overline {PO}} = \mathbf{\frac{\overline {RE}}{\overline {OC}}}[/tex]
Let x represent the distance across the river, we have;
[tex]\overline{PR}[/tex] = x
[tex]\overline{PO}[/tex] = 150 + x
Which gives;
[tex]\displaystyle \frac{x}{150 + x} = \frac{185}{275}[/tex]
275·x = 185 × (150 + x) = 27,750 + 185·x
275·x - 185·x = 27,750
90·x = 27,750
[tex]\displaystyle x = \frac{27,750}{90} = \mathbf{308.\overline 3}[/tex]
Therefore, the distance across the river, x = [tex]\mathbf{\overline{PR}}[/tex] = [tex]\underline{308.\overline 3 \ feet}[/tex]
Learn more about similar triangles here
https://brainly.com/question/10703692
