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Question
in the figure below, \\( \overline{em} \\) is the width of a river. the length of \\( \overline{rw} = 3 \\) ft, \\( \overline{ms} = 32 \\) ft, \\( \overline{sw} = 4 \\) ft, and \\( \angle mer = \angle erw \\). what is the width of the river?
Step1: Prove triangles similar
Since \(\angle MER=\angle ERW\) and \(\angle MSE = \angle RSW\) (vertical angles are equal), by the AA (Angle - Angle) similarity criterion, \(\triangle MSE\sim\triangle RSW\).
Step2: Set up proportion
For similar triangles \(\triangle MSE\) and \(\triangle RSW\), the ratios of corresponding sides are equal. That is \(\frac{EM}{RW}=\frac{MS}{SW}\).
We know \(RW = 3\) ft, \(MS=32\) ft, \(SW = 4\) ft. Let \(EM=x\). Substitute the values into the proportion: \(\frac{x}{3}=\frac{32}{4}\).
Step3: Solve for \(x\)
Cross - multiply: \(4x=3\times32\). Then \(4x = 96\). Divide both sides by 4: \(x=\frac{96}{4}=24\).
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The width of the river \(EM\) is \(24\) ft.