Question

to indirectly measure the distance across a lake, jeremiah makes use of a couple landmarks at points w and x. he measures vy, yw, and yz as marked. find the distance across the lake (wx), rounding your answer to the nearest hundredth of a meter.
(diagram is not to scale.)

Explanation:

Step1: Prove the similarity of triangles

Since \(\angle VWX=\angle VYZ = 90^{\circ}\) and \(\angle V=\angle V\) (common angle), by the AA (Angle - Angle) similarity criterion, \(\triangle VWX\sim\triangle VYZ\).

Step2: Use the property of similar triangles

For similar triangles \(\triangle VWX\) and \(\triangle VYZ\), the ratios of their corresponding sides are equal. That is \(\frac{VW}{VY}=\frac{WX}{YZ}\).
We know that \(VW = VY+YW=125 + 65=190\) m, \(VY = 125\) m, and \(YZ=108.75\) m.
Substitute the values into the proportion \(\frac{190}{125}=\frac{WX}{108.75}\).

Step3: Solve for \(WX\)

Cross - multiply: \(125\times WX=190\times108.75\).
First, calculate \(190\times108.75 = 190\times\frac{10875}{100}=\frac{190\times10875}{100}=\frac{2066250}{100}=20662.5\).
Then \(WX=\frac{20662.5}{125}\).
\(WX = 165.3\) m.

Answer:

\(165.30\) m