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 similarity of triangles

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

Step2: Set up proportion

For similar triangles \(\triangle VWX\) and \(\triangle VYZ\), the ratios of 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 \(YZ = 106.75\) m.
Substituting the values into the proportion \(\frac{190}{125}=\frac{WX}{106.75}\).

Step3: Solve for \(WX\)

Cross - multiply: \(125\times WX=190\times106.75\).
Then \(WX=\frac{190\times106.75}{125}\).
First, calculate \(190\times106.75 = 190\times(100 + 6.75)=190\times100+190\times6.75=19000+1282.5 = 20282.5\).
Now, \(WX=\frac{20282.5}{125}=162.26\) m.

Answer:

\(162.26\) m