Question

1. two common envelope sizes are 3\\(\frac{1}{2}\\) in. \\(\times\\) 6\\(\frac{1}{2}\\) in. and 4 in. \\(\times\\) 9\\(\frac{1}{2}\\) in. are these envelopes similar? explain.
2. challenge \\(\triangle abe\sim\triangle acd\\). what is the value of y in the diagram?

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle ABE\sim\triangle ACD\), the ratios of corresponding sides are equal. That is \(\frac{AB}{AC}=\frac{BE}{CD}\).
From the diagram, \(AB = 5\), \(AC=6\), \(BE = y\), and \(CD = 3.5\) (because \(CD\) is from \(y = 0\) to \(y=3.5\) on the vertical - axis).
So we have the proportion \(\frac{5}{6}=\frac{y}{3.5}\).

Step2: Solve the proportion for \(y\)

Cross - multiply the proportion \(\frac{5}{6}=\frac{y}{3.5}\) to get \(6y=5\times3.5\).
First, calculate \(5\times3.5 = 17.5\). Then the equation is \(6y = 17.5\).
Divide both sides by \(6\): \(y=\frac{17.5}{6}=\frac{175}{60}=\frac{35}{12}\approx2.92\).

Answer:

\(y = \frac{35}{12}\approx2.92\)