QUESTION IMAGE
Question
lesson 2.8
additional practice
- at the same time that a 4.6 - foot - tall trash bin casts a 7 - foot shadow,
a nearby lamppost casts a 17.5 - foot shadow. how
tall is the lamppost?
- in the figure, the triangles are similar. what is the distance d across the pond?
- on a golf course, a flagstick casts a 150 - inch shadow at the
same time that a nearby tee casts a 5 - inch shadow. if the
toe is 2\\( \frac { 3 } { 4 } \\) inches tall, how tall is the flagstick?
- in the figure, the triangles are similar. what is the distance d from the
senior high to the junior high? round to the nearest tenth.
Step1: Set up proportion
For problem 1, since the triangles are similar (objects and their shadows form similar triangles), we can set up the proportion \(\frac{\text{height of object 1}}{\text{shadow of object 1}}=\frac{\text{height of object 2}}{\text{shadow of object 2}}\). Let \(h\) be the height of the lamppost. We have \(\frac{4.6}{7}=\frac{h}{17.5}\).
Step2: Solve for \(h\)
Cross - multiply: \(7h = 4.6\times17.5\). First, calculate \(4.6\times17.5=80.5\). Then \(h=\frac{80.5}{7}=11.5\) feet.
For problem 2, using the similarity of triangles \(\frac{d}{12}=\frac{72}{117 + 72}\). Simplify the denominator \(117+72 = 189\). So \(\frac{d}{12}=\frac{72}{189}\). Cross - multiply: \(189d=72\times12\). \(72\times12 = 864\). Then \(d=\frac{864}{189}\approx4.6\) m.
For problem 3, convert \(2\frac{3}{4}=\frac{2\times4 + 3}{4}=\frac{11}{4}\) inches. Set up the proportion \(\frac{\text{height of tree}}{\text{shadow of tree}}=\frac{\text{height of flagstick}}{\text{shadow of flagstick}}\), \(\frac{\frac{11}{4}}{5}=\frac{h}{150}\). Cross - multiply: \(5h=\frac{11}{4}\times150\). \(\frac{11}{4}\times150=\frac{1650}{4}\). Then \(h=\frac{1650}{4\times5}=\frac{1650}{20}=82.5\) inches.
For problem 4, using the similarity of triangles \(\frac{d}{35}=\frac{29}{290}\). Cross - multiply: \(290d=29\times35\). \(29\times35 = 1015\). Then \(d=\frac{1015}{290}\approx3.5\) km.
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- \(11.5\) feet
- \(4.6\) m
- \(82.5\) inches
- \(3.5\) km