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Glencoe Math: Course 3, Volume 2

GM

Glencoe Math: Course 3, Volume 2 View details

5. Similar Triangles and Indirect Measurement

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Exercise 4 Page 557

The ratio of heights will be equal to the ratio of the lengths of their shadows.

$21$ feet

Practice makes perfect

We are given the following picture and we want to find the height of the big flagpole.

We can see that this is a shadow problem. We can assume that the angles formed by the sunbeams when the objects are hit by the light are congruent. The two shapes that are created by the sunbeams are right triangles.

We can think about the ground as a transversal and the sunbeams as parallel lines. Then the angles created are corresponding angles. Since two pairs of corresponding angles are congruent, the above triangles are similar by the rule for Angle-Angle (AA) Similarity. This means that the ratio of the heights of the flagpoles will be equal to the ratio of the lengths of their shadows.

\frac{Height _{Big Flagpole}}{Height _{Small Flagpole}} = \frac{Shadow _{Big Flagpole}}{Shadow _{Small Flagpole}} ⇓ \frac{h}{7} = \frac{6}{2}

Now, we will solve this proportion using cross products.

\frac{h}{7} = \frac{6}{2}

Cross multiply

h \cdot 2 = 7 \cdot 6

▼

Solve for

h

Multiply

h \cdot 2 = 42

$LHS / 2 = RHS / 2$

\frac{h \cdot 2}{2} = \frac{4 2}{2}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

h \cdot \frac{2}{2} = \frac{4 2}{2}

$\frac{a}{a} = 1$

h \cdot 1 = \frac{4 2}{2}

Identity Property of Multiplication

h = \frac{4 2}{2}

Calculate quotient

h = 21

The height of the big flagpole is

21

feet.

Subchapter links

Guided Practice p.556

Independent Practice p.557

H.O.T. Problems p.558

Standardized Test Practice p.558-560

Extra Practice p.559

Common Core Review p.560