Exercise 28 Page 133

We are told that a person who is 5 feet tall casts a 4-foot long shadow. A building that is nearby casts a 24-foot long shadow. Our task it to find the height of the building h. First, let's take a look at a picture that illustrates this situation.

Next, we will draw the sun's rays in the picture. The rays create the shadows cast by the person and the building. Recall that the sun's rays are parallel. Now, we will name the two triangles which are created by the sun's rays, the shadows, the person standing, and the building as ABC and DEF. Let's take a look at the picture with the added information.

Notice that the shadows are perpendicular to the objects that cast them. Because of this, ∠ BAC and ∠ EDF are right angles. This means that △ ABC and △ DEF are right triangles.

Next, notice that ∠ BCA and ∠ EFD are corresponding angles. We know that if a transversal intersects parallel lines, then corresponding angles are congruent. As we stated before, the sun's rays are parallel, which means that ∠ BCA and ∠ EFD are congruent.

Two angles in △ ABC are congruent to two angles in △ DEF. Because of this, the third angles must also be congruent and the triangles are similar. Now, we will use the fact that corresponding sides in similar triangles are proportional. This means that we can equate the ratios of the corresponding sides. AB/AC=DE/DF ⇕ 5/4=h/24 Finally, we can solve for the height of the building h.

5/4=h/24

▼

Solve for h

MultEqn

LHS * 24=RHS* 24

5/4* 24 =h/24* 24

MoveRightFacToNum

a/c* b = a* b/c

120/4=h/24* 24

ReduceFrac

a/b=.a /4./.b /4.

30=h/24* 24

FracMultDenomToNumber

a/24* 24 = a

30=h

RearrangeEqn

Rearrange equation

h=30

The building is 30 feet high.