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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

7. Scale Drawings and Models

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Exercise 33 Page 605

If two triangles are similar, then the lengths of their corresponding altitudes are proportional to the lengths of their corresponding sides.

17.5

Practice makes perfect

Let's analyze the given triangles.

First, notice that two angles of the bigger triangle are congruent to two angles of the smaller triangle, one pair of which is vertical angles. Therefore, by the Angle-Angle (AA) Similarity Postulate these triangles are similar. △_(Bigger) ~ △_(Smaller) We are given the lengths of the altitudes of the triangles. Recall that if two triangles are similar, the lengths of their corresponding altitudes are proportional to the lengths of their corresponding sides. Therefore, we can write the following proportion. 8/14 = 10/x Let's solve it!

8/14 = 10/x

▼

Solve for x

a/b=.a /2./.b /2.

4/7 = 10/x

LHS * x=RHS* x

4/7 * x = 10/x * x

FracMultDenomToNumber

a/x* x = a

4/7 * x = 10

LHS * 7/4=RHS* 7/4

7/4 * 4/7 * x = 7/4 * 10

MultFracByInverse

a/b* b/a=1

1 * x = 7/4 * 10

1* a=a

x = 7/4 * 10

MoveRightFacToNum

a/c* b = a* b/c

x = 7 * 10/4

SplitIntoFactors

Split into factors

x = 7 * 5 * 2/2 * 2

Cross out common factors

x = 7 * 5 * 2/2 * 2

CancelCommonFac

Cancel out common factors

x = 7 * 5/2

Multiply

x = 35/2

Calculate quotient

x = 17.5

Subchapter links

Exercises p.602-605