# Download 13. The ratio of the measures of the three sides of a triangle is 9: 7: 5

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```So the measures of the three angles are 3(18) or 54,
6(18) or 108, and 1(18) or 18.
7-1 Ratios and Proportions
13. The ratio of the measures of the three sides of a
triangle is 9: 7: 5. Its perimeter is 191.1 inches. Find
the measure of each side
19. The ratio of the measures of the three angles is
10:8:6.
SOLUTION: SOLUTION: Just as the ratio
Just as the ratio
or 9:7 is equivalent to
or
or 10:8 is equivalent to
or
10x:8x , the extended ratio can be written as
10x:8x:6x. 9x:7x , the extended ratio can be written as 9x:7x:5x. The perimeter is 191.1 inches so the sum of the
lengths of the sides is 191.1. Solve for x. We know that sum of the measures of all interior
angles in a triangle is 180. Set the sum of the
extended ratio equal to 180 and solve for x.
So the measures of the three sides are 9(9.1) or 81.9
in., 7(9.1) or 63.7 in., and 5(9.1) or 45.5 in..
So the measures of the three angles are 10(7.5) or
75, 8(7.5) or 60, and 6(7.5) or 45.
Find the measures of the angles of each
triangle.
17. The ratio of the measures of the three angles is 3:6:1.
SOLUTION: Just as the ratio
Solve each proportion.
28. or 3:6 is equivalent to
or
SOLUTION: 3x:6x , the extended ratio can be written as 3x:6x:1x. We know that sum of the measures of all interior
angles in a triangle is 180 degrees. Set the sum of the
extended ratios equal to 180 and solve for x. Cross multiply.
Solve for x.
So the measures of the three angles are 3(18) or 54,
6(18) or 108, and 1(18) or 18.
ALGEBRA Solve each proportion. Round to
the nearest tenth.
19. The ratio of the measures of the three angles is
10:8:6.
31. SOLUTION: Just as the ratio
or 10:8 is equivalent to
or
SOLUTION: 10x:8x , the extended ratio can be written as
10x:8x:6x. We know that sum of the measures of all interior
angles in a triangle is 180. Set the sum of the
extended ratio equal to 180 and solve for x.
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43. CHANGING DIMENSIONS A rectangle has a
Solve for x.
7-1 Ratios and Proportions
ALGEBRA Solve each proportion. Round to
the nearest tenth.
31. Therefore, the area of the new rectangle
c.
SOLUTION: d. The increase in the length to the width of the
rectangle in part b is proportional, i.e.
. The
increase in the length to the width of the rectangle in
43. CHANGING DIMENSIONS A rectangle has a
part c is not, i.e.
. length and a width w.
a. Find the perimeter and area of the rectangle. b. Suppose the length and width of the rectangle are
doubled. What is the ratio of the new perimeter to
the original perimeter? new area to the original
area?
c. Suppose the length and width increase by 2 units.
What is the ratio of the new perimeter to the original
perimeter? new area to the original area?
d. What fact about the operations explains the
differences between the results in part b and part c?
SOLUTION: a. b.
Therefore, the perimeter of the new rectangle
doubles. Therefore, the area of the new rectangle