#1: Is it possible to form a triangle with the given
side lengths? If not, explain why not.
1. 5 cm, 7 cm, 10 cm
SOLUTION:
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
ANSWER:
Yes; 5 + 7 > 10, 5 + 10 > 7, and 7 + 10 > 5
2. 3 in., 4 in., 8 in.
SOLUTION:
No; . The sum of the lengths of any two
sides of a triangle must be greater than the length of
the third side.
ANSWER: 3+4*8
3. 6 m, 14 m, 10 m
SOLUTION:
Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6 .
The sum of the lengths of any two sides of a triangle
must be greater than the length of the third side.
ANSWER:
Yes; 6 + 14 > 10, 6 + 10 > 14, and 10 + 14 > 6
4. MULTIPLE CHOICE If the measures of two
sides of a triangle are 5 yards and 9 yards, what is
the least possible measure of the third side if the
measure is an integer?
A 4 yd
B 5 yd
C 6 yd
D 14 yd
SOLUTION:
Let x represents the length of the third side. Next, set
up and solve each of the three triangle inequalities.
5 + 9 > x, 5 + x > 9, and 9 + x > 5
That is, 14 > x, x > 4, and x > –4.
Notice that x > –4 is always true for any whole
number measure for x. Combining the two remaining
inequalities, the range of values that fit both
inequalities is x > 4 and x < 14, which can be written
as 4 < x < 14. So, the least possible measure of the
third side could be 5 yd.
The correct option is B.
ANSWER:
B
Yes they are because if you divide 18 by 3 you get six and if you divide 15 by 3 you get 5, so it simplifies to 6:3 which is the same as the ratio for rings
