In a triangle BCD, BC = 25 m and CD = 10 m. The perimeter of the triangle may be
a. 69 m
b. 70 m
c. 71 m
d. 72 m
Note:
The sum of the lengths of any two sides of the triangle is greater than the length of the third side.
BC + CD > DB
25 + 10 > DB
35 > DB
The difference of the lengths of any two sides of the triangle is less than the length of the third side.
BC - CD < DB
25 - 10 < DB
15 < DB
Therefore 15 < DB < 35
a. 69 m
BC + CD + DB = 69
25 + 10 + DB = 69
DB = 34 m (the third side of the triangle)
The perimeter of the triangle is 69 m
b. 70 m
BC + CD + DB = 70
25 + 10 + DB = 70
DB = 35 m
c. 71 m
BC + CD + DB = 71
25 + 10 + DB = 71
DB = 36 m
d. 72 m
BC + CD + DB = 72
25 + 10 + DB = 72
DB = 37 m
