Find the unknown side length of the triangle given the perimeter P. Then classify the triangle by its side lengths.
Problem 1 :
If the perimeter of the given triangle is 49 inches, find the missing side.
Solution :
Perimeter of a triangle = Sum of the length of all its sides.
Let s_{1}, s_{2}_{ }and s_{3 }be the sum of the length of sides respectively.
The perimeter of a triangle = 49 in.
Let s_{1} = 18, s_{2} = 12 and s_{3 }= x
49 = s_{1} + s_{2} + s_{3 }
18 + 12 + x = 49
30 + x = 49
x = 49 - 30
x = 19 in
Every side is having different measures, so it is a scalene triangle.
Problem 2 :
Solution :
The perimeter of a triangle = The sum of the length of sides respectively.
Let s_{1}, s_{2} and s_{3 }be the sum of the length of all its sides.
The perimeter of a triangle = 22.5 yd.
Let s_{1} = 6.3, s_{2} = x and s_{3 }= x
22.5 = s_{1} + s_{2} + s_{3}
6.3 + x + x = 22.5
6.3 + 2x = 22.5
2x = 22.5 – 6.3
2x = 16.2
x = 16.2/2
x = 8.1 yd
Two sides are equal. So it is isosceles triangle.
Problem 3 :
Solution :
The perimeter of a triangle = Sum of the length of sides
Let s_{1}, s_{2} and s_{3 }be the sum of the length of all its sides respectively.
The perimeter of a triangle = 84.3 cm.
Let s_{1} = x, s_{2} = x and s_{3 }= x
84.3 = s_{1} + s_{2} + s_{3 }
x + x + x = 84.3
3x = 84.3
x = 84.3/3
x = 28.1 cm
Every side is equal, so it is equilateral triangle.
Problem 4 :
The perimeter of a triangle is 29 millimeters. The length of the first side is twice the length of the second side. The length of the third side is 5 more than the length of the second side. Find the side lengths of the triangle. Then classify the triangle by its side lengths.
Solution :
The perimeter of a triangle = Sum of the length of all its sides.
Length of the sides be s_{1, }s_{2 }and s_{3 }respectively.
s_{1} = 2s_{2}
The length of the third side is 5 more than the length of the second side.
s_{3} = s_{2} + 5
s_{1} + s_{2} + s_{3} = 29
2s_{2 }+ s_{2 }+ s_{2} + 5 = 29
4s_{2} + 5 = 29
4x = 29 – 5
4x = 24
x = 24/4
x = 6
Problem 5 :
The perimeter of a triangular window is 141 inches. The ratio of the side lengths of the window is 11 : 18 :18. Draw and label a diagram of the window. What are the side lengths of the window? Classify the window by its side lengths.
Solution :
The perimeter of a triangle = The sum of the length of all its sides.
Let s_{1} , s_{2} and s_{3 }be the sum of the length of all its sides.
The perimeter of a triangular window = 141 inches.
The ratio of the side lengths of the window = 11 : 18 : 18
s_{1} + s_{2} + s_{3} = 141
Let s_{1} = 11 x, s_{2} = 18x and s_{3} = 18x
11x +18x +18x =141
47x = 141
x = 141/47
x = 3
11(3) = 33 ; 18(3) = 54 ; 18(3) = 54
Therefore the
side lengths of the window are 33 in, 54in, 54 in.
Two equal sides and two equal angles. So it is isosceles triangle.
Problem 6 :
The ratio of the angle measures of a triangle is 7 : 16 : 22. Find the angle measures. Then classify the triangle by its angle measures.
Solution :
The ratio of the angle measures of a triangle are 7 : 16 : 22.
The angle of a triangle = 180°
7x +16x +22x =180°
45x = 180°
x = 180°/45
x = 4
The angle measures are,
7(4) = 28° ; 16(4) = 64° ; 22(4) = 88°
Problem 7 :
The ratio of the side lengths of a triangle is 7 : 24 : 25. The perimeter of the triangle is 392 inches.
a) Find the side lengths. Then classify the triangle by its side lengths.
b) is the triangle a right triangle ? How do you know?
Solution :
a) The ratio of the side lengths of a triangle are in the ratio
7 : 24 : 25
The perimeter of the triangle = 392 inches
7x + 24x +25x = 392
56x = 392
x = 392/56
x = 7
Therefore the side lengths are,
7(7) = 49 ; 24(7) = 168 ; 25(7) = 175
= 49 in ;168 in; 175 in
Every side is having different measures, so it is scalene triangle.
b) yes it is right triangle. It is used Pythagorean theorem.
49^{2} + 168^{2} = 175^{2}
2401 + 28224 = 30625
Since it satisfies Pythagorean theorem, it is right triangle.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM