Properties and Formulae of Triangles
Perimeter (P) of a Triangle:
The perimeter of a triangle is the sum of the lengths of its three sides.
P = a + b + c
Example Problem: Find the perimeter of a triangle with sides of lengths 5 cm, 8 cm, and 6 cm.
P = 5 cm + 8 cm + 6 cm = 19 cm
Area (A) of a Triangle:
The area of a triangle can be calculated using Heron's formula or by knowing the base and height.
- Heron's Formula (with sides a, b, and c):
A = √(s * (s - a) * (s - b) * (s - c))
- Base and Height Method (with base 'b' and corresponding height 'h'):
A = (1/2) * b * h
Example Problem 1: Find the area of a triangle with sides of lengths 6 cm, 8 cm, and 10 cm.
s = (6 cm + 8 cm + 10 cm) / 2 = 12 cm
A = √(12 cm * (12 cm - 6 cm) * (12 cm - 8 cm) * (12 cm - 10 cm))
A ≈ 24 cm2
Example Problem 2: Find the area of a triangle with base length 12 cm and height 5 cm.
A = (1/2) * 12 cm * 5 cm = 30 cm2
Semiperimeter (s) of a Triangle:
The semiperimeter of a triangle is half of its perimeter.
s = (a + b + c) / 2
Example Problem: Find the semiperimeter of a triangle with sides of lengths 4 cm, 7 cm, and 9 cm.
s = (4 cm + 7 cm + 9 cm) / 2 = 10 cm
Pythagorean Theorem (Right-Angled Triangle):
In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
c^2 = a^2 + b^2
Example Problem: Find the length of the hypotenuse in a right-angled triangle with sides of lengths 5 cm and 12 cm.
c^2 = 5 cm^2 + 12 cm^2
c ≈ √(25 cm^2 + 144 cm^2) ≈ √169 cm^2 ≈ 13 cm
Law of Sines:
The law of sines relates the lengths of the sides of any triangle to the sines of its angles.
a / sin(A) = b / sin(B) = c / sin(C)
Example Problem: Find the length of side 'c' in a triangle with angles of 40°, 60°, and side lengths 'a' = 8 cm and 'b' = 10 cm.
sin(40°) ≈ 0.6428, sin(60°) = √3/2 ≈ 0.8660
c / 0.6428 = 10 cm / 0.8660
c ≈ (10 cm * 0.6428) / 0.8660 ≈ 7.43 cm
Law of Cosines:
The law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles.
a^2 = b^2 + c^2 - 2bc * cos(A)
Example Problem: Find the length of side 'a' in a triangle with angles of 45°, 70°, and side lengths 'b' = 8 cm and 'c' = 6 cm.
cos(45°) = √2/2 ≈ 0.7071
a^2 = 8 cm^2 + 6 cm^2 - 2 * 8 cm * 6 cm * 0.7071
a ≈ √(64 cm^2 + 36 cm^2 - 96 cm^2 * 0.7071) ≈ √(100 cm^2 - 67.41 cm^2) ≈ √32.59 cm^2 ≈ 5.71 cm
Angle Bisector Theorem:
In a triangle, if an angle bisector divides the opposite side into segments of lengths 'p' and 'q', then p/q = b/c, where 'b' and 'c' are the other two sides.
p / q = b / c
Example Problem: In triangle ABC, angle bisector AD divides side BC into segments of lengths BD = 4 cm and DC = 6 cm. If side AB = 8 cm and side AC = 10 cm, find the length of side BC.
BD / DC = AB / AC
4 cm / 6 cm = 8 cm / 10 cm
4 / 6 = 8 / 10
2/3 = 4/5
2 * 5 = 3 * 4
10 = 12
Median of a Triangle:
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Length of median from vertex A = (1/2) * √(2b^2 + 2c^2 - a^2)
Example Problem: Find the length of the median from vertex A in a triangle with sides of lengths 'a' = 7 cm, 'b' = 10 cm, and 'c' = 8 cm.
Length of median from vertex A = (1/2) * √(2 * (10 cm)^2 + 2 * (8 cm)^2 - (7 cm)^2)
Length of median from vertex A = (1/2) * √(200 cm^2 + 128 cm^2 - 49 cm^2)
Length of median from vertex A = (1/2) * √279 cm^2 ≈ (1/2) * 16.70 cm ≈ 8.35 cm
Altitude of a Triangle:
An altitude of a triangle is a perpendicular line segment from a vertex to the opposite side or its extension.
Length of altitude from vertex A = (2 / a) * √(s * (s - a) * (s - b) * (s - c))
Example Problem: Find the length of the altitude from vertex A in a triangle with sides of lengths 'a' = 12 cm, 'b' = 15 cm, and 'c' = 9 cm.
s = (12 cm + 15 cm + 9 cm) / 2 = 18 cm
Length of altitude from vertex A = (2 / 12 cm) * √(18 cm * (18 cm - 12 cm) * (18 cm - 15 cm) * (18 cm - 9 cm))
Length of altitude from vertex A = (1/6) * √(18 cm * 6 cm * 3 cm * 9 cm) = (1/6) * √8748 cm^2 ≈ (1/6) * 93.40 cm ≈ 15.57 cm