Area of Triangles - Sixth Grade Mathematics

The area of any triangle is exactly equal to one-half the product of its base and its perpendicular height, reflecting its fundamental identity as half of a parallelogram.

Derivation of the Triangle Area Formula

Any triangle can be duplicated, rotated 180 degrees, and joined along one side to form a parallelogram sharing the exact same base and height.

Duplicating a Triangle to Make a Parallelogram:

       Original Triangle:                     Two Triangles Joined:
               B                                      B --------------- C
              /|                                     /|\               /
             / |                                    / | \  Triangle 2 /
            /  | h                                 /  |  \ (Rotated) /
           /   |                                  /   |   \         /
          +----+----------+                      +----+----+-------+
          A         D     C                      A         D       E
                Base b                                 Base b

Parallelogram Area = base * height
Since the original triangle is exactly ONE-HALF of the parallelogram:
Area of Triangle = (1 / 2) * base * height   or   Area = (b * h) / 2

Locating the Perpendicular Height

The height (altitude) of a triangle is the perpendicular segment drawn from any vertex to the line containing the opposite side (base). Depending on the triangle type, the height falls in different positions:

Position of Height in Different Triangle Types:

1. Acute Triangle: Height falls INSIDE the triangle.
             /\
            / | \   Height h
           /  |  \
          +---+---+
            Base b

2. Right Triangle: Height IS one of the legs (falls ALONG the boundary).
          |\
          | \
        h |  \   Hypotenuse
          |   \
          +----+
           Base b  Area = (leg1 * leg2) / 2

3. Obtuse Triangle: Height falls OUTSIDE the triangle on an extended base!
               /\
              /  \
          h  /    \
          : /      \
          :+--------+
           (ext) Base b

Notice that in an obtuse triangle, the dashed line representing the extended base is only used to measure the height; the base length b only includes the actual side of the triangle.

Step-by-Step Triangle Area Calculations

Working through standard examples demonstrates the proper application of the formula:

Example 1: Acute Triangle with base 14 cm and height 9 cm
           A = (1 / 2) * b * h
           A = (1 / 2) * 14 * 9 = 7 * 9 = 63 cm^2.

Example 2: Right Triangle with legs 8 inches and 15 inches
           A = (1 / 2) * 8 * 15 = 4 * 15 = 60 in.^2.

Example 3: Obtuse Triangle with base 7 m and external height 12 m
           A = (1 / 2) * 7 * 12 = 7 * 6 = 42 m^2.

Finding Missing Dimensions of a Triangle

If the area and one dimension are known, we can isolate the unknown variable.

Problem: A triangle has an area of 48 square inches and a base of 12 inches. Find its height.
Formula: A = (1 / 2) * b * h
Substitute: 48 = (1 / 2) * 12 * h
            48 = 6 * h
Divide by 6: h = 48 / 6 = 8 inches.
Check: (1 / 2) * 12 * 8 = 6 * 8 = 48 in.^2. (Confirmed!)

To quickly find the unknown dimension: double the area first (2 * A), then divide by the known dimension! h = (2 * A) / b and b = (2 * A) / h.

Exercises

Exercise 1. Calculate the area of each triangle: (a) Base = 18 cm, Height = 7 cm; (b) A right triangle with legs measuring 9 meters and 16 meters; (c) Base = 24 inches, Height = 13.5 inches; (d) An obtuse triangle with base = 6.2 cm and an external height = 8 cm.

Exercise 2. A triangular pennant has an area of 135 square inches and a base of 15 inches. What is the height of the pennant?

Exercise 3. A triangle has an area of 54 square meters. If its height is 9 meters, what is the length of its base?

Exercise 4. A right triangle has side lengths of 5 cm, 12 cm, and 13 cm. (a) Identify which sides represent the base, height, and hypotenuse; (b) Calculate the area of the triangle; (c) Find the height perpendicular to the 13 cm hypotenuse.

Solutions

Solution to Exercise 1. For (a), A = (1/2) * 18 * 7 = 9 * 7 = 63 cm^2. For (b), A = (1/2) * 9 * 16 = 9 * 8 = 72 m^2. For (c), A = (1/2) * 24 * 13.5 = 12 * 13.5 = 162 in.^2. For (d), A = (1/2) * 6.2 * 8 = 3.1 * 8 = 24.8 cm^2.

Solution to Exercise 2. Using A = (1/2) * b * h: 135 = (1/2) * 15 * h. Multiply both sides by 2: 270 = 15 * h. Divide by 15: h = 270 / 15 = 18 inches. The height of the pennant is 18 inches.

Solution to Exercise 3. Using A = (1/2) * b * h: 54 = (1/2) * b * 9. Multiply both sides by 2: 108 = 9 * b. Divide by 9: b = 108 / 9 = 12 meters. The base is 12 meters long.

Solution to Exercise 4. For (a), in any right triangle, the hypotenuse is the longest side (13 cm), and the two shorter sides (5 cm and 12 cm) are perpendicular legs acting as base and height. For (b), Area = (1/2) * 5 * 12 = 30 cm^2. For (c), using the 13 cm side as the base: Area = (1/2) * 13 * h = 30. So 13 * h = 60, giving h = 60 / 13 = approximately 4.62 cm.