# Area of Triangles Problems with Solutions

A set of problems on how to calculate the area of triangles using different formulas are presented along with detailed solutions.

## Formulas for Area of Triangles
We first recall some of the most widely used formulas used to calculate the area of a triangle.
## Formula 1 - Base and Height of a triangle are knownWhen a base and the corresponding height are known, the area A of the triangle may be calculated as follows:Area = (1 / 2) × base × height = (1 / 2) × AB × CD ## Formula 2 - Two sides of a triangle and the angle between them are knownWhen two sides and the angle between them are known, the area A of the triangle may be calculated as follows:
Area = (1 / 2) × CA × CB × sin (C)## Formula 3 - Three sides of a triangle are knownWhen the lengths a, b and c of all three sides of a triangle are known, the area A of a triangle may be calculated using Heron's formula:Area = √[ s × (s - a) × (s - b) × (s - c) ]
where s in the above formula is given by s = (1 / 2)(a + b + c) ## Formula 4 - The three vertices of the triangle are known by their coordinatesIf A(x are the three vertices of a triangle given by their coordinates, then the formula for the area of the triangle defined by the three vertices A, B and C is given by:
_{A} , y_{A}), B(x_{B} , y_{B}) and C(x_{C} , y_{C})where det is the determinant of the three by three matrix. ## Problems on Areas
## Solutions to the Above Problems
We next find the determinant of matrix M Det (M) = 18 The are of triangle in part d) = ( 1/2) |Det (M)| = 9 unit ^{ 2}
AH ^{ 2} + 3^{ 2} = 6^{ 2}AH = √(27) = 3 √3 Use formula 1 to find the area of triangle ABC Area of triangle AHC = (1 / 2) × AH × BC = (1 / 2) × 3 √3 × 6 = 15.6 unit ^{2}
FG = 60 - 10 - 20 = 30 The length of the height EH from vertex E of triangle EFG is equal to the width of the rectangle. Area of triangle EFG = (1 / 2) × EH × FG = (1 / 2) × 30 × 30 = 450 unit ^{2}The area of the shaded region is obtained by subtracting the area of the triangle EFG from the area of the rectangle as follows Area of the shaded region = 60 × 30 - 450 = 1350 unit ^{2}
Area of triangles FEG = (1 / 2) base × height = (1 / 2) EG × FF' = (1 / 2) 100 × 25 = 1250 Area of triangles EHG = Area of triangles FEG = 1250 Area of the quadrilateral FGHE = 1250 + 1250 = 2500 unit ^{2}
We now have to find the vertices A, B and C as the intersection of the pairs of lines 1) Find A as the intersection of y = 2x + 2 and y = - x + 4 Solve the system y = 2x + 2 and y = - x + 4 by substitution 2x + 2 = - x + 4 gives the coordinates of A as: x = 2 / 3 and y = 5 / 2 2) Find B as the intersection of y = 2x + 2 and y = x Solve the system y = 2x + 2 and y = x by substitution 2x + 2 = x gives the coordinates of B as: x = - 2 and y = - 2 3) Find C as the intersection of y = x and y = - x + 4 Solve the system y = x and y = - x + 4 by substitution x = - x + 4 gives the coordinates of C as: x = 2 and y = 2 Since we have the coordinates of the three vertices of the triangle, we may use formula 4 as follows Area of triangle ABC =
## More References and LinksSine Law to Solve Triangle Problems.Pythagorean Theorem and Problems with Solutions. Cosine Law Problems. Solve a Triangle Given its Vertices. formulas Heron's Formula for Area of a Triangle Geometry Tutorials, Problems and Interactive Applets. |