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    Math Deficiency - I
    MD-001
    Progress0 / 38 topics
    Topics
    1. Sets: Definition, Representation, and Operations2. Relation and Function: Graphical Transformation of Functions3. Properties of Functions4. Composition and Inverses of Functions5. Domain and Range of Functions6. Maximum and Minimum Values of Functions7. Increasing and Decreasing Functions8. Zeros and Intercepts of Functions9. Piecewise Functions10. Continuity and Discontinuity of Functions11. Polynomials and Rational Functions12. Polynomial Long Division and Synthetic Division13. Solution of Rational Functions14. Absolute Valued Functions and Their Properties15. Asymptotes: Horizontal, Vertical, and Oblique16. Exponential Functions and Their Properties17. Logarithmic Functions and Their Properties18. Systems of Equations: Two Equations and Two Unknowns19. Systems of Equations: Three Equations and Three Unknowns20. Matrix Algebra: Addition, Subtraction, and Multiplication21. Row Operations and Row Echelon Forms22. Augmented Matrices23. Determinant of Matrices: 2x2 and Higher Order24. Cramer's Rule25. Inverse Matrices26. Series and Sequences27. Trigonometry: Angles in Radians and Degrees28. Right Triangle Trigonometry29. Law of Cosines and Sines30. Area of a Triangle31. Graphs of Trigonometric Functions32. Graphs of Inverse Trigonometric Functions33. Basic Trigonometric Identities34. Trigonometric Equations35. General Form of a Conic: Parabolas, Circles, Ellipses, and Hyperbolas36. Degenerate Conics37. Polar and Parametric Equations38. Polar and Rectangular Coordinates
    MD-001›Area of a Triangle
    Math Deficiency - ITopic 30 of 38

    Area of a Triangle

    10 minread
    1,732words
    Intermediatelevel

    Area of a Triangle

    The area of a triangle is the amount of space enclosed within its three sides. Several formulas can be used to calculate the area of a triangle, depending on the information available. Below are the main formulas for calculating the area:


    1. Basic Formula (using base and height)

    The most straightforward formula for calculating the area of a triangle is:

    Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}Area=21​×base×height

    Where:

    • Base: Any one side of the triangle (usually the bottom side).
    • Height: The perpendicular distance from the base to the opposite vertex (the apex of the triangle).

    Example 1:

    If a triangle has a base of 6 units and a height of 4 units, the area is:

    Area=12×6×4=12 square units\text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \text{ square units}Area=21​×6×4=12 square units

    2. Heron's Formula

    When you know the lengths of all three sides of the triangle, you can use Heron's Formula to calculate the area. Heron’s formula is especially useful when the height is not known, and the triangle's dimensions are given by its sides.

    For a triangle with sides aaa, bbb, and ccc, the area AAA is given by:

    A=s(s−a)(s−b)(s−c)A = \sqrt{s(s - a)(s - b)(s - c)}A=s(s−a)(s−b)(s−c)​

    Where:

    • aaa, bbb, and ccc are the lengths of the sides of the triangle.
    • sss is the semi-perimeter of the triangle, which is calculated as:
    s=a+b+c2s = \frac{a + b + c}{2}s=2a+b+c​

    Example 2:

    Given a triangle with sides a=5a = 5a=5, b=6b = 6b=6, and c=7c = 7c=7, calculate the area.

    First, calculate the semi-perimeter:

    s=5+6+72=9s = \frac{5 + 6 + 7}{2} = 9s=25+6+7​=9

    Now apply Heron's Formula:

    A=9(9−5)(9−6)(9−7)=9×4×3×2A = \sqrt{9(9 - 5)(9 - 6)(9 - 7)} = \sqrt{9 \times 4 \times 3 \times 2}A=9(9−5)(9−6)(9−7)​=9×4×3×2​ A=216≈14.7 square unitsA = \sqrt{216} \approx 14.7 \text{ square units}A=216​≈14.7 square units

    3. Area Using Trigonometry (for non-right triangles)

    If you know two sides and the included angle, you can use a trigonometric formula to find the area of the triangle. The formula is:

    A=12×a×b×sin⁡(C)A = \frac{1}{2} \times a \times b \times \sin(C)A=21​×a×b×sin(C)

    Where:

    • aaa and bbb are the lengths of two sides.
    • CCC is the included angle between the sides aaa and bbb.

    Example 3:

    Given a triangle with sides a=8a = 8a=8, b=10b = 10b=10, and the included angle C=45∘C = 45^\circC=45∘, calculate the area.

    Using the formula:

    A=12×8×10×sin⁡(45∘)A = \frac{1}{2} \times 8 \times 10 \times \sin(45^\circ)A=21​×8×10×sin(45∘)

    Since sin⁡(45∘)=22≈0.707\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707sin(45∘)=22​​≈0.707, we have:

    A=12×8×10×0.707=28.28 square unitsA = \frac{1}{2} \times 8 \times 10 \times 0.707 = 28.28 \text{ square units}A=21​×8×10×0.707=28.28 square units

    4. Special Cases: Equilateral Triangle

    An equilateral triangle is a triangle in which all three sides are of equal length. The area of an equilateral triangle can be calculated using a specific formula:

    A=s234A = \frac{s^2 \sqrt{3}}{4}A=4s23​​

    Where:

    • sss is the length of a side of the equilateral triangle.

    Example 4:

    For an equilateral triangle with side length s=6s = 6s=6, the area is:

    A=6234=3634=93≈15.59 square unitsA = \frac{6^2 \sqrt{3}}{4} = \frac{36 \sqrt{3}}{4} = 9\sqrt{3} \approx 15.59 \text{ square units}A=4623​​=4363​​=93​≈15.59 square units

    5. Area of a Triangle Given Coordinates (Coordinate Geometry)

    If the triangle's vertices are given by their coordinates in a 2D plane, you can use the coordinate geometry formula to find the area of the triangle. The formula is:

    A=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|A=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

    Where:

    • (x1,y1)(x_1, y_1)(x1​,y1​), (x2,y2)(x_2, y_2)(x2​,y2​), and (x3,y3)(x_3, y_3)(x3​,y3​) are the coordinates of the three vertices of the triangle.

    Example 5:

    Given a triangle with vertices at (1,2)(1, 2)(1,2), (4,6)(4, 6)(4,6), and (7,3)(7, 3)(7,3), calculate the area.

    Using the formula:

    A=12∣1(6−3)+4(3−2)+7(2−6)∣A = \frac{1}{2} \left| 1(6 - 3) + 4(3 - 2) + 7(2 - 6) \right|A=21​∣1(6−3)+4(3−2)+7(2−6)∣ A=12∣1(3)+4(1)+7(−4)∣A = \frac{1}{2} \left| 1(3) + 4(1) + 7(-4) \right|A=21​∣1(3)+4(1)+7(−4)∣ A=12∣3+4−28∣=12∣−21∣=212=10.5 square unitsA = \frac{1}{2} \left| 3 + 4 - 28 \right| = \frac{1}{2} \left| -21 \right| = \frac{21}{2} = 10.5 \text{ square units}A=21​∣3+4−28∣=21​∣−21∣=221​=10.5 square units

    Summary

    The area of a triangle can be calculated using different methods depending on the given information:

    1. Basic Formula (base and height):

      A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}A=21​×base×height
    2. Heron's Formula (all three sides):

      A=s(s−a)(s−b)(s−c)A = \sqrt{s(s - a)(s - b)(s - c)}A=s(s−a)(s−b)(s−c)​
    3. Trigonometric Formula (two sides and the included angle):

      A=12×a×b×sin⁡(C)A = \frac{1}{2} \times a \times b \times \sin(C)A=21​×a×b×sin(C)
    4. Equilateral Triangle (side length sss):

      A=s234A = \frac{s^2 \sqrt{3}}{4}A=4s23​​
    5. Coordinate Geometry Formula (given vertices (x1,y1)(x_1, y_1)(x1​,y1​), (x2,y2)(x_2, y_2)(x2​,y2​), and (x3,y3)(x_3, y_3)(x3​,y3​)):

      A=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|A=21​∣x1​(y2​−y3​)+x2​(y3​−y1​)+x3​(y1​−y2​)∣

    These formulas are essential tools in geometry and help solve a wide range of triangle-related problems.

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    Law of Cosines and Sines
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    Graphs of Trigonometric Functions

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