Math Problem Statement

Find the area of a triangle with sides of lengths 7 cm, 8 cm, and 9 cm.

Question 2

A triangle has sides of lengths 13 m, 14 m, and 15 m. Calculate its area.

Question 3

Determine the area of a triangle with side lengths 5 units, 12 units, and 13 units.

Question 4

A triangle has sides measuring 10 cm, 24 cm, and 26 cm. What is the area of the triangle?

Question 5

Find the area of a triangle with sides of lengths 6 cm, 8 cm, and 10 cm.

Can you give worked solutions to these questions

Solution

To find the area of a triangle when the lengths of all three sides are known, we can use Heron's formula. This formula states that the area AA of a triangle can be found using the following steps:

  1. Calculate the semi-perimeter ss of the triangle: s=a+b+c2s = \frac{a + b + c}{2} where aa, bb, and cc are the lengths of the sides of the triangle.

  2. Use Heron's formula to find the area AA: A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)}

Let's apply this method to each of the questions.

Question 1

Sides: a=7cm,b=8cm,c=9cma = 7 \, \text{cm}, \, b = 8 \, \text{cm}, \, c = 9 \, \text{cm}

  1. Calculate the semi-perimeter: s=7+8+92=242=12cms = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12 \, \text{cm}

  2. Calculate the area: A=12(127)(128)(129)=12×5×4×3A = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \times 5 \times 4 \times 3} A=720=144×5=125cm226.83cm2A = \sqrt{720} = \sqrt{144 \times 5} = 12\sqrt{5} \, \text{cm}^2 \approx 26.83 \, \text{cm}^2

Question 2

Sides: a=13m,b=14m,c=15ma = 13 \, \text{m}, \, b = 14 \, \text{m}, \, c = 15 \, \text{m}

  1. Calculate the semi-perimeter: s=13+14+152=422=21ms = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \, \text{m}

  2. Calculate the area: A=21(2113)(2114)(2115)=21×8×7×6A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} A=7056=84m2A = \sqrt{7056} = 84 \, \text{m}^2

Question 3

Sides: a=5units,b=12units,c=13unitsa = 5 \, \text{units}, \, b = 12 \, \text{units}, \, c = 13 \, \text{units}

  1. Calculate the semi-perimeter: s=5+12+132=302=15unitss = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \, \text{units}

  2. Calculate the area: A=15(155)(1512)(1513)=15×10×3×2A = \sqrt{15(15-5)(15-12)(15-13)} = \sqrt{15 \times 10 \times 3 \times 2} A=900=30units2A = \sqrt{900} = 30 \, \text{units}^2

Question 4

Sides: a=10cm,b=24cm,c=26cma = 10 \, \text{cm}, \, b = 24 \, \text{cm}, \, c = 26 \, \text{cm}

  1. Calculate the semi-perimeter: s=10+24+262=602=30cms = \frac{10 + 24 + 26}{2} = \frac{60}{2} = 30 \, \text{cm}

  2. Calculate the area: A=30(3010)(3024)(3026)=30×20×6×4A = \sqrt{30(30-10)(30-24)(30-26)} = \sqrt{30 \times 20 \times 6 \times 4} A=14400=120cm2A = \sqrt{14400} = 120 \, \text{cm}^2

Question 5

Sides: a=6cm,b=8cm,c=10cma = 6 \, \text{cm}, \, b = 8 \, \text{cm}, \, c = 10 \, \text{cm}

  1. Calculate the semi-perimeter: s=6+8+102=242=12cms = \frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 \, \text{cm}

  2. Calculate the area: A=12(126)(128)(1210)=12×6×4×2A = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \times 6 \times 4 \times 2} A=576=24cm2A = \sqrt{576} = 24 \, \text{cm}^2

These solutions should provide you with a clear understanding of how to apply Heron's formula to find the area of a triangle when given the side lengths.


Would you like more details on any of these calculations, or do you have any questions?

Related Questions

  1. What is the formula for the area of a triangle when the height is known?
  2. How do you determine if three side lengths can form a valid triangle?
  3. Can Heron's formula be used for all types of triangles, including equilateral?
  4. What are some common mistakes to avoid when using Heron's formula?
  5. How does Heron's formula relate to the Pythagorean theorem?

Tip: Always check that the sum of any two sides of a triangle is greater than the third side to ensure the side lengths form a valid triangle.

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Math Problem Analysis

Mathematical Concepts

Geometry
Triangle properties
Heron's formula

Formulas

Heron's formula for triangle area

Theorems

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Suitable Grade Level

High School