10.1.3 - Verification of areas using Heron’s formula

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Introduction to Heron's Formula

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Teacher
Teacher

Today, we are going to explore Heron's formula. Can anyone already tell me what this formula helps us find?

Student 1
Student 1

The area of a triangle?

Teacher
Teacher

That's correct! It allows us to calculate the area, even when we only know the lengths of the sides. Let's define it. Heron's formula states that the area of a triangle can be calculated from its sides a, b, and c by first calculating the semi-perimeter s, which is \(s = \frac{a + b + c}{2}\).

Student 2
Student 2

What do we do with 's' after we find it?

Teacher
Teacher

Once we have 's', we can plug it into the formula: \(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\). Each part of that expression uses a side length!

Student 3
Student 3

Can you give us an example?

Teacher
Teacher

Absolutely! Let's consider a triangle with sides 40 meters, 32 meters, and 24 meters. First, we find \(s\) which equals 48 meters, then we check how Heron's formula gives us the area.

Calculating Area with Examples

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Teacher
Teacher

Using our triangle with sides 40m, 32m, and 24m, let's calculate the area together. First, recapping, what is \(s\)?

Student 4
Student 4

The semi-perimeter would be 48m.

Teacher
Teacher

Correct! So now let’s plug these into Heron's formula. What do we get?

Student 1
Student 1

We get \(\sqrt{48(48-40)(48-32)(48-24)}\).

Teacher
Teacher

Exactly! Now simplify that expression to find the area.

Student 2
Student 2

It simplifies to 384 square meters!

Teacher
Teacher

Right! And how can we visualize this in comparison to using the base-height method?

Student 3
Student 3

The area is the same! We can check by calculating it with '1/2 × base × height'!

Teacher
Teacher

Great observation! This illustrates the effectiveness of Heron’s formula.

Application in Different Triangle Types

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Teacher
Teacher

Now, let's apply Heron's formula on different triangle types. How about an equilateral triangle with each side measuring 10 cm?

Student 1
Student 1

For this triangle, we'd find \(s = 15\) cm first!

Teacher
Teacher

Good work! Now can you calculate the area from there?

Student 4
Student 4

Sure! The area would be \(\sqrt{15(15-10)(15-10)(15-10)} = 25\sqrt{3}\) cm².

Teacher
Teacher

Excellent. Now what about an isosceles triangle with sides 5 cm and 8 cm, having unequal sides of a and b?

Student 2
Student 2

The semi-perimeter \(s\) would be 9 cm, and I can calculate the area there as well!

Teacher
Teacher

Perfect! Everyone is applying Heron's formula wonderfully.

Verification and Cross-Checking

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Teacher
Teacher

Finally, verification is key. When verifying Heron's formula, what can we contrast it with?

Student 3
Student 3

We could check the area using basic geometry methods!

Teacher
Teacher

Exactly! By comparing Heron’s area to that calculated using base and height, we ensure our computations are correct. Shall we practice another?

Student 4
Student 4

Yes! Let's do a triangle with sides summing to a perimeter of 32 cm.

Teacher
Teacher

Great teamwork everyone, keep practicing applying Heron's formula!

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

Heron's formula provides a method to calculate the area of a triangle when the lengths of all three sides are known, without needing the height.

Standard

This section introduces Heron's formula for finding the area of a triangle using the lengths of its sides. It illustrates how to apply the formula through various triangle examples, emphasizing verification of computed areas.

Detailed

Verification of Areas Using Heron’s Formula

In this section, we explore the application of Heron’s formula, which enables the calculation of the area of a triangle given the lengths of its sides. The formula states that

\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \(s\) is the semi-perimeter defined as \(s = \frac{a + b + c}{2}\), with \(a\), \(b\), and \(c\) representing the sides of the triangle.

We begin with an example triangle park with sides measuring 40m, 32m, and 24m. Calculating the semi-perimeter gives \(s = 48m\) and using Heron's formula, we find the area matches our verification through the traditional base-height method, showcasing the consistency and effectiveness of the formula. We then apply Heron’s formula to additional triangles, examining an equilateral triangle with a side of 10 cm, and an isosceles triangle with sides measuring 5 cm and 8 cm, thus verifying versatility in various triangle types.

This section reinforces the significance of Heron’s formula in practical applications where traditional height-based area calculations may not be feasible.

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Audio Book

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Understanding Heron’s Formula

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The formula given by Heron about the area of a triangle, is also known as Hero’s formula. It is stated as:

Area of a triangle = \( \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c)} \)

where a, b and c are the sides of the triangle, and s = semi-perimeter, i.e., half the perimeter of the triangle = \( \frac{a + b + c}{2} \).

Detailed Explanation

Heron’s formula allows us to find the area of a triangle when the lengths of all three sides are known, without needing to calculate the height. To apply this formula, you first calculate the semi-perimeter (s), which is half of the triangle's perimeter. Then you substitute the values of a, b, and c (the sides) along with s into the formula to find the area.

Examples & Analogies

Imagine you have a triangular garden divided into three sections with sides measuring 40 m, 32 m, and 24 m. You can find the area of this garden without having to measure how high it is by using Heron’s formula.

Calculating Area of a Triangular Park

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Let us apply it to calculate the area of the triangular park ABC, mentioned above (see Fig. 10.2).

Let us take a = 40 m, b = 24 m, c = 32 m,
so that we have s = \( \frac{40 + 24 + 32}{2} = 48 \ m.\)

s – a = (48 – 40) m = 8 m,
s – b = (48 – 24) m = 24 m,
s – c = (48 – 32) m = 16 m.

Therefore, area of the park ABC = \( \sqrt{s \cdot (s - a) \cdot (s - b) \cdot (s - c)} \ = \sqrt{48 \cdot 8 \cdot 24 \cdot 16} = 384 \ m^2. \)

Detailed Explanation

In this example, we start by defining the sides of the triangle. By calculating the semi-perimeter using the formula \( s = \frac{a + b + c}{2} \), we find that the semi-perimeter is 48 m. We then find the values for \( s - a \), \( s - b \), and \( s - c \). Substituting these values back into Heron’s formula gives us the area of the triangular park as 384 m².

Examples & Analogies

Think of the triangular park like slicing a piece of pizza into three distinct slices. Each slice represents a side of the triangle. By knowing the lengths of the slices (sides), you can find out the entire area of the pizza using Heron’s formula without needing to measure how tall the pizza is at a point.

Verification with Right Triangle Properties

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We see that 32² + 24² = 1024 + 576 = 1600 = 40². Therefore, the sides of the park make a right triangle. The largest side, i.e., BC which is 40 m will be the hypotenuse and the angle between the sides AB and AC will be 90°.

We can check that the area of the park is \( \frac{1}{2} \times 32 \times 24 \ m^2 = 384 m^2. \) We find that the area we have got is the same as we found by using Heron’s formula.

Detailed Explanation

Here we confirm that the triangle formed by the sides of the park is a right triangle because the Pythagorean theorem holds true (the square of the longest side equals the sum of the squares of the other two sides). Additionally, we verify the area calculated using basic geometry (1/2 * base * height) matches the area calculated using Heron’s formula, illustrating its accuracy.

Examples & Analogies

Imagine a garden with a right triangle shape, where the base and height are the flower beds. By figuring out the area from both the height formula and Heron’s formula, you can cross-check your calculations to make sure you have enough soil to fill both beds correctly.

Example Applications of Heron’s Formula

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Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz.,
(i) equilateral triangle with side 10 cm.

(ii) isosceles triangle with unequal side as 8 cm and each equal side as 5 cm.

You will see that...
For (i), we have s = \( \frac{10 + 10 + 10}{2} = 15 \ cm. \)
Area of triangle = \( \sqrt{15 \cdot (15 - 10) \cdot (15 - 10) \cdot (15 - 10)} = 25 \sqrt{3} \ cm². \)
For (ii), we have s = \( \frac{8 + 5 + 5}{2} = 9 \ cm. \)
Area of triangle = \( \sqrt{9 \cdot (9 - 8) \cdot (9 - 5) \cdot (9 - 5)} = 12 \ cm².

Detailed Explanation

In this section, we apply Heron’s formula to different types of triangles. For the equilateral triangle, we find the semi-perimeter and use the formula to calculate the area. For the isosceles triangle, we follow the same process, showing how versatile Heron’s formula is for different shapes of triangles.

Examples & Analogies

Consider building structures like shelters or gazebos, which can be triangular in shape. Knowing how to calculate their area using Heron’s formula can help in calculating materials and costs more efficiently, ensuring you purchase the right amount for each triangular section.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Heron's Formula: A method to calculate the area of a triangle using its side lengths.

  • Semi-perimeter: The half of the perimeter of a triangle, critical for using Heron's formula.

  • Right Triangle: A triangle in which one angle is exactly 90 degrees.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • {'example': 'Find the area of a triangle with sides 40m, 32m, and 24m.', 'solution': '\[ s = \frac{40 + 32 + 24}{2} = 48m \] \n\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{48\cdot(48-40)(48-32)(48-24)} = \sqrt{48\cdot8\cdot16\cdot24} = 384 m^2 \]'}

  • {'example': 'Calculate the area of an equilateral triangle with side 10cm.', 'solution': '\[ s = \frac{10 + 10 + 10}{2} = 15cm \] \n\[ \text{Area} = \sqrt{15\cdot(15-10)(15-10)(15-10)} = \sqrt{15\cdot5\cdot5\cdot5} = 25\sqrt{3} cm^2 \]'}

  • {'example': 'Area of an isosceles triangle with side lengths 5cm, 5cm, and 8cm.', 'solution': '\[ s = \frac{5 + 5 + 8}{2} = 9cm \] \n\[ \text{Area} = \sqrt{9\cdot(9-5)(9-5)(9-8)} = \sqrt{9\cdot4\cdot4\cdot1} = 12 cm^2 \]'}

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

🎵 Rhymes Time

  • Heron’s way to find the space, use the sides to set the pace.

📖 Fascinating Stories

  • Imagine a triangle trying to find a way to show off its area by revealing the secrets of its sides to calculate the space inside.

🧠 Other Memory Gems

  • Remember the formula: Sway! Area = √(s(s-a)(s-b)(s-c))!

🎯 Super Acronyms

HAVE

  • Heights aren't vital
  • all sides equal.

Flash Cards

Review key concepts with flashcards.

Glossary of Terms

Review the Definitions for terms.

  • Term: Heron's Formula

    Definition:

    A formula that gives the area of a triangle when the lengths of all three sides are known.

  • Term: Semiperimeter

    Definition:

    Half of the perimeter of a triangle, calculated as s = (a + b + c) / 2.

  • Term: Area

    Definition:

    The amount of space inside a two-dimensional shape, measured in square units.