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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Deriving the Triangle Area Formula

Deck 1 of 2: Where A = ½ab sin(C) Comes From

In this deck:

  • Why the base-height formula sometimes falls short
  • Derive A = ½ab sin(C) from scratch
  • Prove it works for acute, right, and obtuse triangles
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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Learning Objectives

By the end of this deck, you should be able to:

  1. Derive the formula by drawing an auxiliary altitude
  2. Explain why the formula works for acute, right, and obtuse triangles
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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

The Surveying Problem

Surveying scenario: triangular plot with two measured sides and the angle between them, height unknown

A surveyor measures two boundary lengths and the angle between them. Can you find the area without measuring the height?

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

The Base-Height Formula Has a Gap

The standard formula works — but only when height is known:

  • What surveyors measure: two boundary lengths + angle between them
  • What they don't measure: the perpendicular height
  • The goal: express in terms of a side and an angle

If we can write , substitution gives us a new formula.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Setting Up the Derivation

  • Angles , , are opposite their respective sides
  • We choose side (which is ) as the base
  • We need the altitude — the height from vertex perpendicular to

Case 1: the triangle is acute (all angles less than 90°)

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 1: Acute Triangle — Drawing the Altitude

Acute triangle ABC with altitude h drawn from B perpendicular to AC; foot of altitude labeled inside segment AC

Step 1: Draw altitude from perpendicular to .

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 1: Acute Triangle — Applying Sine

Same acute triangle; right triangle at C highlighted, with labels: hypotenuse a, opposite leg h, angle C

In the right triangle at : , so

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 1: Acute Triangle — The Formula

From the previous slide:

Substitute into :

This is the formula. The entire derivation fits in three lines.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Quick Check: What Did We Derive?

What expression for the height is the key step in the derivation?

  • (a)
  • (b)
  • (c)
  • (d)

Choose before advancing.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Answer: Height Equals Side Times Sine

Correct answer: (b)

Why: In the right triangle at vertex :

  • Hypotenuse = (side )
  • Opposite leg = (the altitude)

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 2: Right Triangle (A Quick Verification)

What if angle ?

  • Sides and serve directly as base and height
  • This matches the formula you already knew for right triangles
  • The formula passes the sanity check
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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 3: Obtuse Triangle — The Setup

What if angle ?

  • The altitude from now falls outside the triangle
  • The foot of the altitude is on the extension of beyond

Obtuse triangle ABC; altitude h drawn from B to the extension of AC beyond C, foot labeled outside the triangle

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 3: Obtuse Triangle — Sine of a Supplement

In the obtuse case, the relevant angle at in the right triangle is :

Therefore:

The height expression is identical — the supplement identity saves us.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Case 3: Obtuse Triangle — Same Formula

From the previous slide:

Substitute into :

A = ½ab sin(C) holds for all three triangle types.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Quick Check: Why Does the Obtuse Case Work?

For an obtuse triangle, the altitude falls outside — yet the formula still holds.

In one sentence: why does remain valid when ?

  • (a) Because sin(C) is always positive, the formula can't fail
  • (b) Because , so still holds
  • (c) Because obtuse triangles have larger altitudes, which compensate
  • (d) The formula does not actually work for obtuse triangles

Choose before advancing.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Why Supplementary Angles Have Equal Sines

This identity is the hinge of the obtuse case. Here's why it holds:

  • By the supplementary angle identity from the unit circle
  • Or: in a right triangle, for supplementary pairs
  • Example:

⚠️ Watch out: This does NOT hold for cosine —

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Three Equivalent Forms of the Formula

On triangle , the area can be written three ways:

Rule: Use the form that matches your given information.

The angle must always be the included angle — between the two sides used.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Numerical Example: Using the Formula

Triangle with , , . Find the area.

Step 1: Identify the form — sides and , included angle

Step 2: — verify your calculator is in degree mode

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Your Turn: Apply the Formula

Triangle with , , included angle .

Step 1: Write the formula for this case
Step 2: Substitute the values
Step 3: Evaluate — leave your answer to 2 decimal places

Set up and calculate before advancing.

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Answer: Worked-Out Area for Practice Problem

Check: . Since , area

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

Key Takeaways: Deck 1

derives from by writing

✓ The derivation works for all triangle types: acute, right, and obtuse

✓ For obtuse triangles, preserves the formula

⚠️ The angle must be included — between the two sides you use

⚠️ Don't drop the ½ — it comes from the base-height formula

⚠️ Check degree mode must equal

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HSG.SRT.D.9 · Derive the Triangle Area Formula | Deck 1 of 2

What Comes Next in Deck Two

In Deck 2 (Triangle Area — Applications and Interpretation):

  • Using all three equivalent forms — choosing the right one
  • Geometric interpretation: how angle controls area for fixed sides
  • Why area is maximized when
  • Real-world applications: surveying, parallelograms, land plots
  • Formula selection: when to use vs. vs. Heron's formula

Forward connections: HSG.SRT.D.10 (Law of Sines), physics —

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