3. Area of Triangles

Session 3 Learning Plan

Learning objective: Calculate triangle area from a base and perpendicular height.

Key vocabulary: base, height, perpendicular, triangle

Success criterion: Identify a matching base-height pair and explain why the factor \(\tfrac12\) is required.

Learn and Practice

A triangle is exactly half of a rectangle (or parallelogram) that surrounds it, so its area is
\[ A = \tfrac{1}{2} \times b \times h, \]
where \(b\) is the base and \(h\) is the height. The height is the straight-up distance from the base to the opposite point -- it must be perpendicular (at a right angle) to the base.

Worked example: a triangle with base 10 cm and height 6 cm has area
\[ A = \tfrac{1}{2} \times 10 \times 6 = \tfrac{1}{2} \times 60 = 30 \text{ cm}^2. \]

Case study - a triangular sail: a sail with a 4 m base and 9 m height has area
\[ \tfrac{1}{2} \times 4 \times 9 = 18 \text{ m}^2. \]

Common mistake: forgetting the \(\tfrac{1}{2}\). Multiplying base times height alone gives the area of the whole rectangle, which is twice the triangle.

Why a Triangle Uses One-Half

Duplicate a triangle, rotate the copy, and fit the two together. They form a parallelogram with the same base \(b\) and perpendicular height \(h\). Because the two triangles are congruent, each occupies half the parallelogram. Therefore
\[ A_{\text{triangle}}=\tfrac12 bh. \]
The factor \(\tfrac12\) is not a trick to memorize; it describes the triangle as half of a matching parallelogram.

You can multiply in any efficient order. For \(b=18\) and \(h=7\), halve the even base first:
\[ A=\tfrac12(18)(7)=9(7)=63\text{ units}^2. \]
This reduces calculator work and lowers the chance of forgetting the half.

The Height Must Be Perpendicular

The height is the shortest distance from the chosen base to the opposite vertex. It meets the line containing the base at \(90^\circ\). In an acute triangle, the height may lie inside. In a right triangle, one leg can serve as the base and the other as the height. In an obtuse triangle, the perpendicular height may fall outside the visible triangle on an extended base line.

A slanted side is not automatically a height. If a triangle has base \(10\), perpendicular height \(6\), and another side \(8\), use \(10\) and \(6\):
\[ A=\tfrac12(10)(6)=30. \]
The value \(40\), produced by \(\tfrac12(10)(8)\), uses a side that does not measure perpendicular distance.

Any Side Can Be a Base

A triangle does not have a permanent bottom. Any side can be selected as the base, but it must be paired with the perpendicular height measured to that side. Different valid base-height pairs produce the same area. This fact is especially useful when a diagram is rotated.

Suppose one base-height pair is \(12\) and \(5\). The area is \(30\). If another side of the same triangle is chosen as a base of \(10\), its corresponding height must satisfy
\[ 30=\tfrac12(10)h, \]
so \(h=6\). The height changes when the base changes, but the enclosed region does not.

Right Triangles and Rectangles

A rectangle cut along a diagonal creates two congruent right triangles. A \(9\)-by-\(4\) rectangle has area \(36\), so each triangular half has area \(18\). The formula confirms it:
\[ \tfrac12(9)(4)=18. \]
This visual shortcut can solve problems that shade one side of a rectangle's diagonal.

Multi-Step Example: Triangular Sign

A triangular road sign has base \(30\) inches and perpendicular height \(26\) inches. A can of reflective coating covers \(325\text{ in}^2\). How many cans are required for two coats on the front?

One coat covers
\[ A=\tfrac12(30)(26)=390\text{ in}^2. \]
Two coats require \(2(390)=780\text{ in}^2\). Divide by the coverage:
\[ 780\div325=2.4. \]
Because partial cans cannot supply the remaining coating, round up to \(3\) cans. The common trap is stopping at \(390\), ignoring the second coat, or rounding \(2.4\) down.

Same Base and Same Height

Triangles with the same base length and the same perpendicular height have equal area, even if their top vertices lean left or right. Picture several triangles whose bases lie on the same horizontal segment and whose third vertices lie anywhere on a parallel line above it. Their shapes differ, but \(\tfrac12 bh\) is unchanged. This can solve visual comparison questions without calculating every slanted side.

Algebraic Example

A triangle has area \(54\text{ cm}^2\), base \(12\text{ cm}\), and height \(h\). Set up the relationship:
\[ 54=\tfrac12(12)h=6h. \]
Then \(h=9\text{ cm}\). Verify by substituting: \(\tfrac12(12)(9)=54\). Notice the answer for a missing height uses centimeters, not square centimeters; it is a length.

Session Checkpoint

You are ready to continue when you can explain the factor \(\tfrac12\), identify a perpendicular height in rotated and obtuse diagrams, use the legs of a right triangle efficiently, solve a coverage problem, and work backward from area to a missing base or height.

GED® Coaching: Build a Repeatable Solution

The GED® rewards a dependable process more than memorizing isolated tricks. On every area problem, pause before touching the calculator. Read the final question first and underline the requested quantity. Then label the figure with only the measurements that matter. If the drawing contains extra numbers, ask what role each number plays in the formula. Write the formula in symbols, substitute the measurements with their units, and calculate. This order makes your work easy to check and prevents a calculator entry from becoming an unexplained guess.

Use the GED® formula sheet as a tool, not as a substitute for understanding. The sheet can remind you that a triangle uses \(\tfrac12 bh\), but you must decide which segment is the base and which is the perpendicular height. When a problem is unfamiliar, sketch a simpler version. A rough rectangle, circle, or set of component shapes often reveals the correct operation faster than rereading a dense paragraph.

Estimation is your built-in error detector. Round the dimensions to friendly values and predict the approximate size of the answer. If a rectangle is about \(10\) by \(5\), its area should be near \(50\), not \(5\) or \(500\). Check the unit too: area requires square units. A choice with feet instead of square feet may be testing whether you understand the target rather than whether you can multiply.

Build a visual record as you solve. Shade the region whose area is requested, circle measurements that belong in the formula, and cross out measurements that are irrelevant. For composite figures, draw the cut lines yourself and give each component a short name such as \(A_1\) or “window.” For coordinate figures, write each derived length beside its segment. These small annotations reduce working-memory load and leave evidence you can inspect if the answer looks wrong. They also make it easier to resume a marked problem during a second test pass.

When checking an answer choice, work backward conceptually: ask what calculation could have produced it. If a distractor equals the sum of two dimensions, it probably represents a perimeter-related mistake. If a circle answer is four times your result, diameter may have been used as radius. Recognizing the source of a distractor should confirm sound work, never replace it.

After practice, do not merely record a score. For every miss, write one sentence naming the error and one sentence describing the prevention habit. Examples include: “I used diameter as radius; next time I will mark \(r=d/2\),” or “I added the rectangles twice; next time I will shade non-overlapping parts.” This correction routine turns mistakes into durable GED® skills.

Session Practice Routine

1. State what measurement the problem requests.
2. Mark only the dimensions needed for the selected formula.
3. Write the formula before substituting values.
4. Show the calculation and attach a squared unit.
5. Check the answer by estimation or substitution.

Exit Reflection

Without looking back, write the session's key formula or strategy, name one common error, and explain how you will detect that error on a GED® question.