Area of Triangles and Quadrilaterals

📐 Area of Triangles and Quadrilaterals

Part 1 of 5 — Rectangles, Squares & Parallelograms

Topics in This Part

🔑 Key Concept: Area is the number of unit squares that fit inside a flat shape. Every formula in this lesson is really just a clever way of counting those squares without drawing them.

What "Area" Really Means

Area measures the 2-dimensional space inside a closed figure. We report it in square units: $\text{cm}^2$, $\text{in}^2$, $\text{m}^2$, and so on.

Picture a $3 \times 4$ grid of unit squares:

$\underbrace{4 \text{ columns}}_{\text{base}} \times \underbrace{3 \text{ rows}}_{\text{height}} = 12 \text{ unit squares}$

That single picture — counting rows of squares — is the seed of every area formula. A rectangle is the easiest case because the squares line up perfectly.

💡 Units matter. A length is measured in units (like cm); an area is measured in square units (like $\text{cm}^2$). If your answer to an area problem has no little "2", something went wrong.

Rectangles and Squares

Example: a rectangle

A rectangle is $9\text{ cm}$ long and $4\text{ cm}$ tall:

$A = b \cdot h = 9 \cdot 4 = 36 \text{ cm}^2$

Example: a square

A square tile has side $7\text{ in}$:

$A = s^2 = 7^2 = 49 \text{ in}^2$

🔑 For a rectangle the base and height are just the two side lengths — they are already perpendicular, so no extra work is needed.

Concept Check 🎯

Parallelograms: Base × Height

A parallelogram is a quadrilateral with two pairs of parallel sides. It looks like a "pushed-over" rectangle.

$A = b \cdot h$

Here $b$ is a base (any one side) and $h$ is the height — the perpendicular distance to the opposite side, not the length of the slanted side.

⚠️ The #1 mistake: using the slanted side length instead of the perpendicular height. The slanted side is always longer than the height, so it would give too big an answer.

Why it works

If you slice a right triangle off one end of a parallelogram and slide it to the other end, you get a rectangle with the same base and the same height. Same squares, just rearranged — so the area must be identical: $b \cdot h$.

Example

A parallelogram has base $10\text{ cm}$, perpendicular height $6\text{ cm}$, and a slanted side of $8\text{ cm}$:

$A = b \cdot h = 10 \cdot 6 = 60 \text{ cm}^2$

The $8\text{ cm}$ slanted side is a distractor — it is not the height.

Find the Area 🧮

Compute each area. Enter just the number (square units understood).

1) Rectangle: base $15$, height $8$ → $A = \,?$ 2) Square: side $11$ → $A = \,?$ 3) Parallelogram: base $9$, perpendicular height $7$ → $A = \,?$

Base vs. Height 🔽

Decide which measurement to use as the height in $A = b \cdot h$.

📐 Area of Triangles and Quadrilaterals

Part 2 of 5 — Triangles

🔑 The Idea: A triangle is exactly half of a parallelogram. Two identical triangles snap together into a parallelogram with the same base and height — so a triangle's area is half of $b \cdot h$.

The Half-Base-Height Formula

$A = \frac{1}{2} \, b \, h$

• $b$ = the length of any one side you choose as the base
• $h$ = the perpendicular height from that base to the opposite vertex

The same warning from parallelograms applies: $h$ is the perpendicular height, not a slanted side.

Example: a right triangle

The two legs of a right triangle are perpendicular, so one leg is the base and the other is the height. With legs $6$ and $8$:

$A = \frac{1}{2}(6)(8) = \frac{1}{2}(48) = 24$

Example: any triangle

A triangle has base $14\text{ cm}$ and perpendicular height $5\text{ cm}$:

$A = \frac{1}{2}(14)(5) = \frac{1}{2}(70) = 35 \text{ cm}^2$

💡 Trick: multiply $b \cdot h$ first, then halve. With $b=14, h=5$: $70 \to 35$. It avoids fractions in the middle of your work.

Concept Check 🎯

Working Backwards for a Missing Length

Because $A = \frac{1}{2}bh$, if you know the area and one of $b$ or $h$, you can solve for the other. Double the area first to undo the $\frac{1}{2}$:

$2A = b \, h \quad\Longrightarrow\quad b = \frac{2A}{h}, \qquad h = \frac{2A}{b}$

Example

A triangle has area $48\text{ in}^2$ and height $8\text{ in}$. Find the base.

$b = \frac{2A}{h} = \frac{2(48)}{8} = \frac{96}{8} = 12 \text{ in}$

✅ Check: $\frac{1}{2}(12)(8) = \frac{1}{2}(96) = 48$ ✓

Triangle Practice 🧮

Enter just the number.

1) Triangle, base $16$, height $9$ → $A = \,?$ 2) Right triangle, legs $5$ and $12$ → $A = \,?$ 3) Triangle has area $40$ and base $10$. Perpendicular height $= \,?$

When You Only Know the Three Sides: Heron's Formula

Sometimes you are given all three side lengths $a, b, c$ but no height. Heron's formula handles this. First find the semiperimeter $s$ (half the perimeter):

$s = \frac{a + b + c}{2}, \qquad A = \sqrt{s(s-a)(s-b)(s-c)}$

Example: sides $5, 6, 7$

$s = \frac{5+6+7}{2} = \frac{18}{2} = 9$

$A = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{9 \cdot 4 \cdot 3 \cdot 2} = \sqrt{216} \approx 14.7$

Example: a $3\text{-}4\text{-}5$ triangle

$s = \frac{3+4+5}{2} = 6, \quad A = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = \sqrt{36} = 6$

✅ A $3\text{-}4\text{-}5$ triangle is a right triangle, so we can check: $\frac{1}{2}(3)(4) = 6$ ✓ — Heron's formula agrees.

Heron Step-by-Step 🔽

Use Heron's formula on a triangle with sides $a=9$, $b=10$, $c=11$.

📐 Area of Triangles and Quadrilaterals

Part 3 of 5 — Trapezoids, Rhombuses & Kites

🔑 The Idea: The remaining quadrilaterals each have their own shortcut, but every one traces back to triangles and rectangles hiding inside.

Trapezoids: Average the Parallel Sides

A trapezoid has exactly one pair of parallel sides, called the bases $b_1$ and $b_2$. The height $h$ is the perpendicular distance between those two parallel bases.

$A = \frac{1}{2}\,(b_1 + b_2)\,h$

In words: average the two parallel sides, then multiply by the height.

Example

A trapezoid has parallel sides $6$ and $10$ and height $4$:

$A = \frac{1}{2}(6 + 10)(4) = \frac{1}{2}(16)(4) = \frac{1}{2}(64) = 32$

💡 Sanity check: the average of $6$ and $10$ is $8$, and $8 \cdot 4 = 32$. The trapezoid acts like a rectangle whose width is the average of the two bases.

⚠️ Only the two parallel sides go into $b_1 + b_2$. The slanted (non-parallel) legs are not used in the area formula.

Concept Check 🎯

Rhombuses & Kites: Use the Diagonals

A rhombus (all four sides equal) and a kite (two pairs of adjacent equal sides) share a neat formula based on their diagonals $d_1$ and $d_2$:

$A = \frac{1}{2}\,d_1\,d_2$

This works because the diagonals of both shapes are perpendicular, slicing the figure into right triangles that recombine into a rectangle of half the diagonal area.

Example: a rhombus

A rhombus has diagonals $8$ and $6$:

$A = \frac{1}{2}(8)(6) = \frac{1}{2}(48) = 24$

Example: a kite

A kite has diagonals $10$ and $7$:

$A = \frac{1}{2}(10)(7) = \frac{1}{2}(70) = 35$

🔑 A square is a special rhombus, so you can also use $\frac{1}{2}d_1 d_2$ on a square. With diagonal $d$ (both equal): $A = \frac{1}{2}d^2$.

Quadrilateral Practice 🧮

Enter just the number.

1) Trapezoid: parallel sides $7$ and $13$, height $6$ → $A = \,?$ 2) Rhombus: diagonals $12$ and $5$ → $A = \,?$ 3) Kite: diagonals $9$ and $8$ → $A = \,?$

Match the Formula 🔽

Pick the correct area formula for each shape.

📐 Area of Triangles and Quadrilaterals

Part 4 of 5 — Composite Figures & the Coordinate Plane

🔑 Big Idea: Any complicated flat shape can be broken into triangles and quadrilaterals you already know. Find each piece, then add (or subtract) the areas.

Decomposing Composite Figures

A composite figure is built from simpler shapes. The strategy:

1. Cut the figure into rectangles, triangles, and trapezoids.
2. Find each piece's area separately.
3. Add them up (or subtract a piece that has been cut out).

Example: an "L" shape

An L-shaped room is a $10 \times 8$ rectangle with a $4 \times 3$ rectangle removed from one corner.

$A = (10 \cdot 8) - (4 \cdot 3) = 80 - 12 = 68$

Example: house pentagon (rectangle + triangle)

A "house" shape is a $6 \times 4$ rectangle with a triangular roof of base $6$ and height $3$ on top:

$A = \underbrace{(6 \cdot 4)}_{\text{walls}} + \underbrace{\tfrac{1}{2}(6)(3)}_{\text{roof}} = 24 + 9 = 33$

💡 Subtract when a region is hollow or cut out; add when pieces sit side by side.

Concept Check 🎯

Area on the Coordinate Plane

When a shape sits on a grid, you can read base and height straight off the coordinates.

• A horizontal segment's length is the difference in $x$-values.
• A vertical segment's length is the difference in $y$-values.

Example: a triangle with a horizontal base

A triangle has vertices $A(1, 2)$, $B(7, 2)$, and $C(4, 8)$.

The base $AB$ is horizontal: length $= 7 - 1 = 6$. The third vertex $C$ is at height $y = 8$, so the perpendicular height up from the base ($y = 2$) is $8 - 2 = 6$.

$A = \frac{1}{2}(6)(6) = 18$

💡 Choosing a horizontal or vertical side as the base makes the height trivial — it is just a difference of coordinates, no distance formula needed.

The Shoelace Formula (any polygon)

When no side is conveniently horizontal, the shoelace formula finds the area from the vertices directly. For a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$:

$A = \frac{1}{2}\,\bigl|\,x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\,\bigr|$

The bars $|\;|$ mean absolute value — area is never negative.

Example: vertices $(0,0)$, $(4,0)$, $(2,5)$

$A = \frac{1}{2}\,\bigl|\,0(0 - 5) + 4(5 - 0) + 2(0 - 0)\,\bigr| = \frac{1}{2}\,|0 + 20 + 0| = \frac{1}{2}(20) = 10$

✅ Check with base × height: base from $(0,0)$ to $(4,0)$ is $4$; height up to $y=5$ is $5$. $\frac{1}{2}(4)(5) = 10$ ✓.

Coordinate Practice 🧮

Enter just the number.

1) Triangle with horizontal base from $(2,1)$ to $(10,1)$ and apex at $(5,7)$. Base length $= \,?$ 2) Same triangle: perpendicular height $= \,?$ 3) Same triangle: area $= \,?$

Shoelace Step-by-Step 🔽

Apply the shoelace formula to the triangle with vertices $(1,1)$, $(5,1)$, $(3,6)$.

📐 Area of Triangles and Quadrilaterals

Part 5 of 5 — Mixed Practice & Mastery Check

You can now find the area of rectangles, squares, parallelograms, triangles (with a height or with Heron's formula), trapezoids, rhombuses, kites, composite figures, and shapes on a coordinate grid. Time to put it all together.

Quick Reference

⚠️ Two recurring traps: (1) using a slanted side instead of the perpendicular height, and (2) forgetting the square unit on your answer.

Mixed Practice 🎯

Mixed Numeric Practice 🧮

Enter just the number.

1) Parallelogram: base $13$, perpendicular height $6$ → $A = \,?$ 2) Rhombus: diagonals $14$ and $10$ → $A = \,?$ 3) Composite: a $12 \times 5$ rectangle with a triangle (base $12$, height $4$) on top → $A = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.