Lesson 6 ~25 min Unit 3 · Measurement & Geometry +85 XP

Area of Parallelograms and Trapezia

Unlock $A = bh$ and $A = \frac{1}{2}(a+b)h$ — and never confuse slant height with perpendicular height again.

Today's hook: A solar farm covers a trapezoidal field. The parallel sides are 80 m and 120 m, the perpendicular height is 60 m. How many 2 m² panels fit on the field?

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Think First

warm-up

A parallelogram has base 9 cm, slant side 7 cm, and perpendicular height 5 cm. Without using a formula yet — can you estimate its area? Could you rearrange it into a rectangle?

Record your answer in your workbook.

The Big Idea

+5 XP

A parallelogram has two pairs of parallel sides. Slice a triangle off one end and reattach it to the other — you get a rectangle with the same base $b$ and the same perpendicular height $h$. So the area is simply $b \times h$. A trapezium has exactly one pair of parallel sides $a$ and $b$. Its area is the average of those sides multiplied by the perpendicular height $h$ between them.

Parallelogram: $A = bh$ Trapezium: $A = \tfrac{1}{2}(a+b)h$

Critical Rule

$h$ must be the perpendicular height — the shortest distance between the parallel sides. Never use the slant side.

Trapezium Memory

Think of it as the average of the two parallel sides times the height: $\dfrac{a+b}{2} \times h$.

Units

Area is always in square units: cm², m², mm², etc.

What You'll Master

objectives

Know

Parallelogram area: $A = bh$
Trapezium area: $A = \frac{1}{2}(a+b)h$
Perpendicular height $\neq$ slant side

Understand

Why rearranging a parallelogram gives a rectangle
Why the trapezium formula averages the two parallel sides
How to identify perpendicular height from a diagram

Can Do

Calculate area of parallelograms from base and perpendicular height
Calculate area of trapezia from parallel sides and height
Find an unknown side when area is given
Solve composite-shape problems

Words You Need

vocabulary

ParallelogramA quadrilateral with two pairs of parallel sides. Opposite sides are equal in length.

TrapeziumA quadrilateral with exactly one pair of parallel sides, labelled $a$ and $b$.

Perpendicular heightThe shortest distance between two parallel lines, measured at 90° to them. Also called altitude.

Slant sideThe non-perpendicular side of a parallelogram or trapezium. Always longer than the perpendicular height. Never use for area.

Base ($b$)Any side chosen as the reference side. Area = base × perpendicular height regardless of which side you call the base.

Composite shapeA shape made from two or more basic shapes. Find each area separately, then add or subtract.

Spot the Trap

heads-up

The #1 mistake is using the slant side instead of the perpendicular height. Below: parallelogram $b = 9$ cm, slant = 7 cm, $h = 5$ cm.

Wrong

$A = 9 \times 7 = 63$ cm² ✗

Correct

$A = 9 \times 5 = 45$ cm² ✓

Look for the right-angle marker in the diagram — that shows you where the perpendicular height is measured.

Parallelogram Area — Derived from a Rectangle

+5 XP

Slice the triangle from the left end of a parallelogram and attach it to the right end. The result is a rectangle with the same base $b$ and the same perpendicular height $h$. Therefore $A = b \times h$. The slant side plays no role — only the base and perpendicular height matter.

$$A = b \times h$$

Any side can be the base

Either pair of parallel sides can be the base — just make sure $h$ is perpendicular to your chosen base.

Trapezium Area — Average of Parallel Sides

+5 XP

Place two identical trapezia together to form a parallelogram. That parallelogram has base $(a + b)$ and height $h$. Since two trapezia formed it, one trapezium has half that area:

$$A = \frac{1}{2}(a+b)h$$

where $a$ and $b$ are the two parallel sides and $h$ is the perpendicular distance between them.

$$A = \frac{1}{2}(a+b)h$$

Order doesn't matter

$a + b = b + a$ — it makes no difference which parallel side you label $a$ and which you label $b$.

Identifying Perpendicular Height in Diagrams

+5 XP

When perpendicular height is shown inside the shape, look for a dashed line with a right-angle square at the foot. When the parallelogram leans far over, the height may be drawn outside the shape. Key signals:

A dashed line labelled $h$
A small square at the base (right-angle marker)
The label "height" or "altitude" in the problem text

No marker? Check the text

If no right-angle marker appears, the height is usually stated explicitly in the question text.

Composite Areas Using Parallelograms and Trapezia

+5 XP

Real-world shapes rarely fit one formula. To find composite area:

Split into simple parts (rectangle, parallelogram, trapezium, triangle).
Calculate each part's area separately.
Add areas together (or subtract cut-out regions).

$A_{\text{total}} = A_1 + A_2 + \ldots$

Watch Me Solve It · 3 examples

WE 1 — Parallelogram Area

+10 XP

PROBLEM

Find the area of a parallelogram with base $b = 9$ cm and perpendicular height $h = 6$ cm.

1
Write the formula
$$A = bh$$
2
Substitute values
$A = 9 \times 6$
3
Calculate
$A = 54$ cm²
4
State with units
$A = 54$ cm²

Answer$A = 54$ cm²

WE 2 — Trapezium Area

+10 XP

PROBLEM

Find the area of a trapezium with parallel sides $a = 5$ cm, $b = 11$ cm, and perpendicular height $h = 7$ cm.

1
Write the formula
$$A = \frac{1}{2}(a+b)h$$
2
Add the parallel sides
$a + b = 5 + 11 = 16$ cm
3
Substitute and calculate
$A = \frac{1}{2} \times 16 \times 7 = 8 \times 7 = 56$ cm²
4
State with units
$A = 56$ cm²

Answer$A = 56$ cm²

WE 3 — Find the Height

+10 XP

PROBLEM

A trapezium has area $48$ cm², parallel sides $a = 6$ cm and $b = 10$ cm. Find the perpendicular height $h$.

1
Substitute known values
$48 = \frac{1}{2}(6+10)h = \frac{1}{2} \times 16 \times h = 8h$
2
Solve for $h$
$h = 48 \div 8 = 6$ cm
3
Check by substituting back
$A = \frac{1}{2}(6+10)(6) = 8 \times 6 = 48$ cm² ✓

Answer$h = 6$ cm

Common Pitfalls

heads-up

Using the Slant Side as the Height

Mistake: $A = 9 \times 7 = 63$ cm² (using slant = 7 instead of $h$ = 6).

Fix: Always use the perpendicular height. Look for the right-angle marker.

Forgetting the $\frac{1}{2}$ in the Trapezium Formula

Mistake: $A = (a+b)h$ instead of $\frac{1}{2}(a+b)h$.

Fix: Always halve — the trapezium is half the parallelogram formed by two trapezia.

Using Only One Parallel Side for a Trapezium

Mistake: $A = \frac{1}{2} \times b \times h$ (ignoring side $a$).

Fix: You must add both parallel sides: $a + b$, then multiply by $\frac{1}{2}h$.

Copy Into Your Books

Parallelogram: $A = bh$ ($h$ = perpendicular height, not slant side)

Trapezium: $A = \frac{1}{2}(a+b)h$ ($a$, $b$ = parallel sides)

Find height: Rearrange — $h = A \div b$ (parallelogram) or $h = 2A \div (a+b)$ (trapezium)

Composite: $A_{\text{total}} = A_1 + A_2 + \ldots$ — never use slant side for height

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Parallelograms & Trapezia

4 problems

Set a timer for 4 minutes. Show all working.

1 Parallelogram $b = 15$ cm, $h = 8$ cm. Find $A$.

$A = 15 \times 8 = 120$ cm²
2 Trapezium $a = 9$ m, $b = 15$ m, $h = 6$ m. Find $A$.

$A = \frac{1}{2}(9+15)(6) = \frac{1}{2} \times 24 \times 6 = 72$ m²
3 Trapezium area $60$ cm², $a = 8$ cm, $b = 12$ cm. Find $h$.

$60 = \frac{1}{2}(8+12)h = 10h$, so $h = 60 \div 10 = 6$ cm.
4 Solar farm hook: trapezium $a = 80$ m, $b = 120$ m, $h = 60$ m. Area, then number of 2 m² panels.

$A = \frac{1}{2}(80+120)(60) = \frac{1}{2} \times 200 \times 60 = 6000$ m². Panels $= 6000 \div 2 = 3000$ panels.

Complete in your workbook.

Quick Check · 5 questions

Which formula gives the area of a parallelogram?

+10 XP

Parallelogram: $b = 14$ cm, $h = 5$ cm. What is the area?

+10 XP

Trapezium: $a = 7$ cm, $b = 13$ cm, $h = 8$ cm. Find the area.

+10 XP

Which height should you use in the parallelogram area formula?

+10 XP

Trapezium area $= 60$ cm², $a = 8$ cm, $b = 12$ cm. Find $h$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. A parallelogram has base $15$ cm and perpendicular height $8$ cm. Calculate its area. Show all working.

Show full working in your book.

UnderstandEasy2 MARKS

Q7. A trapezoidal garden has parallel sides $9$ m and $15$ m, and height $6$ m. Find the area.

Show full working in your book.

ReasonHard4 MARKS

Q8. A kite has diagonals $12$ cm and $8$ cm. Show that its area $= \frac{1}{2}d_1 d_2$ by splitting it into two triangles using one diagonal. Find each triangle's area, then the total.

Show full working in your book.

Comprehensive Answers

MC: 1-A, 2-B, 3-C, 4-A, 5-B

Q6: $A = bh = 15 \times 8 = 120$ cm².

Q7: $A = \frac{1}{2}(9+15)(6) = \frac{1}{2} \times 24 \times 6 = 72$ m².

Q8: The longer diagonal ($12$ cm) splits the kite into two triangles. Each triangle has base $12$ cm and height $= 8 \div 2 = 4$ cm. Area of each $= \frac{1}{2} \times 12 \times 4 = 24$ cm². Total $= 24 + 24 = 48$ cm². Formula check: $\frac{1}{2} \times 12 \times 8 = 48$ cm² ✓.

Stretch Challenge · +25 XP

Trapezoidal Roof Solar Cost

A trapezoidal roof section has parallel sides $6$ m and $10$ m, and perpendicular height $3.5$ m. Solar panels are 1 m² each and cost $450 each. Find the total cost to cover the roof.

Reveal solution

$A = \frac{1}{2}(6+10)(3.5) = \frac{1}{2} \times 16 \times 3.5 = 28$ m². Panels needed $= 28$. Total cost $= 28 \times \$450 = \$12\,600$.

Quick Review

Parallelogram

$A = bh$

Trapezium

$A = \frac{1}{2}(a+b)h$

Height rule

Perpendicular $h$ only — never the slant side

Find height

$h = A \div b$ (parallelogram)

Composite

Split → find each area → add

Units

Always cm², m², mm²

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