The area of a parallelogram and a trapezium
🎯 In this topic you will
- Derive and use the formulae for the area of a parallelogram and a trapezium.
🧠 Key Words
- trapezia
Show Definitions
- trapezia: The plural of trapezium — quadrilaterals with exactly one pair of parallel sides (UK usage).
Look at this parallelogram.
Imagine you cut off a triangle from the left end of the parallelogram and move it to the right end.
You can see that you have made a rectangle.
So the area of the parallelogram is the same as the area of the rectangle with the same base and perpendicular height.
You can write the formula for the area of a parallelogram as:
$\text{area}=\text{base}\times\text{height}$
or simply $A=bh$
Now look at this trapezium.
The lengths of its parallel sides are $a$ and $b$.
Its perpendicular height is $h$.
Two trapezia can be put together like this to make a parallelogram with a base length of $(a+b)$ and a height $h$.
The area of the parallelogram is:
$\text{area}=\text{base}\times\text{height}=(a+b)\times h$
The area of one trapezium is half the area of the parallelogram.
So, the area of a trapezium is:
$A=\dfrac{1}{2}\times (a+b)\times h$
❓ EXERCISES
1. Copy and complete the workings to find the area of each parallelogram.
a.
$A = bh = 8 \times 4 = \_\_\_\_\ \text{cm}^2$
b.
$A = bh = \_\_\_\_\ \times \_\_\_\_ = \_\_\_\_\ \text{m}^2$
👀 Show answer
1a.$A = 8 \times 4 = 32\ \text{cm}^2$.
1b.$A = 6 \times 1.5 = 9\ \text{m}^2$.
2. Copy and complete the workings to find the area of each trapezium.
a.
$A=\dfrac{1}{2}\times(a+b)\times h=\dfrac{1}{2}\times(6+8)\times 5$
$=\dfrac{1}{2}\times \_\_\_\_\times 5$
$=\_\_\_\_\times 5$
$=\_\_\_\_\ \text{cm}^2$
b.
$A=\dfrac{1}{2}\times(a+b)\times h=\dfrac{1}{2}\times(\_\_\_\_+\_\_\_\_)\times 7$
$=\dfrac{1}{2}\times \_\_\_\_\times 7$
$=\_\_\_\_\times 7$
$=\_\_\_\_\ \text{mm}^2$
👀 Show answer
2a.$6+8=14$, so $\dfrac{1}{2}\times 14=7$, then $7\times 5=35\ \text{cm}^2$.
2b.$4+12=16$, so $\dfrac{1}{2}\times 16=8$, then $8\times 7=56\ \text{mm}^2$.
❓ EXERCISES
3. This is part of Bembe’s homework.
a. Explain the mistake Bembe has made.
b. Work out the correct answer.
👀 Show answer
3a. Bembe used the slanted side $5\ \text{cm}$ instead of the perpendicular height$4\ \text{cm}$ in $A = bh$.
3b. Correct area: $A = 7 \times 4 = 28\ \text{cm}^2$.
4. Sofia, Marcus and Zara are discussing the methods they use to find the area of a trapezium.
“I work out $(a+b)$, then divide by $2$, then multiply by $h$.”
“I work out half of $h$, then work out $(a+b)$, then multiply my two answers together.”
“I work out $(a+b)$, then multiply by $h$, then divide by $2$.”
Will they all get the same answer? Explain why.
👀 Show answer
• $\dfrac{(a+b)}{2}\times h=\dfrac{1}{2}(a+b)h$
• $\left(\dfrac{1}{2}h\right)\times(a+b)=\dfrac{1}{2}(a+b)h$
• $(a+b)\times h \div 2=\dfrac{(a+b)h}{2}$. By commutativity and associativity of multiplication, they give the same result.
🧠 Think like a Mathematician
Question 5. Look back at the methods used by Sofia, Marcus and Zara in Question 4. For each trapezium below, decide whose method is best to use and explain why. Then find the area.
Methods from Q4 (all equivalent to$A=\dfrac{1}{2}(a+b)h$):
- Sofia:$\dfrac{(a+b)}{2}\times h$ (average first).
- Marcus:$\left(\dfrac{h}{2}\right)\times(a+b)$ (half the height first).
- Zara:$\dfrac{(a+b)\,h}{2}$ (multiply then halve).
a.$a=6\ \text{cm},\ b=4\ \text{cm},\ h=3\ \text{cm}$
b.$a=7\ \text{cm},\ b=4\ \text{cm},\ h=6\ \text{cm}$
c.$a=2\ \text{m},\ b=3\ \text{m},\ h=5\ \text{m}$
d.$a=16\ \text{mm},\ b=14\ \text{mm},\ h=12\ \text{mm}$
👀 Show Answer
- a. Best: Sofia $(\dfrac{a+b}{2}\times h)$ or Zara $(\dfrac{(a+b)h}{2})$ — both keep integers. Work: $\dfrac{6+4}{2}\times 3=5\times 3=15\ \text{cm}^2$.
- b. Best: Marcus — halve the even height first. Work: $\left(\dfrac{6}{2}\right)\times(7+4)=3\times 11=33\ \text{cm}^2$.
- c. Best: Zara — multiply then halve (easy numbers). Work: $\dfrac{(2+3)\times 5}{2}=\dfrac{25}{2}=12.5\ \text{m}^2$.
- d. Best: Marcus — halve the even height, then multiply. Work: $\left(\dfrac{12}{2}\right)\times(16+14)=6\times 30=180\ \text{mm}^2$. (Sofia or Zara also give the same result; choose the order that keeps arithmetic simple.)
❓ EXERCISES
6. Work out the area of each trapezium using the method shown.
Look back at Question 4 to check the method.
a. Sofia’s method
b. Marcus’s method
c. Zara’s method
👀 Show answer
6a. Sofia: average first. $\dfrac{a+b}{2}\times h=\dfrac{3+7}{2}\times 5=5\times 5=25\ \text{m}^2$.
6b. Marcus: half the height first. $\left(\dfrac{h}{2}\right)\!(a+b)=\left(\dfrac{6}{2}\right)(8+9)=3\times17=51\ \text{cm}^2$.
6c. Zara: multiply then halve. $\dfrac{(a+b)h}{2}=\dfrac{(6+9)\times 5}{2}=\dfrac{75}{2}=37.5\ \text{mm}^2$.
7. This is part of Zalika’s homework.
a. Explain the mistake Zalika has made.
b. Work out the correct answer.
👀 Show answer
7a. She mixed units: Shape A is in $\text{mm}^2$ but Shape B is in $\text{cm}^2$, then she subtracted the numbers directly.
7b. Convert to the same unit before subtracting. Convert B to mm first: $6\ \text{cm}=60\ \text{mm}$, $10\ \text{cm}=100\ \text{mm}$, $9\ \text{cm}=90\ \text{mm}$.
$\text{Area B}=\dfrac{1}{2}(60+100)\times 90=\dfrac{1}{2}\times160\times90=80\times90=7200\ \text{mm}^2$.
$\text{Area A}=15\times12=180\ \text{mm}^2$.
Difference: $7200-180=7020\ \text{mm}^2$ (equivalently $70.2\ \text{cm}^2$).
🧠 Think like a Mathematician
Question 8. Work on your own to decide whether the following statement is correct. Explain your reasoning clearly.
Zara says: “If you double the base length of a parallelogram and double the height of the parallelogram, the area of the parallelogram will be doubled.”
Method (suggested):
- Let the base be $b$ and the perpendicular height be $h$. The area is $A = bh$.
- Double both dimensions: new base $2b$, new height $2h$. Compute the new area $A'=(2b)(2h)$.
- Compare $A'$ with $A$. Optionally check with numbers (e.g., $b=3$, $h=4$).
Follow-up Questions:
👀 Show Answer
- Main claim: Not correct. Since $A=bh$, doubling both gives $A'=(2b)(2h)=4bh$, so the area is quadrupled, not doubled. Example: $3\times4=12$ becomes $6\times8=48$ (four times bigger).
- 1. Doubling only the base doubles the area: $(2b)h=2(bh)$.
- 2. Doubling only the height doubles the area: $b(2h)=2(bh)$.
- 3. Doubling the base and halving the height leaves the area unchanged: $(2b)\!\left(\dfrac{h}{2}\right)=bh$.
- 4. Scaling both by $k$ multiplies the area by $k^2$: $(kb)(kh)=k^2 bh$.
❓ EXERCISES
💡 Tip
To work out an estimate, round all the numbers to one significant figure.
9. The diagram shows a trapezium.
a. Work out an estimate of the area of the trapezium.
b. Use a calculator to work out the accurate area of the trapezium.
👀 Show answer
a. Round to one s.f.: $2.3\to 2$, $9.8\to 10$, $4.6\to 5$.
Estimate: $A\approx\dfrac{1}{2}(2+10)\times 5=\dfrac{1}{2}\times 12\times 5=30\ \text{cm}^2$.
b. Accurate: $A=\dfrac{1}{2}(2.3+9.8)\times 4.6=\dfrac{1}{2}\times 12.1\times 4.6=27.83\ \text{cm}^2$.
10. Here are four shapes, A, B, C and D.
Here are five area cards.
a. Using estimation only, match each shape with the correct area card.
b. Use a calculator to check your answers to part a.
c. Sketch a shape that has an area equal to the area on the card you have not matched.
👀 Show answer
a (estimation only).
A (trapezium): round $4.5\!\to\!5$, $5.4\!\to\!5$, $3.8\!\to\!4$ ⇒ $\dfrac{1}{2}(5+5)\times 4=20$ ⇒ closest to $18.81$ → card ii.
B (rectangle): $4.2\!\to\!4$, $3.7\!\to\!4$ ⇒ $4\times 4=16$ ⇒ closest to $15.54$ → card iv.
C (parallelogram): $3.4\!\to\!3$, $2.9\!\to\!3$ ⇒ $3\times 3=9$ ⇒ closest to $9.86$ → card i.
D (triangle): $8.2\!\to\!8$, $2.7\!\to\!3$ ⇒ $\dfrac{1}{2}\times 8\times 3=12$ ⇒ closest to $11.07$ → card v.
b (accurate check).
A: $\dfrac{1}{2}(4.5+5.4)\times 3.8=\dfrac{1}{2}\times 9.9\times 3.8=18.81\ \text{cm}^2$ → card ii.
B: $4.2\times 3.7=15.54\ \text{cm}^2$ → card iv.
C: $3.4\times 2.9=9.86\ \text{cm}^2$ → card i.
D: $\dfrac{1}{2}\times 8.2\times 2.7=11.07\ \text{cm}^2$ → card v.
c. One possible sketch: a parallelogram with base $6.8\ \text{cm}$ and height $3.6\ \text{cm}$ so that $A=6.8\times 3.6=24.48\ \text{cm}^2$ (matches card iii).
💡 Tip
Be careful with the units.
11. A parallelogram has an area of $832\ \text{mm}^2$.
It has a perpendicular height of $2.6\ \text{cm}$.
What is the length of the base of the parallelogram?
👀 Show answer
With $A=bh$, base $b=\dfrac{A}{h}=\dfrac{832}{26}=32\ \text{mm}$ (i.e., $3.2\ \text{cm}$).