MathsICSE Class 8 · Area of Plane Figures

Area of a Trapezium and of Polygons by Splitting (ICSE Class 8 Maths)

Updated 17 October 2026 · Fact-checked

Area of a trapezium $= \frac{1}{2}(a + b) h$, where $a, b$ are the parallel sides and $h$ the distance between them. Area of a quadrilateral with diagonal $d$ and perpendicular offsets $h_1, h_2$ is $\frac{1}{2} d (h_1 + h_2)$. Any polygon is found by splitting into triangles, rectangles and trapeziums and adding.

Two equal trapeziums make a parallelogram of area (a + b) h, so one trapezium is half of it: ½(a + b) h.

Illustration drawn for QuestionBuddy.

Understand Area of a Trapezium and Polygons

The CISCE curriculum asks for the area of a trapezium, a polygon and a semicircle (the semicircle has its own page).

Trapezium. A quadrilateral with one pair of parallel sides $a$ and $b$ and perpendicular distance $h$ between them:
$ $A = \frac{1}{2} (a + b) h .$ $
Reason: two copies of the trapezium (one turned upside down) form a parallelogram of base $(a + b)$ and height $h$ , so one copy has half its area.

Working backwards. From $A$ , $h$ and one parallel side, find the other: $a + b = \frac{2 A}{h}$ . With a ratio such as $3 : 5$ , write the sides as $3 k$ and $5 k$ .

Quadrilateral by a diagonal. Draw a diagonal $d$ and perpendiculars $h_{1}, h_{2}$ from the other two vertices. The area is the sum of two triangles: $\frac{1}{2} d h_{1} + \frac{1}{2} d h_{2} = \frac{1}{2} d (h_{1} + h_{2})$ .

Polygon. Split into triangles, rectangles or trapeziums using diagonals or perpendiculars; compute each area; add. A "house-shape" plot is a rectangle with a triangle on one side.

Common trap. A line joining the midpoints of the non-parallel sides has length $\frac{a + b}{2}$ but does not halve the area: the upper and lower trapeziums have different parallel sides, so their areas differ even though their heights are equal.

Units. Area in square units; do not mix cm and m.

Area of a Trapezium and Polygons: formulas to remember

Trapezium: $A = \frac{1}{2} (a + b) h$; $a, b$ parallel sides; $h$ perpendicular distance.
Quadrilateral by diagonal: $A = \frac{1}{2} d (h_{1} + h_{2})$; Offsets are perpendicular to the diagonal.
Missing parallel side: $b = \frac{2 A}{h} - a$; Reverse of the area formula.

How to solve Area of a Trapezium and Polygons questions

Identify the parallel sides and the perpendicular height, then split complex shapes.

1Mark the parallel sides and the perpendicular height.
2Substitute into the trapezium formula (units consistent).
3For a polygon, split it into triangles, rectangles or trapeziums.
4Add the parts and state the units.
5Check by an alternative split if time allows.

Area of a Trapezium and Polygons in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Sloping field

Suppose a field is shaped like a trapezium with parallel sides 25 m and 20 m, 8 m apart, and the owner wants to know how much seed is needed at 1 kg per 30 square metres.

Find the area and the seed needed.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} (25 + 20) (8) = 180$ m $^{2}$ .
Seed $= \frac{180}{30} = 6$ kg.

Answer: 180 m $^{2}$ ; 6 kg.

Memory hook: Average of the parallel sides times the height.

Tricks to remember Area of a Trapezium and Polygons

Short memory aids. Say each one aloud once, then use it on a question the same day.

Average width times height
(a + b)/2 is the average width; multiply by h.
Use it when: Remembering the trapezium formula.
Mid-line is not an area halver
The mid-line cuts the height in half, not the area.
Use it when: Avoiding a classic mistake.

Area of a Trapezium and Polygons: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Area of a trapezium · 1 mark · MCQBoard level

A trapezium has parallel sides 12 cm and 8 cm and height 5 cm. Its area is:
(a) 40 cm $^{2}$ (b) 50 cm $^{2}$ (c) 60 cm $^{2}$ (d) 100 cm $^{2}$

Try it first, then show the solutionSolution

$\frac{1}{2} (12 + 8) \times 5 = 50$ .

Answer: (b)

Example 2: Finding the height · 3 marks · Short answerBoard level

The area of a trapezium is 180 cm $^{2}$ and its parallel sides are 25 cm and 20 cm. Find the distance between them.

Try it first, then show the solutionSolution

$180 = \frac{1}{2} (25 + 20) h = 22.5 h$ .
$h = \frac{180}{22.5} = 8$ cm.

Answer: 8 cm

Example 3: Quadrilateral by offsets · 3 marks · Short answerHarder board level

In quadrilateral $A B C D$ the diagonal $A C = 20$ cm and the perpendiculars from $B$ and $D$ to $A C$ are 7 cm and 9 cm. Find its area.

Try it first, then show the solutionSolution

$△ A B C = \frac{1}{2} \times 20 \times 7 = 70$ .
$△ A C D = \frac{1}{2} \times 20 \times 9 = 90$ .
Area $= 70 + 90 = 160$ cm $^{2}$ .

Answer: 160 cm $^{2}$

Example 4: Parallel sides in a ratio · 4 marks · Short answerHarder board level

The parallel sides of a trapezium-shaped plot are in the ratio $3 : 5$ and are 12 m apart. The area is 384 m $^{2}$ . Find the parallel sides.

Try it first, then show the solutionSolution

Let the sides be $3 k$ and $5 k$ .
$\frac{1}{2} (3 k + 5 k) (12) = 48 k = 384 \Rightarrow k = 8$ .
Sides: $24$ m and $40$ m. Check: $\frac{1}{2} (64) (12) = 384$ .

Answer: 24 m and 40 m

Example 5: The mid-line trap · 4 marks · Long answerHOTS / competency

In trapezium $A B C D$ , $A B ∥ D C$ , $A B = 14$ cm, $D C = 8$ cm and the height is 6 cm. A line parallel to $A B$ is drawn at half the height. Find the length of this line, the areas of the two parts, and say whether the line halves the area.

Try it first, then show the solutionSolution

Mid-line length $= \frac{14 + 8}{2} = 11$ cm.
Lower part: parallel sides $14$ and $11$ , height $3$ : $\frac{1}{2} (25) (3) = 37.5$ cm $^{2}$ .
Upper part: sides $11$ and $8$ , height $3$ : $\frac{1}{2} (19) (3) = 28.5$ cm $^{2}$ . Total $66 = \frac{1}{2} (22) (6)$ .
$37.5 \neq = 28.5$ : the line does not halve the area.

Answer: 11 cm; 37.5 and 28.5 cm $^{2}$ ; it does not halve the area.

Where students lose marks in Area of a Trapezium and Polygons

The slips examiners see most often, and what to write instead.

Mistake: Using the slant side instead of the perpendicular height
The formula needs the perpendicular distance between the parallel sides.
Wrong working
Slant side 13 used as $h$ .
Right working
Use the perpendicular height (for example 12 from a 5-12-13 triangle).
Do this: Mark the right angle on the figure.
Usually costs: Answer mark.
Mistake: Forgetting the half in the formula
The factor $\frac{1}{2}$ is essential.
Wrong working
$A = (12 + 8) \times 5 = 100$ .
Right working
$A = \frac{1}{2} (20) (5) = 50$ .
Do this: Say "half the sum times the height".
Usually costs: Answer mark.

Exam tips

Mark the parallel sides and the perpendicular height on your sketch.
State units as square units.
For composite plots, give each part's area on a separate line.

Practice questions on Area of a Trapezium and Polygons

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The area of a trapezium with parallel sides 9 cm and 11 cm and height 4 cm is:
(a) 20 cm $^{2}$ (b) 36 cm $^{2}$ (c) 40 cm $^{2}$ (d) 80 cm $^{2}$
Hint
Average of 9 and 11 is 10.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
Find the area of a trapezium with parallel sides 15 cm and 9 cm, and height 7 cm.
Hint
Half of the sum times height.
Answer
$\frac{1}{2} (24) (7) = 84$ cm $^{2}$ .
Q3 · 3 marks · Short answerHarder board level
The area of a trapezium is 330 cm $^{2}$ , its height is 15 cm and one parallel side is 18 cm. Find the other parallel side.
Hint
Sum of parallel sides = 2A/h.
Answer
$a + b = \frac{660}{15} = 44$ , so the other side is $44 - 18 = 26$ cm.
Q4 · 3 marks · Short answerHarder board level
A quadrilateral has a diagonal of 18 cm and perpendicular offsets of 6.5 cm and 8.5 cm. Find its area.
Hint
Add the offsets.
Answer
$\frac{1}{2} \times 18 \times (6.5 + 8.5) = 9 \times 15 = 135$ cm $^{2}$ .
Q5 · 4 marks · Case-basedHOTS / competency
A plot is shaped like a rectangle $A B C D$ with $A B = 20$ m and $B C = 12$ m, with a triangle $C D E$ added on side $C D$ (apex $E$ ) of height 8 m from $E$ to $C D$ . Find its area and the cost of tiling at Rs 15 per square metre.
Hint
Rectangle plus triangle.
Answer
Rectangle $= 20 \times 12 = 240$ ; triangle $= \frac{1}{2} \times 20 \times 8 = 80$ . Total $320$ m $^{2}$ . Cost $= 320 \times 15 =$ Rs 4,800.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Area of a trapezium?Show answer
1/2 (sum of parallel sides) x height.
2.Area of a quadrilateral with diagonal d and offsets h1, h2?Show answer
1/2 d (h1 + h2).
3.Does the mid-line halve the area?Show answer
No.
4.Parallel sides 10 and 6, height 4: area?Show answer
32.

Area of a Trapezium and Polygons: frequently asked questions

Do the non-parallel sides matter?

Only through the height; the formula uses the perpendicular distance.

Can I split a trapezium into triangles?

Yes; the two parallel sides each give a triangle of height h, which gives the same formula.