MathsICSE Class 8 · Chapter 20

Area of Plane Figures ICSE Class 8: Trapezium, Polygon and Semicircle

Updated 15 October 2026 · Fact-checked

The area of a trapezium is half the sum of its parallel sides times the distance between them: ½(a + b)h. ICSE Class 8 finds the area of a polygon by splitting it into triangles, rectangles and trapeziums, and the area of a semicircle as ½πr², with perimeter πr + 2r.

What Area of Plane Figures is about

Area is the amount of flat surface a shape covers, measured in square units (cm², m²). Perimeter is the length of its boundary.

The CISCE Class VIII curriculum asks for the area of:

a trapezium (a quadrilateral with one pair of parallel sides),
a polygon (by splitting it into shapes whose areas you know),
a semicircle (half a circle).

You already know: rectangle $l \times b$ , square $a^{2}$ , triangle $\frac{1}{2} \times base \times height$ , parallelogram base × height, circle $π r^{2}$ and circumference $2 π r$ . Take $π = \frac{22}{7}$ or $3.14$ as the question says.

In a school exam you will find areas, work backwards to a missing side or height, find areas and perimeters of shapes made of rectangles and semicircles, and solve costing problems (fencing per metre, tiling or turfing per square metre). Always write the unit (cm² for area, cm for length).

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Area of Plane Figures sits in the Mensuration theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Area of a trapezium

Draw diagonal $A C$ in trapezium $A B C D$ with $A B ∥ D C$ . It makes two triangles with the same height $h$ (the distance between the parallel sides):
$Area = \frac{1}{2} \times A B \times h + \frac{1}{2} \times D C \times h = \frac{1}{2} (a + b) h$ ,
where $a$ and $b$ are the parallel sides.

In words: average of the parallel sides × height. The height must be measured at right angles to the parallel sides, not along a slanting side.

Worked example

Find the area of a trapezium whose parallel sides are 12 cm and 8 cm and whose height is 5 cm.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} (12 + 8) \times 5$ .
$= \frac{1}{2} \times 20 \times 5 = 50$ cm².

Answer: 50 cm²

Working backwards in a trapezium

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

height $h = \frac{2 A}{a + b}$ ,
sum of parallel sides $a + b = \frac{2 A}{h}$ , so a missing side $= \frac{2 A}{h} - (known side)$ .

If the sides are in a ratio, let them be $k x$ and $m x$ .

Worked example

A trapezium has area 72 cm² and parallel sides 10 cm and 14 cm. Find its height.

Try it first, then show the solutionSolution

$h = \frac{2 \times 72}{10 + 14}$ .
$= \frac{144}{24} = 6$ cm.

Answer: 6 cm

Area of a polygon by splitting

Split the polygon into rectangles, triangles and trapeziums, find each area and add (or find a big shape and subtract the missing piece).

Diagonal and offsets: for a quadrilateral with diagonal $d$ and perpendiculars (offsets) $h_{1}$ and $h_{2}$ from the other two vertices to the diagonal:
$Area = \frac{1}{2} \times d \times (h_{1} + h_{2})$ .

Rhombus (or any quadrilateral whose diagonals cross at right angles): $Area = \frac{1}{2} d_{1} d_{2}$ .

Draw-and-label checklist: sketch the polygon, draw the dividing lines, label each part (I, II, III) with its measurements, and show each area separately.

Worked example

A pentagon is made of a rectangle 10 cm × 6 cm with a triangle on top whose base is the 10 cm side and whose height is 4 cm. Find its area.

Try it first, then show the solutionSolution

Rectangle: $10 \times 6 = 60$ cm².
Triangle: $\frac{1}{2} \times 10 \times 4 = 20$ cm².
Total $= 80$ cm².

Answer: 80 cm²

Area of a rhombus

The diagonals of a rhombus cut each other at right angles and divide it into four right-angled triangles. Adding their areas gives
$Area = \frac{1}{2} \times d_{1} \times d_{2}$ .

If you know a side and the height instead, the rhombus is a parallelogram: area $=$ base × height.

Worked example

Find the area of a rhombus with diagonals 16 cm and 12 cm.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times 16 \times 12$ .
$= 96$ cm².

Answer: 96 cm²

Semicircle: area and perimeter

A semicircle is half a circle, cut along a diameter.

Area $= \frac{1}{2} π r^{2}$ .
Perimeter $=$ curved part + diameter $= π r + 2 r$ .

The curved part alone is half the circumference: $\frac{1}{2} \times 2 π r = π r$ . Do not forget the straight edge (the diameter) when the question asks for the perimeter of the semicircle.

If the diameter is given, halve it first to get $r$ .

Worked example

Find the area and perimeter of a semicircle of radius 7 cm ( $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77$ cm².
Curved part $= \frac{22}{7} \times 7 = 22$ cm.
Perimeter $= 22 + 14 = 36$ cm.

Answer: Area 77 cm²; perimeter 36 cm

Combined shapes

Many real shapes are rectangles with semicircles on the ends (running tracks, windows, tables).

Area: add (or subtract) the parts.
Perimeter: add only the outside edges. A side that is covered by a semicircle is replaced by the curved edge and is not counted.

Two semicircles on opposite ends of the same diameter make a full circle: their areas add to $π r^{2}$ .

Worked example

A shape is a rectangle 14 cm × 10 cm with a semicircle on one 10 cm side. Find its area ( $π = 3.14$ ).

Try it first, then show the solutionSolution

Rectangle: $14 \times 10 = 140$ cm².
Semicircle: $r = 5$ , area $= \frac{1}{2} \times 3.14 \times 25 = 39.25$ cm².
Total $= 179.25$ cm².

Answer: 179.25 cm²

Maths in real life: Area of Plane Figures

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.

Trapezium

Suppose a table top is a trapezium with parallel sides 1.2 m and 0.8 m, which are 0.6 m apart.

Find its area.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} (1.2 + 0.8) \times 0.6$ .
$= \frac{1}{2} \times 2 \times 0.6 = 0.6$ m².

Answer: 0.6 m²

Memory hook: Average of the parallel sides × height.

Semicircle

Imagine a semicircular glass panel of radius 0.35 m above a door.

How much glass is needed? ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times \frac{22}{7} \times 0.35 \times 0.35$ .
$= \frac{1}{2} \times 0.385 = 0.1925$ m².

Answer: 0.1925 m²

Memory hook: A semicircle is half of πr².

Rhombus

Suppose a poster is shaped like a rhombus with diagonals 60 cm and 40 cm.

How much paper does it cover?

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times 60 \times 40$ .
$= 1200$ cm².

Answer: 1200 cm²

Memory hook: Diagonals multiplied, then halved.

Topics in this chapter

Area of Plane Figures formula sheet

Trapezium: $A = \frac{1}{2} (a + b) h$; a, b parallel; h perpendicular distance.
Quadrilateral with offsets: $A = \frac{1}{2} d (h_{1} + h_{2})$; d = diagonal.
Rhombus: $A = \frac{1}{2} d_{1} d_{2}$; Product of diagonals ÷ 2.
Semicircle area: $A = \frac{1}{2} π r^{2}$; Half the circle.
Semicircle perimeter: $P = π r + 2 r$; Curved part + diameter.
Triangle: $A = \frac{1}{2} bh$; Used when splitting polygons.

How the school exam asks Area of Plane Figures

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Trapezium area	Find the area; or find the height or a parallel side from the area.	½(a + b)h, or rearrange it.
Polygon by splitting	Find the area of an L-shape, a house shape or a field with offsets.	Split into rectangles, triangles and trapeziums; add or subtract.
Rhombus	Find the area of a rhombus from its diagonals.	½ d₁d₂.
Semicircle	Area and perimeter of a semicircle; radius from area.	½πr² and πr + 2r.
Combined shapes and cost	Running tracks, windows, gardens; cost of fencing or turfing.	Areas for turfing and tiling; perimeters for fencing.

Tricks to remember Area of Plane Figures

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

Average × height
Trapezium area = (average of the parallel sides) × height.
Use it when: Remembering ½(a + b)h.
Half the circle, plus the straight bit
Semicircle perimeter: half the circumference (πr) plus the diameter (2r).
Use it when: Not missing the 2r.
Diagonals × ÷ 2
Rhombus area: multiply the diagonals, then halve.
Use it when: Rhombus questions.
Fence the edge, turf the inside
Fencing, borders and ribbons use perimeter; carpets, grass and tiles use area.
Use it when: Choosing area or perimeter in costing.

Area of Plane Figures: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Trapezium · 1 mark · MCQExam level

The area of a trapezium with parallel sides 7 cm and 9 cm and height 4 cm is:
(a) 16 cm²
(b) 32 cm²
(c) 64 cm²
(d) 63 cm²

Try it first, then show the solutionSolution

$\frac{1}{2} (7 + 9) \times 4$ .
$= 8 \times 4 = 32$ cm².

Answer: (b)

Example 2: Semicircle · 3 marks · Short answerExam level

Find the area and the perimeter of a semicircle of radius 14 cm ( $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308$ cm².
Curved edge $= \frac{22}{7} \times 14 = 44$ cm.
Perimeter $= 44 + 28 = 72$ cm.

Answer: 308 cm²; 72 cm

Example 3: Diagonal and offsets · 3 marks · Short answerExam level

In quadrilateral $A B C D$ , diagonal $A C = 18$ cm. The perpendiculars from $B$ and $D$ to $A C$ are 7 cm and 5 cm. Find the area.

Try it first, then show the solutionSolution

The diagonal splits $A B C D$ into triangles $A B C$ and $A C D$ on the same base $A C$ .
Area $= \frac{1}{2} \times 18 \times (7 + 5)$ .
$= 108$ cm².

Answer: 108 cm²

Example 4: Missing parallel side · 3 marks · Short answerHarder exam level

The area of a trapezium is 180 cm² and its height is 12 cm. One parallel side is 18 cm. Find the other.

Try it first, then show the solutionSolution

$a + b = \frac{2 \times 180}{12} = 30$ cm.
Other side $= 30 - 18 = 12$ cm.
Check: $\frac{1}{2} (18 + 12) \times 12 = 180$ ✓.

Answer: 12 cm

Example 5: A window · 4 marks · Long answerHarder exam level

A window is a rectangle 1.4 m wide and 1 m high with a semicircle on top, whose diameter is the 1.4 m top edge. Find (i) the area of glass (ii) the length of the outer frame ( $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Rectangle area $= 1.4 \times 1 = 1.4$ m². Semicircle: $r = 0.7$ m, area $= \frac{1}{2} \times \frac{22}{7} \times 0.49 = 0.77$ m².
(i) Glass $= 1.4 + 0.77 = 2.17$ m².
Frame: two sides $2 \times 1 = 2$ m, bottom 1.4 m, curved top $\frac{22}{7} \times 0.7 = 2.2$ m.
(ii) Frame $= 2 + 1.4 + 2.2 = 5.6$ m (the top straight edge is inside, so it is not counted).

Answer: (i) 2.17 m² (ii) 5.6 m

Example 6: An L-shaped room · 4 marks · Long answerHarder exam level

An L-shaped floor is a 12 m × 10 m rectangle with a 5 m × 4 m rectangle missing from one corner. Find its area in two different ways.

Try it first, then show the solutionSolution

Method 1 (subtract): $12 \times 10 - 5 \times 4 = 120 - 20$ .
$= 100$ m².
Method 2 (split): a $12 \times 6$ strip plus a $7 \times 4$ strip $= 72 + 28$ .
$= 100$ m² ✓.

Answer: 100 m²

Example 7: A park with semicircular ends · 5 marks · Long answerHOTS / competency

A park is a rectangle 40 m × 28 m with a semicircle on each 28 m side. Find (i) its area (ii) its perimeter (iii) the cost of fencing it at ₹25 per metre ( $π = \frac{22}{7}$ ).

Try it first, then show the solutionSolution

Two semicircles of radius 14 m make one full circle: area $= \frac{22}{7} \times 14 \times 14 = 616$ m².
(i) Area $= 40 \times 28 + 616 = 1120 + 616 = 1736$ m².
Curved edges: one full circumference $= 2 \times \frac{22}{7} \times 14 = 88$ m.
(ii) Perimeter $= 2 \times 40 + 88 = 168$ m (the 28 m sides are covered by the semicircles).
(iii) Cost $= 168 \times 25 = ₹4200$ .

Answer: (i) 1736 m² (ii) 168 m (iii) ₹4,200

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

The school garden

Suppose a school garden bed is a trapezium with parallel sides 16 m and 10 m, which are 6 m apart. A semicircular pond of diameter 7 m is dug inside the bed. The rest of the bed is to be planted with grass costing ₹40 per m². Take $π = \frac{22}{7}$ .

(i) · 3 marks
Find the area of the bed.
Answer
Answer: 78 m²
$\frac{1}{2} (16 + 10) \times 6 = 78$ m².
(ii) · 3 marks
Find the area of the pond.
Answer
Answer: 19.25 m²
$r = 3.5$ m. Area $= \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5 = 19.25$ m².
(iii) · 4 marks
Find the area to be planted with grass and its cost.
Answer
Answer: 58.75 m²; ₹2,350
Grass $= 78 - 19.25 = 58.75$ m². Cost $= 58.75 \times 40 = ₹2350$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the area of a trapezium with parallel sides 12 cm and 8 cm, 7 cm apart. (3 marks)

Step	Marks
Formula $A = \frac{1}{2} (a + b) h$	1
$= \frac{1}{2} (12 + 8) \times 7$	1
$= 70$ cm²	1

Find the area and perimeter of a semicircle of diameter 21 cm ( $π = \frac{22}{7}$ ). (4 marks)

Step	Marks
$r = 10.5$ cm	1
Area $= \frac{1}{2} \times \frac{22}{7} \times 10.5 \times 10.5 = 173.25$ cm²	1
Curved edge $= \frac{22}{7} \times 10.5 = 33$ cm	1
Perimeter $= 33 + 21 = 54$ cm	1

A metal plate is a rectangle 20 cm × 12 cm with a semicircle of diameter 12 cm cut out of one shorter side. Find the area of the plate ( $π = 3.14$ ). (5 marks)

Step	Marks
Rectangle area $= 20 \times 12 = 240$ cm²	1
Semicircle radius $= 6$ cm	1
Semicircle area $= \frac{1}{2} \times 3.14 \times 36 = 56.52$ cm²	1
Area of plate $= 240 - 56.52$	1
$= 183.48$ cm²	1

Where students lose marks in Area of Plane Figures

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

Mistake: Using a slanting side as the height
The height of a trapezium is the perpendicular distance between the parallel sides.
Do this: Look for the right-angle mark or the word 'distance between'.
Usually costs: The answer.
Mistake: Leaving out the diameter in the semicircle's perimeter
The boundary includes the straight edge.
Wrong working
Perimeter $= π r = 22$ cm
Right working
$π r + 2 r = 22 + 14 = 36$ cm
Do this: Ask: what would I walk round?
Usually costs: 1 mark.
Mistake: Using the diameter as the radius
Formulas use $r$ .
Wrong working
$d = 7$ : area $= \frac{1}{2} \times \frac{22}{7} \times 49$
Right working
$r = 3.5$ : area $= 19.25$
Do this: Halve the diameter first.
Usually costs: The whole calculation.
Mistake: Counting inside edges in a perimeter
Edges covered by another part of the shape are not on the boundary.
Do this: Trace the outline with your finger and add only those edges.
Usually costs: 1 to 2 marks.
Mistake: Wrong units
Area is in square units.
Wrong working
Area $= 70$ cm
Right working
Area $= 70$ cm²
Do this: Write cm² or m² for every area.
Usually costs: 1 mark.

Practice questions on Area of Plane Figures

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 · MCQExam level
The area of a semicircle of radius 7 cm ( $π = \frac{22}{7}$ ) is:
(a) 154 cm²
(b) 77 cm²
(c) 22 cm²
(d) 36 cm²
Hint
Half of πr².
Answer
(b)
Q2 · 2 marks · Short answerExam level
Find the area of a trapezium with parallel sides 15 cm and 9 cm and height 8 cm.
Hint
½(a + b)h.
Answer
$\frac{1}{2} \times 24 \times 8 = 96$ cm².
Q3 · 2 marks · Short answerExam level
Find the area of a rhombus whose diagonals are 10 cm and 24 cm.
Hint
½d₁d₂.
Answer
120 cm².
Q4 · 2 marks · Short answerHarder exam level
Find the perimeter of a semicircle of diameter 35 cm ( $π = \frac{22}{7}$ ).
Hint
$r = 17.5$ .
Answer
$\frac{22}{7} \times 17.5 + 35 = 55 + 35 = 90$ cm.
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): The area of a trapezium equals its height multiplied by the average of its parallel sides.
Reason (R): A diagonal divides a trapezium into two triangles with the same height.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
How is the formula derived?
Answer
(c)
Q6 · 3 marks · Short answerHarder exam level
A trapezium has area 96 cm² and height 8 cm. Its parallel sides are in the ratio $1 : 3$ . Find them.
Hint
Sum $= \frac{2 A}{h}$ .
Answer
Sum $= 24$ cm, so the sides are 6 cm and 18 cm.
Q7 · 2 marks · Short answerHarder exam level
In quadrilateral $P QR S$ , diagonal $P R = 24$ cm and the perpendiculars from $Q$ and $S$ to $P R$ are 9.5 cm and 6.5 cm. Find its area.
Hint
½d(h₁ + h₂).
Answer
$\frac{1}{2} \times 24 \times 16 = 192$ cm².
Q8 · 3 marks · Short answerHarder exam level
A shape is a square of side 14 cm with a semicircle drawn outwards on one side. Find its area ( $π = \frac{22}{7}$ ).
Hint
Semicircle radius 7 cm.
Answer
$196 + 77 = 273$ cm².
Q9 · 3 marks · Short answerHarder exam level
A house-shaped pentagon is a rectangle 8 m wide and 5 m high with a triangular roof on the 8 m side, 3 m high. Find its area.
Hint
Rectangle + triangle.
Answer
$40 + \frac{1}{2} \times 8 \times 3 = 40 + 12 = 52$ m².
Q10 · 3 marks · Short answerHarder exam level
The area of a semicircle is 1232 cm². Find its diameter ( $π = \frac{22}{7}$ ).
Hint
$r^{2} = \frac{2 A}{π}$ .
Answer
$r^{2} = 1232 \times 2 \times \frac{7}{22} = 784$ , $r = 28$ cm, diameter 56 cm.
Q11 · 4 marks · Long answerHOTS / competency
Two semicircles of diameter 14 cm are drawn inside a square of side 14 cm, on two opposite sides. Find the area of the square not covered by the semicircles ( $π = \frac{22}{7}$ ).
Hint
Each semicircle is 77 cm².
Answer
$196 - 2 \times 77 = 42$ cm².
Q12 · 4 marks · Long answerHOTS / competency
Trapezium $A B C D$ has $A B ∥ D C$ , $A B = 20$ cm, $D C = 12$ cm and height 9 cm. Find its area, then find the areas of triangles $A B C$ and $A C D$ and show they add up to the same total.
Hint
Both triangles have height 9 cm.
Answer
Trapezium $= \frac{1}{2} (20 + 12) \times 9 = 144$ cm². $△ A B C = \frac{1}{2} \times 20 \times 9 = 90$ ; $△ A C D = \frac{1}{2} \times 12 \times 9 = 54$ ; $90 + 54 = 144$ ✓.

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.Formula for the area of a trapezium?Show answer
½(a + b)h.
2.Area of a semicircle?Show answer
½πr².
3.Perimeter of a semicircle?Show answer
πr + 2r.
4.Area of a rhombus with diagonals 10 cm and 8 cm?Show answer
40 cm².
5.Two semicircles on the same diameter make what?Show answer
A full circle.
6.Which do you use for fencing: area or perimeter?Show answer
Perimeter.
7.Area of a semicircle of radius 7 cm (π = 22/7)?Show answer
77 cm².
8.Height of a trapezium of area 50 cm² with parallel sides 6 cm and 4 cm?Show answer
10 cm.

Best order to study it

Revise areas of rectangles, triangles, parallelograms and circles.
Derive and use the trapezium formula, including working backwards.
Find areas of polygons by splitting, and the rhombus formula.
Learn the area and perimeter of a semicircle.
Finish with combined shapes and costing problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Is your trapezium height perpendicular to the parallel sides?
Did you halve the diameter to get r?
Did you include 2r in the semicircle perimeter?
Did you split the polygon clearly and label each part?
Did you count only outside edges for a perimeter?
Did you use the value of π given in the question?
Are your units correct (cm or cm²)?

Area of Plane Figures: frequently asked questions

Why is the trapezium area ½(a + b)h?

A diagonal splits the trapezium into two triangles with bases a and b and the same height h, so the total is ½ah + ½bh = ½(a + b)h.

Should I use π = 22/7 or 3.14?

Use the value the question gives. If the radius is a multiple of 7, 22/7 is usually intended and makes the arithmetic easy.

Is the perimeter of a semicircle half the circumference?

No. Half the circumference (πr) is only the curved part; you must add the diameter (2r).

How do I find the area of an irregular polygon?

Split it into rectangles, triangles and trapeziums (or draw a diagonal with offsets), find each area, and add them.

Does the rhombus formula work for a square?

Yes. A square is a rhombus, so its area is also ½ × d × d = ½d².

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII