MathsICSE Class 8 · Area of Plane Figures

Semicircle, Rhombus and Combined Figures: Area and Perimeter (ICSE Class 8 Maths)

Updated 17 October 2026 · Fact-checked

Semicircle of radius $r$: area $= \frac{1}{2}\pi r^2$ and perimeter $= \pi r + 2r$. Rhombus with diagonals $d_1, d_2$: area $= \frac{1}{2} d_1 d_2$. For combined figures, add or subtract areas of simple parts, and count only the outside boundary for the perimeter.

Understand Semicircle, Rhombus and Combined Figures

The curriculum includes the area of a semicircle and of polygons. This page adds the rhombus and combined shapes.

Semicircle. Half a circle of radius $r$ :

area $= \frac{1}{2} π r^{2}$ ,
curved edge $= π r$ ,
perimeter $= π r + 2 r$ (curved edge plus the diameter).

Use $π = \frac{22}{7}$ when $r$ is a multiple of $7$ ; for $r = 3.5$ use $\frac{22}{7}$ as well.

Rhombus. Its diagonals bisect each other at right angles, so it is four congruent right triangles with legs $\frac{d _{1}}{2}$ and $\frac{d _{2}}{2}$ :
$ $A = 4 \times \frac{1}{2} \times \frac{d _{1}}{2} \times \frac{d _{2}}{2} = \frac{1}{2} d_{1} d_{2} .$ $
The side is $(\frac{d _{1}}{2})^{2} + (\frac{d _{2}}{2})^{2}$ .

Combined figures.

Running track / stadium: a rectangle with a semicircle on each of two opposite sides. Area $=$ rectangle $+ π r^{2}$ (two semicircles); perimeter $=$ two straight sides $+ 2 π r$ .
Window: a rectangle with a semicircle on top. The frame length counts the three sides of the rectangle and the arc, but not the base of the semicircle.
Remaining region: big shape minus small shape.

Perimeter versus area. Perimeter is a length (cm, m), area is square units; the cost of fencing uses perimeter, the cost of painting uses area.

Semicircle, Rhombus and Combined Figures: formulas to remember

Semicircle: $A = \frac{1}{2} π r^{2}, P = π r + 2 r$; Perimeter includes the diameter.
Rhombus: $A = \frac{1}{2} d_{1} d_{2}$; Diagonals perpendicular.
Track (two semicircular ends): $A = l b + π r^{2}, P = 2 l + 2 π r$; $r = \frac{b}{2}$ .

How to solve Semicircle, Rhombus and Combined Figures questions

Split the figure, compute each part, then decide whether to add or subtract and what the boundary is.

1Divide the figure into rectangles, triangles and semicircles.
2Find the radius from the diameter given.
3Compute each area with the right formula.
4Add or subtract the parts for the area.
5For perimeter, trace only the outer boundary.

Semicircle, Rhombus and Combined Figures 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.

Arched window

Suppose a window is a rectangle 1.4 m wide and 2 m high, topped by a semicircle on the 1.4 m side. Glass costs Rs 600 per square metre and the frame costs Rs 250 per metre.

Find the area of the glass and the cost of the glass.

Try it first, then show the solutionSolution

Semicircle radius $0.7$ m: area $= \frac{1}{2} \times \frac{22}{7} \times 0.49 = 0.77$ m $^{2}$ . Rectangle $= 1.4 \times 2 = 2.8$ m $^{2}$ .
Total $= 3.57$ m $^{2}$ ; glass cost $= 3.57 \times 600 = 2142$ .

Answer: 3.57 m $^{2}$ ; Rs 2,142.

Memory hook: Rectangle plus half-circle on top.

Tricks to remember Semicircle, Rhombus and Combined Figures

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

Half of a circle
A semicircle has half the area but its perimeter is more than half: add the diameter.
Use it when: Avoiding the missing-diameter mistake.
Two ends make a circle
Two semicircular ends of a track together form one full circle.
Use it when: Track problems.

Semicircle, Rhombus and Combined Figures: solved examples

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

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

The area of a semicircle of radius 14 cm is:
(a) 308 cm $^{2}$ (b) 616 cm $^{2}$ (c) 154 cm $^{2}$ (d) 44 cm $^{2}$

Try it first, then show the solutionSolution

$\frac{1}{2} \times \frac{22}{7} \times 14 \times 14 = 308$ .

Answer: (a)

Example 2: Perimeter and area of a semicircle · 3 marks · Short answerBoard level

A semicircular plate has diameter 28 cm. Find its area and its perimeter.

Try it first, then show the solutionSolution

Radius $= 14$ cm. Area $= \frac{1}{2} \times \frac{22}{7} \times 196 = 308$ cm $^{2}$ .
Curved edge $= \frac{22}{7} \times 14 = 44$ cm.
Perimeter $= 44 + 28 = 72$ cm.

Answer: Area 308 cm $^{2}$ ; perimeter 72 cm

Example 3: Rhombus · 3 marks · Short answerHarder board level

A rhombus has area 240 cm $^{2}$ and one diagonal 30 cm. Find the other diagonal, the side and the perimeter.

Try it first, then show the solutionSolution

$240 = \frac{1}{2} \times 30 \times d_{2} \Rightarrow d_{2} = 16$ cm.
Half-diagonals $15$ and $8$ ; side $= 225 + 64 = 17$ cm.
Perimeter $= 4 \times 17 = 68$ cm.

Answer: 16 cm; 17 cm; 68 cm

Example 4: Running track · 4 marks · Short answerHarder board level

A track is a rectangle 20 m by 14 m with a semicircle on each of the two 14 m sides. Find the area enclosed and the perimeter.

Try it first, then show the solutionSolution

Rectangle $= 20 \times 14 = 280$ m $^{2}$ . The two semicircles form one circle of radius $7$ : $\frac{22}{7} \times 49 = 154$ m $^{2}$ .
Area $= 280 + 154 = 434$ m $^{2}$ .
Perimeter $= 2 \times 20 + 2 \times \frac{22}{7} \times 7 = 40 + 44 = 84$ m.

Answer: 434 m $^{2}$ ; 84 m

Example 5: Window problem · 4 marks · Case-basedHOTS / competency

A window is a rectangle 1.4 m wide and 2 m high with a semicircle on its top, the diameter of the semicircle being the 1.4 m side. Find the area of the glass, its cost at Rs 600 per m $^{2}$ , the length of the frame (outer boundary) and its cost at Rs 250 per metre.

Try it first, then show the solutionSolution

Radius $0.7$ m; semicircle area $= \frac{1}{2} \times \frac{22}{7} \times 0.49 = 0.77$ ; rectangle $= 2.8$ ; total $3.57$ m $^{2}$ .
Glass cost $= 3.57 \times 600 = 2142$ .
Frame: $2 + 2 + 1.4$ for the three rectangle sides, plus the arc $\frac{22}{7} \times 0.7 = 2.2$ : total $7.6$ m.
Frame cost $= 7.6 \times 250 = 1900$ .

Answer: 3.57 m $^{2}$ ; Rs 2,142; 7.6 m; Rs 1,900

Where students lose marks in Semicircle, Rhombus and Combined Figures

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

Mistake: Leaving out the diameter in the perimeter of a semicircle
The perimeter is the full boundary, including the straight side.
Wrong working
Perimeter of semicircle ( $r = 7$ ) $= 22$ cm.
Right working
$22 + 14 = 36$ cm.
Do this: Trace the boundary with your pencil.
Usually costs: Answer mark.
Mistake: Using the diameter as the radius
$\frac{1}{2} π r^{2}$ needs the radius.
Wrong working
Diameter 28: area $= \frac{1}{2} \times \frac{22}{7} \times 2 8^{2} = 1232$ .
Right working
$r = 14$ , area $= 308$ .
Do this: Write "r = d/2" first.
Usually costs: Answer mark.

Exam tips

Write the radius before substituting.
Include the straight edge in semicircle perimeters, but not an internal join in combined shapes.
Convert all lengths to the same unit before costing.

Practice questions on Semicircle, Rhombus and Combined 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 · MCQBoard level
The perimeter of a semicircle of radius 7 cm is:
(a) 22 cm (b) 29 cm (c) 36 cm (d) 44 cm
Hint
Arc 22 plus diameter 14.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
Find the area of a semicircle of radius 3.5 cm.
Hint
Half of pi r squared.
Answer
$\frac{1}{2} \times \frac{22}{7} \times 12.25 = 19.25$ cm $^{2}$ .
Q3 · 3 marks · Short answerHarder board level
A rhombus has perimeter 52 cm and one diagonal 24 cm. Find the other diagonal and the area.
Hint
Side is 13; half-diagonal 12.
Answer
Side $13$ ; half the other diagonal $= 169 - 144 = 5$ , so $d_{2} = 10$ cm. Area $= \frac{1}{2} (24) (10) = 120$ cm $^{2}$ .
Q4 · 3 marks · Short answerHarder board level
A square of side 14 cm has a semicircle drawn outwards on one of its sides as diameter. Find the area of the whole figure.
Hint
Semicircle radius is 7.
Answer
Square $= 196$ ; semicircle $= \frac{1}{2} \times \frac{22}{7} \times 49 = 77$ . Total $273$ cm $^{2}$ .
Q5 · 4 marks · Case-basedHOTS / competency
A rectangular lawn is 14 m by 7 m. A semicircular flower bed of diameter 14 m occupies one end, touching the 14 m side. Find the area left for grass, the length of fencing needed along the curved edge, and its cost at Rs 80 per metre.
Hint
Radius of the bed is 7.
Answer
Rectangle $98$ ; bed $\frac{1}{2} \times \frac{22}{7} \times 49 = 77$ . Grass $= 21$ m $^{2}$ . Curved edge $= \frac{22}{7} \times 7 = 22$ m. Cost $= 22 \times 80 = 1760$ .

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 semicircle?Show answer
1/2 pi r^2.
2.Perimeter of a semicircle?Show answer
pi r + 2r.
3.Area of a rhombus?Show answer
1/2 d1 d2.
4.Two semicircular ends on a track form?Show answer
One full circle.

Semicircle, Rhombus and Combined Figures: frequently asked questions

Which value of pi should I use?

22/7 when the radius is a multiple of 7; otherwise the question will state the value.

Is the join line between two shapes part of the perimeter?

No; only the outer boundary counts.