Year 9: Geometry Problems and
Proof
www.drfrostmaths.com
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 18th January 2016
Geometry Problems: Section 1 –
Lengths of Lines
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Starter
Find the length of the following lines.
3
𝐴
𝐵
𝐴
2
2
1
2
𝐴
𝐵
𝐶
𝐷
1
𝐶
𝐵
𝑀
𝐶
2
𝐴𝐶 = 2?
2 tends to result from finding diagonals
of squares.
! To get the diagonal of a square, × 𝟐
To get the side length of a square from
the diagonal, ÷ 𝟐
𝑟
𝐴𝑀 = 3?
3 tends to result from
equilateral triangles (or half an
equilateral triangle).
! To get the height of an
equilateral triangle, half the
side length then × 𝟑
(In terms of 𝑟)
?𝒓
𝐴𝐵 = 𝟏 +
𝐴𝐶 = 𝟏 −? 𝒓
?𝒓
𝐵𝐶 = 𝟐 −
Further Examples
1
2
1
2
1
1
A regular octagon is inscribed
inside a square. Find the fraction
of the length of the square that is
taken up by the octagon.
By forming a 1 × 1 square, we find
the length of the square is 2 + 2.
?
Thus the fraction is
2
2+ 2
2
A circle is inscribed inside a square inside
a circle. What fraction of the outer circle
is occupied by the smaller one?
Let radius of smaller circle be 1. By
Pythagoras radius of outer circle is 𝟐.
𝑨𝒓𝒆𝒂𝒊𝒏𝒏𝒆𝒓 = 𝝅 × 𝟏𝟐 = 𝝅
?
𝑨𝒓𝒆𝒂𝒐𝒖𝒕𝒆𝒓 = 𝝅 ×
So it’s half.
𝟐
𝟐
= 𝟐𝝅
Test Your Understanding
Easier
[IMC 2006 Q14] A 3 × 8 rectangle is cut
into two pieces along the dotted line
shown. The two pieces are then
rearranged to form a right-angled triangle.
What is the perimeter of the triangle
formed?
Solution: 24
?
Harder
1
3
3
1
[Kangaroo Pink 2012 Q11] Six identical circles
fit together tightly in a rectangle of width 6 cm
as shown. What is the height, in cm, of the
rectangle? Solution: 𝟐 𝟑 + 𝟐
?
Touching Circles
When you have touching circles/semicircles or a circle in contact with other shapes.
a. Connect the centres of any touching circles.
b. Draw a line from the centre of any circle to any points of contact with other shapes.
c. Try to form right-angled triangles from any diagonals (e.g. by adding
vertical/horizontal lines).
1?
2−
?𝑟
𝑟+
?1
[Based on SMC 2014 Q19]
The diagram shows a quarter-circle of
radius 2, and two touching semicircles.
The larger semicircle has radius 1.
Can you form a right-angle triangle and
find expressions for all three sides?
(You may wish to make the radius of the
smaller circle 𝑟)
?
Identify triangle
Test Your Understanding
Easier
1
2𝑥
𝑥
Suppose the radius of the circle is 1. We wish to find
the length of each square. Suppose we called it 𝑥.
Form an appropriate right-angled triangle (using the
tips on the previous slide) and find expressions for the
three sides.
? Reveal
Harder
(We’d then subsequently use
Pythagoras to give us an equation and
solve, but we’ll do this later)
[IMC 2005 Q21] Two circles with radii 1cm and 4cm touch. The point 𝑃 is on
the smaller circle, 𝑄 is on the larger circle and 𝑃𝑄 is a tangent to both circles.
What is the length of 𝑃𝑄?
A 17cm
B 3cm
C 2 3 cm
D 3 2cm
E 4cm
1cm
3cm
1cm
5cm
Using Pythagoras, 𝑷𝑸
?
= 𝟓𝟐 − 𝟑𝟐 = 𝟒
One final example…
2
3
30°
1
An equilateral triangle is
inscribed inside a circle of
radius 2. Determine the
perimeter of the triangle.
As per usual advice, draw a line
from centre of the circle to a point
of contact…
If we add this line, can we notice
anything special about this triangle
formed?
Perimeter = 𝟔 𝟑
(Side note: the above diagram also tells us
that the centre of an equilateral triangle is a
third of the way up the height – very useful!)
If you ever see a 30-60-90 triangle, you should
recognise it as half an equilateral triangle.
This allows us to use Pythagoras to find the
remaining side.
Exercise 1
1
[Kangaroo Pink 2008 Q3] Four unit squares
are placed edge to edge as shown. What is
the length of the line 𝑃𝑄?
A 5
B 13 C 5 + 2
D 5 E 13
Solution: B
?
2
Two identical squares are inscribed in a
semicircle of radius 2. Find the length of each
square.
Solution: 𝟏 ?
3
A square of length 2 is inscribed in a semicircle.
Find the radius of the semicircle.
Solution: 𝟓?
4
[Kangaroo Pink 2011 Q6] The diagram shows a shape
made from a regular hexagon of side one unit, six
triangles and six squares. What is the perimeter of the
shape?
1
A 6(1 + 2)
B 6 1+2 3
D 6+3 2
E 9
C 12
Solution: ?
12
5
1
2
Exercise 1
5
[IMC 2009 Q7] Four touching circles all have radius 1 and their centres
are at the corners of a square. What is the radius of the circle through
the point of contact 𝑋, 𝑌, 𝑍 and 𝑇?
1
1
A 2
B 2 2
C 1
D 2
E 2
?
Exercise 1
6
[Based on IMC 2015 Q24] In square 𝑅𝑆𝑇𝑈 a quarter-circle arc
with centre 𝑆 is drawn from 𝑇 to 𝑅. A point 𝑃 on this arc is 1
unit from 𝑇𝑈 and 8 units from 𝑅𝑈. Suppose we wanted to find
the the length of the side of square 𝑅𝑆𝑇𝑈.
Let the length of the square be 𝑥.
Draw a line in your diagram horizontally right from the point 𝑃
to form a right-angled triangle. Find the lengths of the three
sides of your triangle in terms of 𝑥 (you do not at this stage
need to find the value of 𝑥).
𝑥−8
? 𝑥
𝑥−1
Exercise 1
7
[Kangaroo Pink 2015 Q9] In the grid, each
small square has side of length 1. What is the
minimum distance from ‘Start’ to ‘Finish’
travelling only on edges or diagonals of the
squares?
A 2 2
B 10 + 2
C 2+2 2
D 4 2
E 6
Solution: C
?
8
[IMC 2011 Q19] Harrogate is 23km due north of Leeds, York is 30km due east of
Harrogate, Doncaster is 48km due south of York, and Manchester is 70km due west
of Doncaster. To the nearest kilometre, how far is it from Leeds to Manchester, as the
crow flies?
A 38km B 47km C 56km D 65km E 74km
Solution: B (Manchester is 40km West and 25km south of Leeds. Use Pythagoras)
?
Exercise 1
9
[IMC 2009 Q20] A square, of side two units, is folded in half to form a triangle. A
second fold is made, parallel to the first, so that the apex of this triangle folds
onto a point on its base, thereby forming an isosceles trapezium. What is the
perimeter of this trapezium?
A 4+ 2
B 4+2 2
C 3+2 2
D 2+3 2
Solution: D
E 5
1
2 2
?
2
1
Exercise 1
10 [Kangaroo Pink 2010 Q7] The diagram shows a square
𝑃𝑄𝑅𝑆 and two equilateral triangles 𝑅𝑆𝑈 and 𝑃𝑆𝑇. 𝑃𝑄 has
length 1. What is the length of 𝑇𝑈?
A
2
B
3
2
3
C
D
2
1
5−1
30°
60°
60°
E 6−1
Solution: A
?
11 Find the area of a regular octagon of side length 1.
1
1
1
1
2
2
1
Area: 4 triangles + 4 rectangles + 1 square
𝐴=
1
2
1
𝐴=1
𝐴=
1
4
1
2
?
=𝟏+
=𝟐+
𝟒
𝟐
𝟒
𝟐
+𝟏
1
Exercise 1
12
[Based on Hamilton 2006 Q4] A circle is inscribed in a square and a rectangle is
placed inside the square but outside the circle. Two sides of the rectangle lie
along sides of the square and one vertex lies on the circle, as shown. The
rectangle is twice as high as it is wide. We wish to find the ratio of the area of the
square to the area of the rectangle.
a) Using the information, give suitable numeric lengths to the width and height of
the rectangle.
b) As per the usual advice for diagrams involving circles, add a suitable line to your
diagram, and form a right-angle triangle. If the radius of your circle is 𝑟, give
expressions for the three lengths of your triangle.
? Reveal
𝑟−1
𝑟−2
𝑟
1
2
Exercise 1
13
[IMC 2012 Q21] The parallelogram 𝑃𝑄𝑅𝑆 is formed by joining together
four equilateral triangles of side 1 unit, as shown. What is the length of
the diagonal 𝑆𝑄?
𝑃
𝑄
3
2
𝑆
5
2
𝑅
By forming the triangle above, we know the
𝟓
base is 𝟐 and the height of the triangle (using
the ‘splitting method’) ?
is
again gives:
𝟑
.
𝟐
𝟕
Use Pythagoras
Exercise 1
14 [IMC 2004 Q18] In the triangle 𝑃𝑄𝑅, there is a right angle at 𝑄 and angle 𝑄𝑃𝑅 is
60°. The bisector of the angle 𝑄𝑃𝑅 meets 𝑄𝑅 at 𝑆, as shown. What is the ratio
𝑄𝑆: 𝑆𝑅?
A 1: 1
B 1: 2
C 1: 3 − 3
D 1: 3
E 1: 2
30°
By using the information, we can work
out all the angles as shown. Since
𝚫𝑷𝑺𝑹 is isosceles, then 𝑷𝑺 = 𝑷𝑹.
Suppose we made it 2. We can see
𝚫𝑷𝑸𝑺 is half an equilateral triangle,
𝟏
thus 𝑸𝑺 = 𝟐 𝑷𝑺 = 𝟏.
? The answer is therefore E.
30°
2
60° 120°
1
30°
2
Alternatively we could have used the
Angle Bisector Theorem, which would
mean the ratio of 𝑸𝑺: 𝑸𝑹 is the same
as 𝑷𝑸: 𝑷𝑹.
Exercise 1
15
[Cayley 2014 Q4] The square 𝐴𝐵𝐼𝐽 lies inside the regular octagon
𝐴𝐵𝐶𝐷𝐸𝐹𝐺𝐻. The sides of the octagon have length 1. Find length 𝐶𝐽.
𝐵
𝐴
1
2
𝐶
𝐻
3
𝐽
𝐼
?
By Pythagoras,
𝐷
𝐺
𝐹
𝐸
∠𝑱𝑩𝑪 = 𝟗𝟎° because ∠𝑱𝑩𝑰
= 𝟒𝟓° and
∠𝑰𝑩𝑪 = 𝟏𝟑𝟓° − 𝟗𝟎° = 𝟒𝟓°
(as 𝟏𝟑𝟓° is the interior
angle of an octagon).
𝑪𝑱 =
𝟐
𝟐
+ 𝟏𝟐 = 𝟑
Exercise 1
16
[JMO Mentoring May2012 Q4] A triangle has two angles which measure 30° and
105°. The side between these angles has length 2 cm. What is the perimeter of
the triangle?
1
3
90°
?
30°
90°
1
2
45°
60°
2
45°
By drawing and splitting
into two right-angle
triangle, we can see we
have half a square and
half an equilateral
triangle.
Solution: 𝟑 + 𝟐 + 𝟑
Exercise 1
17
In the diagram, the letter 𝑆 is made from two arcs, 𝐾𝐿 and 𝑀𝑁, which are each
five-eighths of the circumference of a circle of radius 1, and the line segment
𝐿𝑀, which is tangent to both circles. At points 𝐾 and 𝑁, common tangents to the
two circles touch one of the circles. What is the length 𝐿𝑀?
A
3
2
B 3− 2
1
1
1
1
C 2
D
3 2
2
E 1+ 2
Since the angles in the
triangles are 𝟒𝟓°, we can see
that the lines connecting 𝑳
and 𝑴 to the centre of the
?
diagram are both just 1 (as
the triangles are isosceles).
So the answer is just 2.
Exercise 1
18
[IMC 1999 Q25] A rectangular sheet of paper with sides 1 and 2 has been
folded once as shown, so that one corner just meets the opposite long edge.
What is the value of the length 𝑑?
1
7
A 2
B 2−1
C 16
D
3− 2
E
2
3
𝐴
𝐸
2
?
𝐷
𝐶
𝐵
Since the paper is folded, 𝑨𝑪
= 𝑩𝑫 = 𝟐
By Pythagoras, 𝑩𝑪 = 𝟏.
Thus ∠𝑨𝑪𝑩 = 𝟒𝟓°.
Consequently, as ∠𝑬𝑪𝑫 = 𝟒𝟓°,
𝑪𝑫𝑬 is also an isosceles rightangled triangle and so
𝑬𝑫 = 𝑫𝑪 = 𝟐 − 𝟏
Exercise 1
19
[Kangaroo Pink 2008 Q24] In the diagram, 𝐾𝐿𝑀𝑁 is a unit square. Arcs of radius
one unit are drawn using each of the four corners of the squares as centres. The
arcs centred at 𝐾 and 𝐿 intersect at 𝑄; the arcs centred at 𝑀 and 𝑁 intersect at 𝑃.
What is the length of 𝑃𝑄?
A 2− 2
1
B
3
4
C
5− 2
D
3
3
E
3−1
𝚫𝑴𝑷𝑵 is an equilateral triangle of side 1. Its
𝟏
height is 𝟐 𝟑 (recall that we can halve the
side length then × 𝟑).
?
Therefore, observing that 𝑷𝑸 is the overlap of
two equilateral triangles,
𝟏
𝟏
𝑷𝑸 =
𝟑+
𝟑−𝟏 = 𝟑−𝟏
𝟐
𝟐
Geometry Problems: Section 2 –
Finding lengths using algebra
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
(Note to teacher: Please don’t do this lesson
until you’ve taught solving quadratic equations
to your class)
Introducing variables for unknown lengths
We touched upon how it can be useful to introduce variables for unknown lengths.
We then hope to be able to form an equation so we can solve (usually using Pythagoras).
𝑟−1
𝑟−2
𝑟
1
2
[Hamilton 2006 Q4] A circle is inscribed in a
square and a rectangle is placed inside the
square but outside the circle. Two sides of the
rectangle lie along sides of the square and one
vertex lies on the circle, as shown. The
rectangle is twice as high as it is wide.What is
the ratio of the area of the square to the area
of the rectangle?
We formed the following triangle in the previous
exercise. Now we can use Pythagoras to solve!
𝒓 − 𝟏 𝟐 + 𝒓 − 𝟐 𝟐 = 𝒓𝟐
However 1 is not possible as 𝟏 < 𝟐.
𝒓𝟐 − 𝟐𝒓 + 𝟏 + 𝒓𝟐 − 𝟒𝒓 + 𝟒 = 𝒓𝟐
Thus the area of the rectangle is 2
𝒓𝟐 − 𝟔𝒓 + 𝟓 = 𝟎
? and area of the square 𝟏𝟎 × 𝟏𝟎
𝒓−𝟓 𝒓−𝟏 =𝟎
= 𝟏𝟎𝟎. Ratio is 𝟓𝟎: 𝟏.
𝒓 = 𝟓 𝒐𝒓 𝒓 = 𝟏
Test Your Understanding
Find the radius
of the smaller
circle.
1
1
1
1+𝑟
1−𝑟
?
?
Form right-angled triangle (and find
expressions for lengths)
2+ 1−𝑟 2 = 1+𝑟 2
Use1Pythagoras
to get equation
1 + 1 − 2𝑟 + 𝑟 2 = 1 + 2𝑟 + 𝑟 2
4𝑟 = 1
?
1Solve your equation
𝑟=
4
Using variables to represent ratio
Q A rectangle whose length is twice its height is
inscribed inside a circle of radius 1. Find the
height of the rectangle.
2𝑥
𝑥
1
As per usual advice, connect centre
of circle to point of contact.
What could we make width and
height of rectangle?
Ratio is 2:1, so use 𝟒𝒙 and 𝟐𝒙 (so
that we can easily halve them).
𝒙𝟐 + 𝟐𝒙 𝟐 = 𝟏
𝒙𝟐 + 𝟒𝒙𝟐 = 𝟏
𝟓𝒙𝟐 = 𝟏
𝟏
𝟐
𝒙 = ?
𝟓
𝟏
𝒙=
𝟓
𝟐
So height is
𝟓
Exercise 2
1 [SMC 2014 Q19] The diagram shows a quadrant of radius
2, and two touching semicircles. The larger semicircle
2−𝑟
has radius 1. What is the radius of the smaller
semicircle?
A
𝜋
6
B
3
2
C
1
2
D
1
3
E
1
𝑟+1
2
3
𝟏+ 𝟐−𝒓 𝟐 = 𝒓+𝟏 𝟐
𝟏 + 𝟒 − 𝟒𝒓 + 𝒓𝟐 = 𝒓𝟐 + 𝟐𝒓 + 𝟏
𝟐?
𝟔𝒓 = 𝟒
𝒓=
𝟑
2 [IMC 2015 Q24] In square 𝑅𝑆𝑇𝑈 a quarter-circle arc with
centre 𝑆 is drawn from 𝑇 to 𝑅. A point 𝑃 on this arc is 1
unit from 𝑇𝑈 and 8 units from 𝑅𝑈. What is the length of
the side of square 𝑅𝑆𝑇𝑈?
A 9
B 10
C 11
D 12
E 13
𝒓 − 𝟏 𝟐 + 𝒓 − 𝟖 𝟐 = 𝒓𝟐
𝒓𝟐 − 𝟐𝒓 + 𝟏 + 𝒓𝟐?− 𝟏𝟔𝒓 + 𝟔𝟒 = 𝒓𝟐
𝒓𝟐 − 𝟏𝟖𝒓 + 𝟔𝟓 = 𝟎
𝒓 − 𝟏𝟑 𝒓 − 𝟓 = 𝟎 → 𝒓 = 𝟏𝟑
𝑟−8
𝑟
𝑟−1
Exercise 2
3
[IMC 2011 Q23] A window frame in Salt’s Mill consists of
two equal semicircles and a circle inside a large semicircle
with each touching the other three as shown. The width
of the frame is 4m. What is the radius of the circle in
metres?
A
4
2
3
B
2
2
C
3
4
D 2 2−1
E 1
2−𝑟
1+𝑟
1
𝟏+ 𝟐−𝒓 𝟐 = 𝟏+𝒓 𝟐
𝟏 + 𝟒 − 𝟒𝒓 + 𝒓𝟐 = 𝟏 + 𝟐𝒓 + 𝒓𝟐
?
𝟐
𝒓=
𝟑
[IMC 2010 Q23] The diagram shows a pattern of eight
equal shaded squares inside a circle of area 𝜋 square
units. What is the area (in square units) of the shaded
region?
1
3
2
7
A 13
B 15
C 13
D 19
E 2
𝟏
𝒙𝟐 + 𝟐𝒙 𝟐 = 𝟏 → 𝒙 =
𝟓
𝟐
?𝟏
𝟑
𝑨𝒓𝒆𝒂 = 𝟖
=𝟏
𝟓
𝟓
1
2𝑥
𝑥
Exercise 2
5
[Cayley 2010 Q5] A square sheet of paper ABCD is
folded along FG, as shown, so that the corner B is
folded onto the midpoint M of CD. Prove that the sides
of triangle GCM have lengths in the ratio 3: 4: 5.
Let 𝒂 = 𝑪𝑮 and 𝒃 = 𝑪𝑴. Since 𝑮𝑴 = 𝑮𝑩 due to
folding, 𝑴𝑮 = 𝟐𝒃 − 𝒂
Then:
𝒂𝟐 + 𝒃𝟐 = 𝟐𝒃 − 𝒂 𝟐
𝒂𝟐 + 𝒃𝟐 = 𝟒𝒃𝟐 − 𝟒𝒂𝒃 + 𝒂𝟐
𝟑𝒃𝟐 = 𝟒𝒂𝒃
𝟑𝒃 = 𝟒𝒂 𝒂𝒔 𝒃 ≠ 𝟎
𝟒
𝒃= 𝒂 ?
𝟑
Thus ratio of three sides:
𝒂 ∶ 𝒃 ∶ 𝟐𝒃 − 𝒂
𝟒
𝟖
𝒂∶ 𝒂∶ 𝒂−𝒂
𝟑
𝟑
𝟑𝒂: 𝟒𝒂: 𝟓𝒂
𝟑: 𝟒: 𝟓
𝒃
𝟐𝒃 − 𝒂
𝒂
Exercise 2
6
[Cayley 2008 Q5] A kite has sides AB and AD of
length 25cm and sides CB and CD of length
39cm. The perpendicular distance from B to
AD is 24cm. The perpendicular distance from B
to CD is h cm.
Find the value of h.
Pythagoras on 𝚫𝑩𝑬𝑨:
𝑨𝑬 = 𝟐𝟓𝟐 − 𝟐𝟒𝟐 = 𝟕
𝑫𝑬 = 𝟐𝟓 − 𝟕 = 𝟏𝟖
𝑩𝑫 = 𝟐𝟒𝟐 + 𝟏𝟖𝟐 = 𝟑𝟎
(in both cases you should be able to spot
Pythagorean’s triples and multiples of them)
We could continue to?use Pythagoras to find
𝒉. But a smarter way is to realise the area of
𝚫𝑩𝑫𝑪 can be found in two ways, using either
CD as the base or BD:
The perpendicular height of the triangle using
BD as the base is 𝟑𝟗𝟐 − 𝟏𝟓𝟐 = 𝟏𝟐.
𝟏
𝟏
𝟑𝟔𝟎
Thus 𝟐 × 𝟏𝟐 × 𝟑𝟎 = 𝟐 × 𝟑𝟗 × 𝒉 → 𝒉 = 𝟏𝟑
E
F
Exercise 2
7
[Hamilton 2010 Q5] The diagram shows three
touching circles, whose radii are a, b and c, and
whose centres are at the vertices Q, R and S of a
rectangle QRST. The fourth vertex T of the rectangle
lies on the circle with centre S. Find the ratio 𝑎: 𝑏: 𝑐.
By connecting the centres of the circles we get lengths
𝒂 + 𝒃, 𝒃 + 𝒄, 𝒂 + 𝒄. Using Pythagoras:
𝒂+𝒃 𝟐+ 𝒃+𝒄 𝟐 = 𝒂+𝒄 𝟐
𝒂𝟐 + 𝟐𝒂𝒃 + 𝒃𝟐 + 𝒃𝟐 + 𝟐𝒃𝒄 + 𝒄𝟐 = 𝒂𝟐 + 𝟐𝒂𝒄 + 𝒄𝟐
𝟐𝒃𝟐 + 𝟐𝒂𝒃 + 𝟐𝒃𝒄 = 𝟐𝒂𝒄
𝒃𝟐 + 𝒂𝒃 + 𝒃𝒄 = 𝒂𝒄
But we can also see from the diagram that 𝒂 + 𝒃 = 𝒄,
thus:
𝒃𝟐 + 𝒂𝒃 + 𝒃 𝒂 + 𝒃 = 𝒂 𝒂 + 𝒃
𝒃𝟐 + 𝒂𝒃 + 𝒂𝒃 + 𝒃𝟐 = 𝒂𝟐 + 𝒂𝒃
𝟐𝒃𝟐 + 𝒂𝒃 = 𝒂𝟐
𝟐𝒃𝟐 + 𝒂𝒃 − 𝒂𝟐 = 𝟎
𝟐𝒃 − 𝒂 𝒃 + 𝒂 = 𝟎
𝟏
𝒃= 𝒂
𝟐
𝟑
𝒄=𝒂+𝒃= 𝒂
𝟐
Ratio is: 𝟐: 𝟏: 𝟑
?
Exercise 2
8
[Kangaroo Pink 2004 Q24] The shaded area is equal to 2𝜋. What is the length
of 𝑃𝑄?
A 1
B 2
C 3
D 4
E impossible to determine
Let radii of two circles be 𝒂 and 𝒃. Since the
diameter of the large circle is 𝟐𝒂 + 𝟐𝒃, the radius
must be 𝒂 + 𝒃.
𝑺𝒉𝒂𝒅𝒆𝒅 𝒂𝒓𝒆𝒂 = 𝝅 𝒂 + 𝒃 𝟐 − 𝝅𝒂𝟐 − 𝝅𝒃𝟐
= 𝟐𝝅𝒂𝒃 = 𝟐𝝅
Therefore 𝒂𝒃 = 𝟏
Drawing the indicated triangle
? we know two of the
lengths, and can use Pythagoras to find the third.
𝑷𝑸 = 𝟐 𝒂 + 𝒃 𝟐 − 𝒂 − 𝒃 𝟐
= 𝟐 𝟒𝒂𝒃
= 𝟐 𝟒 𝒂𝒃
=𝟒
𝑎
𝑎+𝑏
𝑎−𝑏
𝑏
Exercise 2
9
[Maclaurin 2011 Q5] Three circles touch the
same straight line and touch each other, as
shown. Prove that the radii a, b, and c, where c
is smallest, satisfy the equation:
1
1
1
+
=
𝑎
𝑐
𝑏
Key here is to form a number of right-angled
triangles. Notice the top length of the rectangle is
the same as the bottom length – this allows us to
express the length in two different ways:
?
𝑎−𝑏
𝑎−𝑐
𝑏−𝑐
𝒂+𝒃 𝟐− 𝒂−𝒃 𝟐
= 𝒂+𝒄 𝟐− 𝒂−𝒄 𝟐+
𝟒𝒂𝒃 = 𝟒𝒂𝒄 + 𝟒𝒃𝒄
𝟐 𝒂 𝒃=𝟐 𝒂 𝒄+𝟐 𝒃 𝒄
Then dividing everything by 𝟐
𝟏
𝟏
=
+
𝒄
𝒃
𝒃+𝒄
𝟐
− 𝒃−𝒄
𝒂 𝒃 𝒄:
𝟏
𝒂
𝟐
Geometry Problems: Section 3 –
Areas and Circles
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Starter
In terms of 𝝅, find the area of the shaded regions.
(You may assume any curved line is an arc of a circle)
1
N
2
1
2
2
𝜋
𝐴𝑟𝑒𝑎 = 1 −?
2
1
1
𝜋 1
𝜋−2
𝐴𝑟𝑒𝑎 = − 𝑜𝑟
4 2 ? 4
𝜋
1
𝐴𝑟𝑒𝑎 = + ?
4
2
Hint: You know how
to find the area of a
right-angled triangle!
Could we perhaps
split the shape up?
Common Shapes You’ll See
These are particularly common circle-related shapes you will see in
Maths Challenge/Olympiad questions.
1
1
1
𝝅
?
𝑨=𝟏−
𝟒
𝝅 𝟏
𝑨= −
𝟒 𝟐
𝝅
𝑨 = −?𝟏
𝟐
You can either think of this as double
the previous shape, or as adding two
quarter circles together and discarding
the square to leave the overlap.
Example
[IMC 2012 Q19] The shaded region shown in the diagram is bounded by four arcs,
each of the same radius as that of the surrounding circle. What fraction of the
surrounding circle is shaded?
4
𝜋
1
A 𝜋−1
B 1−4
C 2
D
1
3
E it depends on the radius of the circle
Let radius of the circle be 1 for
simplicity.
Then we can see a quarter of the
shaded region can be formed by
starting with a 𝟏 × 𝟏 square and
cutting out a quarter
? circle. This
𝝅
has area 𝟏 − 𝟒 .
Area of shaded region = 𝟒 − 𝝅
Fraction shaded =
𝟒−𝝅
𝟒
𝝅
=𝟏−𝟒
Exercise 3
1
[JMO 2015 A5] Two circles of radius 1 cm fit
exactly between two parallel lines, as shown in
the diagram. The centres of the circles are 3 cm
apart. What is the area of the shaded region
bounded by the circles and the lines?
Solution: (𝟔 − 𝝅) cm2 (note how the answer is
?
written)
2
[Kangaroo Grey 2005 Q9] In the diagram, the five
circles have the same radii and touch as shown.
The square joins the centres of the four outer
circles. The ratio of the area of the shaded parts
of all five circles to the area of the unshaded
parts of all five circles is
A 5: 4
B 2: 3
C 2: 5
D 1: 4
E 1: 3
Solution: B
?
Exercise 3
3
[IMC 2014 Q20] The diagram shows a regular
pentagon and five circular arcs. The sides of the
pentagon have length 4. The centre of each arc is
a vertex of the pentagon, and the ends of the arc
are the midpoints of the two adjacent edges.
What is the total shaded area?
A 8𝜋
B 10𝜋
C 12𝜋
D 14𝜋
E 16𝜋
Solution: D
?
4
[IMC 2014 Q10] The diagram shows five touching
semicircles, each with radius 2.
What is the length of the perimeter of the
shaded shape?
A 5𝜋
B 6𝜋
C 7𝜋
D 8𝜋
E 9𝜋
Solution: B
?
Exercise 1 Answers
5
[Hamilton 2011 Q4] A square just fits within a circle, which itself just fits within
another square, as shown in the diagram. Find the ratio of the two shaded areas.
Let width of inner square be 2.
Then radius of circle is 𝟐 and length
of outer square 𝟐 𝟐.
2
2
1
Area on left: (cut inner square out of
circle then quarter)
𝝅
𝟐
𝟐 − 𝟒 𝟐𝝅 − 𝟒 𝝅
=
= −𝟏
𝟒
𝟒
𝟐
?
Area bottom-right:
𝟐 𝟐
Ratio:
𝟐
−𝝅
𝟒
𝝅
𝟐
𝟐
𝟐
=
𝝅
𝟖 − 𝟐𝝅
𝝅
=𝟐−
𝟒
𝟐
−𝟏∶𝟐−𝟐
𝒐𝒓 𝝅 − 𝟐 ∶ 𝟒 − 𝝅
Exercise 3
6
[IMC 1997 Q23] The square 𝐴𝐵𝐶𝐷 is inscribed in a
circle of radius 1. Semicircles are drawn with
diameters 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, 𝐷𝐴 as shown, and the parts
of these semicircles which lie outside the original
circle are shaded. What is the total area of these
four shaded regions?
A 3𝜋 + 2
B 2
C 𝜋
𝜋
D 1
E 2
Solution: B (We could start with the area of
everything using the four
semicircles and the
?
square, then cutting out the circle in the middle)
Exercise 3
7
[Cayley 2009 Q2] The boundary of a shaded figure
consists of four semicircular arcs whose radii are
all different. The centre of each arc lies on the line
AB, which is 10cm long.
What is the length of the perimeter of the figure.
Solution: 𝟏𝟎𝝅 cm
?
8
[IMC 2008 Q17] The shaded region is bounded by
eight equal circles with centres at the corners and
midpoints of the sides of a square. The perimeter
of the square has length 8. What is the length of
the perimeter of the shaded region?
A 𝜋
B 2𝜋
C 8𝜋
D 3𝜋
E 4𝜋
Solution: D
?
Exercise 3
9
[Kangaroo Pink 2007 Q18] A coin with diameter 1 cm
rolls around the outside of a regular hexagon with
edges of length 1cm until it returns to its original
position. In centimetres, what is the length of the
path traced out by the centre of the coin?
𝜋
A 6 + 2 B 12 + 𝜋
D 12 + 2𝜋
Solution: C
C 6+𝜋
E 6 + 2𝜋
?
Exercise 1 Answers
10
[IMC 2006 Q21] The diagram shows two semicircular arcs, 𝑃𝑄𝑅𝑆 and
𝑄𝑂𝑅. The diameters, 𝑃𝑆 and 𝑄𝑅, of the two semicircles are parallel; 𝑃𝑆
is of length 4 and is a tangent to semicircular arc 𝑄𝑂𝑅. What is the area
of the shaded region?
A 2𝜋 − 2
B 3𝜋
C 𝜋
D 4
E 2𝜋 − 4
If radius for big semicircle is 2, then
𝟐
radius of smaller one is = 𝟐.
𝟐
The shaded region consists of two
𝝅
2
2
2
= 2
𝟐
𝟐
shapes: a semicircle, of area
𝟐
= 𝝅, and the bit at the top.
The bit at the top is a segment. We can
get the area of a segment by starting
with a quarter circle and cutting out
the triangle.
?
𝝅 × 𝟐𝟐
𝑻𝒐𝒑 𝒃𝒊𝒕 =
−𝟐=𝝅−𝟐
𝟒
So 𝑨𝒓𝒆𝒂 = 𝝅 − 𝟐 + 𝝅 = 𝟐𝝅 − 𝟐
Exercise 3
11
[STMC Regional 2007/08 Q5] The square in the
diagram below has sides of length two units. The
shaded sections are enclosed by 4 semi-circles.
Calculate the exact value of the total area of the
unshaded regions.
Solution: 𝟖 − 𝟐𝝅
(Note that if we added four semicircles and
?
subtracted a square, this would give us the shaded
region)
Exercise 1 Answers
12
[Hamilton 2005 Q2] The region shown shaded in the diagram is bounded by
three touching circles of radius 1 and the tangent to two of the circles.
Calculate the perimeter of the shaded region.
The circles meet at 𝟔𝟎°
angles. So each of the
three central arcs are a
sixth of a circle. We have
an additional two
quarter arcs ?
at the
bottom, making a total
of one full circle!
Perimeter:
𝟐+𝝅
2
Exercise 3
13
[Hamilton 2010 Q4] The diagram shows a quarter-circle
with centre O and two semicircular arcs with diameters
OA and OB. Calculate the ratio of the area of the region
shaded grey to the area of the region shaded black.
Solution: 𝟏: 𝟏
?
14 [Kangaroo Grey 2005 Q18] Two rectangles 𝐴𝐵𝐶𝐷 and
𝐷𝐵𝐸𝐹 are shown in the diagram. What is the area of
the rectangle 𝐷𝐵𝐸𝐹?
A 10cm2
B 12cm2
C 13cm2
D 14cm2
E 16cm2
Solution: B
?
Exercise 3
15
[IMC 2003 Q19] What is the area of the pentagon
shown?
A
C
E
1
𝑎 𝑏−𝑐
2
1
𝑎(𝑏 + 𝑐)
2
1
𝑐 𝑎+𝑏
2
Solution: B
16
B
D
1
𝑏
2
1
𝑏
2
𝑎+𝑐
𝑐−𝑎
?
[IMC 2005 Q18] Three-quarters of the area of the
rectangle has been shaded. What is the value of 𝑥?
A 2
B 2.4
C 3
D 3.6
E 4
Solution: E
?
Exercise 3
17
[Cayley 2015 Q5] The diagram shows a right-angled
triangle and three circles. Each side of the triangle is a
diameter of one of the circles. The shaded region R is the
region inside the two smaller circles but outside the largest
circle. Prove that the area of R is equal to the area of the
triangle.
Let lengths of triangle be 𝟐𝒂, 𝟐𝒃, 𝟐𝒓. As the triangle is
right-angled, 𝒂𝟐 + 𝒃𝟐 = 𝒓𝟐 . The area of the triangle is
𝟐𝒂𝒃.
We can find R by adding the two smaller semicircles and
the triangle in the middle, before subtracting the large
semicircle.
?
𝟏 𝟐 𝟏 𝟐
𝟏 𝟐
𝑨𝒓𝒆𝒂𝑹 = 𝝅𝒂 + 𝝅𝒃 + 𝟐𝒂𝒃 − 𝝅𝒓
𝟐
𝟐
𝟐
𝟏 𝟐 𝟏 𝟐
𝟏
= 𝝅𝒂 + 𝝅𝒃 + 𝟐𝒂𝒃 − 𝝅 𝒂𝟐 + 𝒃𝟐
𝟐
𝟐
𝟐
= 𝟐𝒂𝒃
Thus the areas are the same.
2𝑎
2𝑟
2𝑏
Exercise 3
18 [Kangaroo Grey 2012 Q17] In the diagram, 𝑊𝑋𝑌𝑍 is a
square, 𝑀 is the midpoint of 𝑊𝑍 and 𝑀𝑁 is
perpendicular to 𝑊𝑌. What is the ratio of the area of the
shaded triangle 𝑀𝑁𝑌 to the area of the square?
A 1: 6
B 1: 5
C 7: 36
D 3: 16
E 7: 40
Solution: D
19
?
[IMC 1998 Q24] Each of the sides of this regular octagon
has length 2cm. What is the difference between the area
of the shaded region and the area of the unshaded region
(in cm2)?
A 0
B 1
C 1.5
D 2
E 2 2
Solution: A (we saw how to split up the octagon in
?
Exercise 1)
Geometry Problems: Section 4 –
Cutting and Reforming
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Skill #4: Cutting and Reforming
Frequently, we can chop a shape into bits and put them back together in some
other way, clearly with the same area.
Question: The diagram shows three semicircles, each of radius 1.
What is the size of the total shaded area?
The key is cutting this into shapes we
know the area of. We have a 2x1
rectangle, and the remaining bits can
combine to form a circle of area π.
This avoids us having to calculate the
complicated area in the middle.
A: π + 2
B: 5
D: 4
E: 2𝜋 − 1
C:
3
𝜋
2
+1
Test Your Understanding
The figure shows two shapes that fit
together exactly. Each shape is formed by
four semicircles of radius 1. What is the
total shaded area?
?
We can cut up the shape and
reform it to get a 𝟐 × 𝟒
rectangle with area 8.
Test Your Understanding
The diagram shows a design formed by drawing six lines in a
regular hexagon. The lines divide each edge of the hexagon
into three equal parts. What fraction of the hexagon is shaded?
A
1
5
B
2
9
C
1
4
D
3
10
E
5
16
(Hint: Can you add lines to the
shape such that it’s broken up
into smaller shapes all of the
same area?)
After adding the lines to break
up the shape into small
triangles, we could just count
the total triangles and shaded
triangles.
?
But since the pattern in the
blue region is repeated, we
could just find the fraction
𝟐
shaded in this, which is .
𝟗
Exercise 2 Answers
Q1
The diagram contains six equilateral triangles with sides of
length 2 and a regular hexagon with sides of length 1.
What fraction of the whole shape is shaded?
A
1
8
B
1
7
C
1
6
D
1
5
E
1
4
1
?
5
Exercise 2 Answers
Q2
The diagram shows an equilateral triangle with its corners at the midpoints of alternate sides of a regular hexagon. What fraction of the area of
the hexagon is shaded?
A
1
2
B
1
3
C
3
8
D
4
9
E
7
12
5
?
8
Exercise 2 Answers
Q3
Find the fraction of the shape which is shaded.
?
1
2
Exercise 2 Answers
Q4
The diagram on the right shows a rectangle
with sides of length 5cm and 4cm. All the arcs
are quarter-circles of radius 2cm. What is the
total shaded area in cm2.
A 12 − 2𝜋
D 10
B 8
C 8 + 2𝜋
E 20 − 4𝜋
?
Exercise 2 Answers
Q5
The diagram shows a ceramic design by the Catalan architect Antoni Gaudi. It is
formed by drawing eight lines connecting points which divide the edges of the
outer regular octagon into three equal parts, as shown.
What fraction of the octagon is shaded?
A 1/5
B 2/9
C 1/4
D 3/10
E 5/16
Here we really do
have to focus on
just part of the
shape to keep
things simple.
?
𝟐
𝟗
of the triangles
are shaded. Note
that we ignored
some of the lines.
Exercise 2 Answers
Q6
The diagram shows a square with two lines from a corner to the middle of an
opposite side. The rectangle fits exactly inside these two lines and the square
itself. What fraction of the square is occupied by the shaded rectangle?
A
1
3
B
2
5
C
3
10
D
1
2
E
3
8
We needn’t break up to
non-shaded large
triangles as we know
they make up half the
?
square.
Of the other half, 8
tenths are shaded.
𝟏
𝟖
𝟐
So 𝟐 × 𝟏𝟎 = 𝟓 is shaded.
Geometry Problems: A few
bonus problems…
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Bonus Questions
The radius of the circle is 1.
The arc is formed by a circle
whose centre is the point A.
We wish to find the area shaded.
𝐵
𝐴
1
a Find the length of 𝐴𝐵, i.e. the
radius of the circle centred at 𝐴.
?𝟐
b Hence find the area of segment
area formed by the chord 𝐵𝐶.
𝝅
𝐶
𝟐
𝟒
𝟐
𝝅
−?𝟏 = − 𝟏
𝟐
c Hence find the area of the shaded
region.
𝝅
𝝅
−
−𝟏 =𝟏
𝟐
𝟐 ?
Bonus Questions
Q2
Two unit circles are inscribed
inside a rectangle, such that each
circle passes through the centre of
the other. Determine the shaded
area. (Hint: Think what the angle
of spread of each of the two
segments is)
𝐵
𝐴
𝐶
Since 𝑨𝑩, 𝑩𝑪 and 𝑨𝑪 are all the
radius of the circles, they form an
equilateral triangle. If we split the
shaded region in half and look at
the right half, this segment has an
angle of 𝟏𝟐𝟎°.
Segment area is third of a circle
?
minus indicated triangle (whose
base and height we can find by
Pythagoras).
𝝅
𝟑
−
𝟑
𝟒
𝟐𝝅
𝟑
So area of shaded = 𝟑 − 𝟐
Bonus Questions
QN
Four quarter circles join the corners of a
square with side length 2. Determine the
area of the shaded region.
𝐵
The key is to split the central area into 4.
Each area can be found by starting with
the area of the arc, and subtracting the
two triangles.
Since 𝑨𝑩𝑪 is an equilateral triangle,
∠𝑩𝑨𝑫 = 𝟑𝟎°
𝐷
𝝅
Area of sector: 𝟏𝟐
Area of triangle: 𝟐 ×?𝟏 × 𝟑 − 𝟏
Area of quarter of shaded region:
𝝅
𝟏
−𝟐
𝟑−𝟏
𝟏𝟐
𝟐
𝝅
=
− 𝟑+𝟏
𝟏𝟐
Area of shaded:
𝝅
−𝟒 𝟑+𝟏
𝟑
𝟏
𝐴
𝐶