Geometry Worksheet 1 – Angles
1. [IMC 1998] Determine the angle x.
2. The following shape is a regular decagon (10 sides). Find the angle x. (Hint: use circle
theorems)
3. [Cayley 2006] Rectangles ABCD and AZYX are congruent (i.e. same size and shape).
Determine angle BMX.
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4. [IMC 2007/08] The line RM bisects angle PRQ. What is angle MRN?
A)
180−(𝛼−𝛽)
2
B)
𝛽−𝛼
2
C)
𝛼+2𝛽
2
D)
360−𝛼−2𝛽
2
E)
𝛼+𝛽
2
5. [Cayley 2003/04] This beautiful star pattern is perfectly symmetrical. Find the angle x. (Note,
there’s a great many ways you can do this!)
6. [IMC] The diagram shows a regular pentagon and a regular hexagon which overlap. What is
the value of x?
x°
A) 82
B) 84
C) 85
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D) 87
E) 91
Geometry Worksheet 2 – Lengths and Areas
1. [IMC 1998] What fraction of the following shape is shaded?
2. [IMC] The point O is the centre of a circle of radius 1 unit, OA, OC are radii, and OABC is a
square. What is the area of the shaded region (in square units)?
O
C
π
4
B)
1
π
2
E)
2
A)
1
D)
2
A
π
2
C)
(1  π )
4
π
4
B
3. [IMC 1999] 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 of d?
A)
1
2
B) √2 − 1
C)
7
16
D) √3 − √2
E)
√3
2
4. [IMC]The diagram shows two circles and four equal semi-circular arcs. The area of the inner
shaded circle is 1.
What is the area of the outer circle?
A)
D)
2
π
2
B)
E)
2
9
4
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C)
1+ 2
5. [Macclaurin 2005/06] Two circles are drawn in a rectangle of 6 by 4, such that the larger
circle touches three sides of the rectangle, whereas the smaller one only touches 2.
Determine the radius of the smaller circle.
Some hints: Remember the tip from the slides to add lines for the radii of the circles, in such a way that they
might help us solve the problem (e.g. a radius towards the point where the circles touch will surely help!).
Introduce a variable, say r, to represent the radius of the smaller circle. This led me to the following diagram
below (you should label it with the lengths involved): by forming a triangle, we can now exploit Pythagoras
Theorem to form a quadratic equation involving r, which we can solve.
6. [Hamilton 2008] A regular octagon with sides of length a is inscribed in a square with sides of
length 1, as shown. Prove that a2 + 2a = 1. (Hint: remember the tip about comparing two
identical lengths or areas to form an equation; in this case it’s easier to form one comparing
lengths).
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Geometry Tips
Angles
1. Make sure you can quickly apply the basics, i.e. interior angles of triangle add up to 180°,
interior angles of quadrilateral add up to 360°. The one below is often easy to miss:
2. Exterior angle of a regular polygon with n sides (between an edge and another edge
‘extended out’):
360
𝑛
Therefore interior angle = 180 −
360
𝑛
3. Circle theorems:
4. Look for opportunities to draw a circle in the diagram (e.g. around a regular polygon). This
allows you to take advantage of circle theorems.
Areas and Lengths
1. In many problems you’ll have a right-angled isosceles triangle, i.e. with angles 90°, 45°, 45°.
a. If you know the length of the hypotenuse: Divide by √2 to get the length of the
smaller sides.
b. If you know the length of the shorter sides: Multiply by √2 to get the length of the
hypotenuse
2. Look for opportunities to cut up the area of a shape, to either divide it into shapes you can
calculate the area for, or rearrange the bits to form a new simpler shape.
𝑎
𝑐
3. For two similar triangles: 𝑏 = 𝑑
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4. Introduce variables for unknown lengths (or angles).
5. In general, look out for opportunities to form equations when you have more complicated
expressions involving unknown quantities. For example, this might be because you formed a
right-angled triangle and want to use Pythagoras Theorem.
6. Particularly for problems involving multiple circles or circles inscribed in other shapes, add
extra lines to your diagram, e.g. the radius of your circle, in such a way that it helps you form
an expression or equation.
7. And in general add extra lines to your diagram! These might be lengths that you already
know, ones which you can give some expression for in terms of existing variables (e.g. √2𝑟)
or ones that you don’t yet know – in which case introduce a variable.
𝜃
8. Area of sector: 360 × 𝜋𝑟 2
9. For any problem involving the area of a shape with an arc for one edge and the other edges
straight (e.g. a circle segment), start with a sector of the corresponding circle, then ‘cut out’
any polygonal areas as necessary. For example, in the problem below where we need to
establish the area of the shaded region, we might divide it up into four so each region has
only one curved edge, then subtract areas as follows:
10. When areas are stated to be equal, be methodical in working out the two areas, equating
them, and simplifying.
11. Be careful about using radius vs diameter, and remembering to half your area for semicircles, etc.
12. It’s worthwhile memorising this table:
sin
45°
1
cos
√2
1
tan
√2
1
30°
1
2
√3
2
1
√3
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60°
√3
2
1
2
√3
Geometry Answers – Worksheet 1
1. [IMC 1999] Determine the angle x.
Answer: 108° (by using the fact the triangles are isosceles).
2. The following shape is a regular decagon (10 sides). Find the angle x. (Hint: use circle
theorems)
Answer: As in the slides, inscribe the decagon in a circle, and add lines from the ends of
𝟒
the internal diagonals to the centre of the circle. Then the angle between these radii is 𝟏𝟎
of 360°, i.e. 144°. By our circle theorems, angle x is half of this, i.e. 72°.
3. [Cayley 2005/06] Rectangles ABCD and AZYX are congruent (i.e. same size and shape).
Determine angle BMX.
Answer: Since the rectangles are congruent, angle AXZ = 70°, and since triangle ABD is
right-angled, ABD = 20°. Angle AXZ is the exterior angle of triangle BMX, so is ABD + BMX.
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So 70° = 20° + BMX, so BMX = 50°.
[Frost’s alternative proof]: If AXM is 70°, then since ADMX is a quadrilateral, DMX = 130°.
Then BMZ = 130°. Then we have 100° left split evenly between BMX and DMZ, which is
50°.
4. [IMC 2007/08] The line RM bisects angle PRQ. What is angle MRN?
A)
180−(𝛼−𝛽)
2
B)
𝛽−𝛼
2
C)
𝛼+2𝛽
2
D)
360−𝛼−2𝛽
2
E)
𝛼+𝛽
2
Answer: The sum of the interior angles of a triangle is
𝜶
𝜷
QRM = 𝟗𝟎 − (𝟐 − 𝟐 ). As RN is perpendicular
𝜶
𝜷
𝜷
𝜶
= (𝟗𝟎 − ( − )) − (𝟗𝟎 − 𝜷) = −
𝟐
𝟐
𝟐
𝟐
PRQ = (180 – α – β)°; hence
to PQ, angle NRQ = (𝟗𝟎 − 𝜷)°. So angle MRN
5. [Cayley 2003/04] This beautiful star pattern is perfectly symmetrical. Find the angle x. (Note,
there’s a great many ways you can do this!)
Answer: 20°. Dividing the star up, e.g. into a square and four isosceles triangles, or
drawing a diamond around the star, and then applying basic laws of angles, will do the
trick.
6. [IMC] The diagram shows a regular pentagon and a regular hexagon which overlap. What is
the value of x?
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x°
A) 82
B) 84
C) 85
D) 87
E) 91
Answer: B. Each interior angle of a regular pentagon is 108°, whilst each interior angle of a
regular hexagon is 120°. The non-regular pentagon in the centre of the diagram contains
two angles which are interior angles of the regular hexagon, two angles which are interior
angles of the regular pentagon and a fifth angle, the one marked x°. So x + 2 × 120 + 2 ×
108 = 5 × 108 = 540. Hence x = 84.
Geometry Answers – Worksheet 2
1. [IMC 1998] What fraction of the following shape is shaded?
Answer: Half. Let the radius of the circle be r. This implies that the radius of the semicircle
𝟏
is 2r. The area of the semicircle is therefore 𝟐 × 𝝅 × (𝟐𝒓)𝟐 = 𝟐𝝅𝒓𝟐 which is twice the area
of the circle.
2. [IMC] The point O is the centre of a circle of radius 1 unit, OA, OC are radii, and OABC is a
square. What is the area of the shaded region (in square units)?
O
A
C
B
π
π
(1  π )
B)
C)
1
4
2
4
π
π
D)
E)
2
2
2
4
Answer: B. The area of the square is 1 unit2 and the area of the whole circle is  × 12 units2.
A)
1
𝟏
So, the shaded area is 𝟏 − 𝟒 𝝅 unit2.
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3. [IMC 1999] A rectangular sheet of paper with sides 1 and r2 has been folded once as shown,
so that one corner just meets the opposite long edge. What is the value of the length of d?
1
B) √2 − 1
A) 2
7
C) 16
D) √3 − √2
E)
√3
2
Answer: The fold is made along BE. A folds onto A′.
A′B = AB = 2  A′C = 1 (by Pythagoras’ theorem). ΔA′BC is therefore a right-angled
isosceles triangle
 BA 'C = 45°  EA' D = 45°
 ED = A’D = 2 – 1
4. [IMC]The diagram shows two circles and four equal semi-circular arcs. The area of the inner
shaded circle is 1.
What is the area of the outer circle?
A)
2
π
2
D)
B)
E)
2
9
4
C)
1+ 2
Answer: B. From the symmetry of the figure, the two circles must be concentric. Let their
centre be O. Let the radius of the semicircles be r. Then the radius of the outer circle is 2r
and, by Pythagoras’ Theorem, the radius of the inner shaded circle is
r 2  r 2 , , that is
2r . So the radii of the two circles are in the ratio
ratio of their areas is 1 : 2.
2 , and hence the
r
r
r r
O
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2 : 2, that is 1 :
5. [Maclaurin 2005/06] Two circles are drawn in a rectangle of 6 by 4, such that the larger circle
touches three sides of the rectangle, whereas the smaller one only touches 2. Determine the
radius of the smaller circle.
Answer: By filling in the recommended diagram, we get this:
This gives us a right-angled triangle, so (2-r)2 + (4-r)2 = (2+r)2. Expanding and simplifying,
this gives us r2 – 16r + 16 = 0. Using the quadratic formula, r = 8  4√3. But the smaller
circle can’t have a radius which exceeds the bigger one (with radius 2), so r = 8 – 4√3.
6. [Hamilton 2008] A regular octagon with sides of length a is inscribed in a square with sides of
length 1, as shown. Prove that a2 + 2a = 1.
Frost’s solution: The usual approach – first divide up the octagon into shapes we can find
lengths for, as follows:
Now form an equation comparing the length of the octagon and the length of the square.
i.e. 𝒂 +
𝟐𝒂
√𝟐
= 𝟏. Then making a the subject of the formula, we get: 𝒂 =
always simplify
𝟐
√𝟐
to
𝟐
√𝟐
×
√𝟐
√𝟐
denominator’), which means
. We could
𝟐
√𝟐
𝟐√𝟐
= √𝟐 (this process is known as ‘rationalising the
𝟐
𝟏
= 𝟏+√𝟐 . Now we just have to show this satisfies a2 + 2a
𝟐
𝟐(𝟏+√𝟐)
𝟏
𝟏
𝟑+𝟐√𝟐
𝟑+𝟐√𝟐
=
𝟏
)
𝟏+√𝟐
Substituting in, we get (
𝟏
𝟏+
+𝟐(
)=
𝟏+√𝟐
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𝟐
(𝟏+√𝟐)
+
𝟐
(𝟏+√𝟐)
=
𝟏+𝟐√𝟐+𝟐
=
𝟑+𝟐√𝟐
= 1.
=𝟏