Geometry Worksheet 2 – Lengths and Areas
1. [Source: IMC] Three equilateral triangles of unit radius are placed side by side. Determine
the distance 𝐴𝐵.
B
A
2. [Source: UKMT Mentoring] ABCD is a square with side length 1. P is the midpoint of CD and
Q is the intersection of AP and BD. Find the area of quadrilateral BCPQ.
3. Find the proportion of the following shapes that are shaded.
a
b
c
d
4. [Source: SMC] The diagram shows a pattern found on a floor tile in the cathedral in Spoleto,
Umbria. A circle of radius 1 surrounds four quarter circles, also of radius 1, which enclose a
square. The pattern has four axes of symmetry. What is the side length of the square?
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5. One large circle and 3 smaller circles are inscribed inside a triangle. If the radius of the larger
circle is 1, determine the radius of the smaller circles.
6. [Source: SMC] The diagram shows two different semicircle inside a square with sides of
length 2. The common centre of the semicircles lies on the diagonal of the square. What is
the total shaded area?
A
𝜋
B
6𝜋(3 − 2√2)
C
𝜋√2
D
3𝜋(2 − √2)
E
8𝜋(2√2 − 3)
7. In the diagram, the circle and the two semicircles have radius 1. What is the
perimeter of the square?
A 6 + 4√2
B 2 + 4√2 + 2√3
D 4 + 2√2 + 2√6
C 3√2 + 4√3
E 12
8. [Source: SMC] [Difficult!] Three circles are contained within two parallel lines. The radii of
the smallest and next biggest circles is 𝑠 and 𝑡 respectively. What is the radius of the big
circle in terms of 𝑠 and 𝑡?
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Geometry Worksheet 2 - ANSWERS
1.
The horizontal distance between the two points is 5/2. Splitting a triangle into two right-angled ones
and using Pythagoras, the height of the triangle is √3/2. Then by forming a triangle using these two
lengths and using Pythagoras, the distance 𝐴𝐵 is √28/2 = √7.
2.
We can split the quadrilateral up into separate shapes:
1
1
4
8
The shaded square clearly has area and the top-right triangle area . The key is finding where the
two diagonal lines intersect. Suppose we imagined the point 𝐷 as the origin, and 𝐷𝐶 the x-axis. Then
the line 𝐴𝑃 has equation 𝑦 = 1 − 2𝑥 and the line 𝐷𝐵 has equation 𝑦 = 𝑥. They intersect when 1 −
1
1
1
1
1
3
2
2
3
6
2𝑥 = 𝑥, so 𝑥 = . Since 𝑃 has an 𝑥 value of , the width of the left shaded triangle is − = and the
1
1
1
1
1
2
6
1
2
1
24
𝟓
4
8
𝟏𝟐
height . So it has an area of × × =
2
The total area therefore is
1
24
+ + =
.
.
The ‘model solutions’ instead used the fact that 𝐷𝑃𝑄 and 𝐴𝐵𝑄 are similar (a justification of why the
angles are the same would be required), with lengths in the ratio 2: 1. This provides an alternative
way of establishing that 𝑄 is a third of the way between lines 𝐷𝐶 and 𝐴𝐵.
3.
A)
Let the radius of the circle be 1. Then we can see using simple trigonometry that
the distance from the centre of the circle to the corners of the triangle is 2. The
two smaller triangles in the diagram combined make an isosceles triangle with
two lengths of 2 and an angle of 120° between them. Thus the area of this is
1
2
1
𝑎𝑏 sin 𝐶 = × 2 × 2 × sin 120 = 2 sin 60 = √3. The big diagram consists of 3
2
of these, so the total area is 3√3. The area of the circle is 𝜋, so the proportion is
𝜋
3√3
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.
B)
We can use a similar approach. Let the radius of the circle be 1. Then a
third of the triangle is one with two sides 1 and an angle of 120°
1
1
√3
2
3√3
2
4
between them. This has the area × 1 × 1 × sin 120 = sin 60 =
that total area of the triangle is
3√3
4𝜋
4
. So
, and the proportion shaded is clearly
.
C)
Let the radius of the larger circle be 1. Then by Pythagoras, the radius of
the smaller circle is ½. The proportion is thus:
D) Let the radius of the big circle be say 1. Then the distance from
the centre of this circle to the corner of the square is √2. Let the
radius of the smaller circle be 𝑟. Then the distance from the centre
of the big circle to the corner can be expressed in terms of 𝑟 as
1 + 𝑟 + 𝑟√2. So if √2 = 1 + 𝑟(1 + √2), then 𝑟 =
√2−1
.
1+√2
Thus the
proportion of the shape shaded is
𝜋𝑟 2
4
2
=
(√2−1)
2
4(1+√2)
(since the square has area 4).
4.
Since 𝑂𝑃 = 1, by Pythagoras, 𝑂𝑄 = √2. Since 𝑄𝑅 is 1, 𝑂𝑅 = √2 − 1. Thus the entire diagonal of the
square will be 2√2 − 2, and hence the side length
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2√2−1
√2
= 2 − √2.
5.
Adding forms a suitable right-angled triangle, and since the large triangle is equilateral, ∠𝐷𝐸𝐹 = 30°.
Let 𝑟 be the radius of the smaller circle.
Using basic trigonometry and the triangle 𝐴𝐵𝐸, we find the length 𝐴𝐸 = 2. But we can also find this
length in terms of 𝑟. Clearly 𝐴𝐸 = 1 + 𝑟 + 𝐷𝐸. Using trigonometry again on the triangle 𝐷𝐸𝐹 (using
the known length 𝐷𝐹), we find that 𝐷𝐸 =
2
√3
𝑟. Thus 2 = 1 + 𝑟 +
2
√3
𝑟, thus rearranging, 𝑟 =
√3
.
√3+2
6.
Let the radius of the smaller semicircle be 𝑎 and the radius of the larger semicircle 𝑏. Then 𝑎 + 𝑏 = 2.
We also know that the length 𝑂𝑋 = 𝑏 = 𝑎√2 (using the “√2 trick”) since 𝑂𝑋 is the diagonal of a
square with length 𝑎. Substituting the second equation into the first, 𝑎 + 𝑎√2 = 2, so 𝑎 =
𝑏=
2√2
1+√2
𝜋
. The total area of the two semicircles is therefore (
6𝜋(3 − 2√2).
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2
2 1+√2
2
𝜋
2√2
2
1+√2
) + (
2
) =
6𝜋
3+2√2
2
1+√2
=
, and
7.
[UKMT model solution] Let the vertices of the square be 𝐴, 𝐵, 𝐶, 𝐷 and the centres of the
circle and two semicircles be 𝑃, 𝑄, 𝑅 as shown. The midpoint of 𝑄𝑅 is 𝑆. By symmetry, 𝑃
and 𝑆 both lie on diagonal 𝐵𝐷 of square 𝐴𝐵𝐶𝐷 and the whole figure is symmetrical
about 𝐵𝐷.
As 𝑃 is distance 1 from both 𝐴𝐷 and 𝐷𝐶, the length of 𝐷𝑃 is √2. As the circles and
semicircles are mutually tangent, 𝑃𝑄𝑅 is an equilateral triangle of side 2, so the length of 𝑃𝑆 is √3. As
angles 𝑄𝐵𝑆 and 𝐵𝑆𝑄 are 45° and 90° respectively, triangle 𝑆𝐵𝑄 is isosceles, so 𝑆𝐵 = 𝑆𝑄 = 1. Hence
the length of 𝐵𝐷 is √2 + √3 + 1. Now the length of the side of the square is 𝐵𝐷 ÷ √2 so the
perimeter of the square is 4 × (𝐵𝐷 ÷ √2), that is 2√2 × 𝐵𝐷.
So the perimeter is 2√2(√2 + √3 + 1), that is 4 + 2√6 + 2√2.
8.
As always with these kinds of problems, the strategy is to add lines to form right-angled triangles,
then use Pythagoras to generate some equations.
Let 𝑥 be the radius of the biggest circle. Then 𝑋𝑌 = 𝑥 + 𝑠, and we can form similar expressions for 𝑋𝐶
and 𝑌𝐶. Drawing a box around the triangle (formed by joining the centres of the circles) creates a
number of handy right-angled triangles.
We can see that 𝐴𝑌 = 𝑥 − 𝑠 and similarly, 𝑌𝐷 = 𝑥 − 𝑡. Using Pythagoras on the triangle 𝐴𝑋𝑌, we
find that 𝐴𝑋 = 2√𝑥𝑠. Similarly, 𝐷𝐶 = 2√𝑥𝑡 by using triangle 𝑌𝐷𝐶. This means that 𝑋𝐵 = 𝐷𝐶 − 𝐴𝑋 =
2√𝑥𝑡 − 2√𝑥𝑠.
We can find 𝐵𝐶 in two different ways. Since 𝐵𝐶 = 𝐴𝐷 = 𝐴𝑌 + 𝑌𝐷 = 2𝑥 − 𝑠 − 𝑡, by using Pythagoras
2
on triangle 𝑋𝐵𝐶 we can form the equation (𝑠 + 𝑡)2 − (2√𝑥𝑡 − 2√𝑥𝑠) = (2𝑥 − 𝑠 − 𝑡)2 . Almost
everything cancels out, leaving 𝑥 = 2√𝑠𝑡.
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