igureMaths
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40 questions at your level
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Level 3
Level 4
Level 5
Simplifying surds
17 questions
Lesson
Not started
Mark as done
150
−
k
=
2
6
\sqrt{150} - \sqrt{k} = 2\sqrt{6}
150
−
k
=
2
6
, where
k
k
k
is an integer. Work out the value of
k
k
k
.
●●●●●
Level 5
3 marks
Start
→
Mark as done
Here is a sphere.
r
Surface area of sphere
=
4
π
r
2
= 4\pi r^2
=
4
π
r
2
5
6
\frac{5}{6}
6
5
of the surface area of this sphere is
160
π
160\pi
160
π
cm²
Find the diameter of the sphere.
Give your answer in the form
a
b
a\sqrt{b}
a
b
where
a
a
a
is an integer and
b
b
b
is a prime number.
(4 marks)
●●●●●
Level 5
4 marks
Start
→
Mark as done
Write
11
5
−
3
−
2
3
\dfrac{11}{5 - \sqrt{3}} - \dfrac{2}{\sqrt{3}}
5
−
3
11
−
3
2
in the form
a
+
b
3
c
\dfrac{a + b\sqrt{3}}{c}
c
a
+
b
3
where
a
a
a
,
b
b
b
and
c
c
c
are integers.
(4 marks)
●●●●●
Level 5
4 marks
Start
→
Mark as done
(a)
Rationalise the denominator of
14
6
\dfrac{14}{\sqrt{6}}
6
14
Give your answer in its simplest form.
(2 marks)
48
−
2
3
+
1
\dfrac{\sqrt{48} - 2}{\sqrt{3} + 1}
3
+
1
48
−
2
can be written in the form
a
+
b
3
a + b\sqrt{3}
a
+
b
3
where
a
a
a
and
b
b
b
are integers.
(b)
Work out the value of
a
a
a
and the value of
b
b
b
.
(4 marks)
●●●●●
Level 5
6 marks
Start
→
Mark as done
Ten identical regular octagons are joined together as shown in the diagram.
a
S
The octagons enclose the shape
S
S
S
.
Each side of each octagon has length
a
a
a
.
Find an expression, in terms of
a
a
a
, for the area of
S
S
S
.
Give your answer in the form
p
(
5
+
2
2
)
a
2
p(5 + 2\sqrt{2})a^2
p
(
5
+
2
2
)
a
2
where
p
p
p
is an integer.
You must show all your working.
(5 marks)
●●●●●
Level 5
5 marks
Start
→
Mark as done
The diagram shows a circle, centre
O
O
O
, radius
r
r
r
cm, and two regular octagons.
r cm
O
The sides of the larger octagon are tangents to the circle.
The vertices of the smaller octagon lie on the circle.
By considering perimeters, show that
4
2
−
2
<
π
<
8
(
2
−
1
)
4\sqrt{2 - \sqrt{2}} < \pi < 8(\sqrt{2} - 1)
4
2
−
2
<
π
<
8
(
2
−
1
)
(4 marks)
●●●●●
Level 5
4 marks
Start
→
Expanding brackets with surds
13 questions
Lesson
Not started
Mark as done
A rectangle has length
(
3
+
5
)
(3 + \sqrt{5})
(
3
+
5
)
cm and width
(
3
−
5
)
(3 - \sqrt{5})
(
3
−
5
)
cm.
(3 + √5) cm
(3 − √5) cm
Diagram NOT accurately drawn
(a)
Work out the area of the rectangle, in cm².
(b)
Work out the perimeter of the rectangle, in cm.
●●●●●
Level 5
4 marks
Start
→
Mark as done
(a)
Rationalise the denominator of
14
6
\dfrac{14}{\sqrt{6}}
6
14
Give your answer in its simplest form.
(2 marks)
48
−
2
3
+
1
\dfrac{\sqrt{48} - 2}{\sqrt{3} + 1}
3
+
1
48
−
2
can be written in the form
a
+
b
3
a + b\sqrt{3}
a
+
b
3
where
a
a
a
and
b
b
b
are integers.
(b)
Work out the value of
a
a
a
and the value of
b
b
b
.
(4 marks)
●●●●●
Level 5
6 marks
Start
→
Rationalising the denominator
16 questions
Lesson
Not started
Mark as done
Write
2
+
3
2
−
3
\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}
2
−
3
2
+
3
in the form
a
+
b
3
a + b\sqrt{3}
a
+
b
3
, where
a
a
a
and
b
b
b
are integers.
●●●●●
Level 5
3 marks
Start
→
Mark as done
Write
1
5
−
2
−
5
\dfrac{1}{\sqrt{5} - 2} - \sqrt{5}
5
−
2
1
−
5
in its simplest form.
●●●●●
Level 5
3 marks
Start
→
Mark as done
Write
5
−
3
1
+
3
\dfrac{5 - \sqrt{3}}{1 + \sqrt{3}}
1
+
3
5
−
3
in the form
a
+
b
3
a + b\sqrt{3}
a
+
b
3
, where
a
a
a
and
b
b
b
are integers.
●●●●●
Level 5
3 marks
Start
→
Mark as done
4
2
+
8
=
k
2
\dfrac{4}{\sqrt{2}} + \sqrt{8} = k\sqrt{2}
2
4
+
8
=
k
2
, where
k
k
k
is an integer.
Work out the value of
k
k
k
.
●●●●●
Level 5
3 marks
Start
→
Mark as done
Write
11
5
−
3
−
2
3
\dfrac{11}{5 - \sqrt{3}} - \dfrac{2}{\sqrt{3}}
5
−
3
11
−
3
2
in the form
a
+
b
3
c
\dfrac{a + b\sqrt{3}}{c}
c
a
+
b
3
where
a
a
a
,
b
b
b
and
c
c
c
are integers.
(4 marks)
●●●●●
Level 5
4 marks
Start
→
Mark as done
Given that
p
p
p
is a prime number,
rationalise the denominator of
4
3
−
p
\dfrac{4}{3 - \sqrt{p}}
3
−
p
4
Give your answer in its simplest form.
(2 marks)
●●●●●
Level 5
2 marks
Start
→
Mark as done
(a)
Rationalise the denominator of
14
6
\dfrac{14}{\sqrt{6}}
6
14
Give your answer in its simplest form.
(2 marks)
48
−
2
3
+
1
\dfrac{\sqrt{48} - 2}{\sqrt{3} + 1}
3
+
1
48
−
2
can be written in the form
a
+
b
3
a + b\sqrt{3}
a
+
b
3
where
a
a
a
and
b
b
b
are integers.
(b)
Work out the value of
a
a
a
and the value of
b
b
b
.
(4 marks)
●●●●●
Level 5
6 marks
Start
→