Geometry 1 Answers - The Grange School Blogs

AQA Level 2 Further mathematics Geometry I
Topic Assessment
1. Find the angles marked with letters in each of the diagrams below.
(i)
(ii)
q
58°
e
b a
31°
O
18°
c
p
d
73°
r
[8]
2. A, B, C and D are points on the circumference of a circle. The line AC bisects angle
BCD.
C
D
F
B
A
E
Prove that BD is parallel to the tangent EF.
[5]
3. Find the perimeter and area of each of these shapes.
(i)
(ii)
4 cm
5 cm
4 cm
6 cm
12 cm
[8]
4. A pyramid has a horizontal square base ABCD of side 10 metres. The vertex, V, is 4
metres vertically above the middle point, M, of the square base.
(i) Calculate the length VA to three significant figures.
[3]
(ii) Calculate the angle VAB in degrees to the nearest degree.
[3]
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AQA FM Geometry I Assessment solutions
5. The sloping rectangular surface of a desk ABCD is such that AB = 100 cm and AB = 70
cm.
The top edge CD is raised 12 cm above the level of the bottom edge.
In the diagram ABEF is a horizontal plane and CE and DF are vertical lines of length 12
cm.
D
70
C
F
12
E
A
100
B
(i) Calculate the lengths AC and CF to three significant figures.
[4]
(ii) Calculate the angle between the plane surface of the desk and the vertical plane
CDFE.
[4]
(iii) Calculate the angle between the line AC and the vertical plane CDFE.
[5]
6. AB is a diameter of the circle.
D lies on AB so that it divides it in the ratio 1 :
x.
The shaded area has a perimeter made of three
semicircles.
Prove that the perimeter of the shaded area is
equal to the circumference of the circle with
diameter AB.
[4]
7. The diagram shows a regular octagon of side 4 cm. The circle passes through all the
vertices of the octagon.
What percentage of the area of the circle is taken up by the octagon?
[6]
Total 50 marks
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AQA FM Geometry I Assessment solutions
Topic Assessment Solutions
1. (i) Angle a corresponds to the 73° angle, so a = 73°.
58°
31°
e
d
31°
b 73°
31°
c
73°
Using angles on a straight line, b = 180° - 73° - 31°
so b = 76°.
c corresponds to 31°, so c = 31°.
d + 31° corresponds to 58°, so d = 27°.
Using angles on a straight line, e + 31° + 27° = 180°
so e = 122°.
[5]
(ii) Since triangle OAB is isosceles (OA and OB are both radii),
p = 180° - 18° - 18° = 144°.
q
A
O
18° 144°
18°
r
B
The angle subtended by AB at the circumference is half the angle
subtended at the centre, so q = 21  144  72 .
Opposite angles in a cyclic quadrilateral add up to 180°,
so r = 180° - 72° = 108°.
[3]
C
2. Let angle BAE = x°
x
D
x
By the alternate segment theorem,
BCA = x and BDA = x.
F
B
Since AC bisects BCD, angle ACD = angle BCA = x.
Angle ACD is subtended by AD, so angle ABD = x,
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x
A
E
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C
AQA FM Geometry I Assessment solutions
xx
and by the alternate segment theorem DAF = x.
x
D
F
Now DAF and ADB are equal, and since they form
alternate angles, BD and EF must be parallel.
B
x
x
x
A
E
[5]
3. (i)
4
c
4
b
5
4
4
12
a
Using Pythagoras on right-hand triangle: a 2  42  5 2
a 2  25  16  9
a 3
b = 12 – 3 – 4 = 5
Using Pythagoras on left-hand triangle: c 2  42  b 2  16  25  41
c  41
Perimeter  4  5  12  41
 27.4 cm (3 s.f.)
Area  21 (4  12)  4  32 cm²
[4]
(ii)Radius = 3 cm
Length of curve  21  2 r
 3
Perimeter  6  3
6 cm
 15.4 (3 s.f)
Area  21  r 2  21   9
 14.1 (3 s.f.)
[4]
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AQA FM Geometry I Assessment solutions
4. (i)
A
B
M
10
D
10
C
AC 2  10 2  10 2  200
AC  200
AM 
1
2
200  50
V
4
A

50
M
VA 2  50  42  66
VA  66  8.12 m (3 s.f.)
[3]
(ii)
V
5
cos  
66
  52 (to nearest degree)
66
A

[3]
5
B
5. (i) AC 2  100 2  70 2  14900
AC  14900  122 cm (3 s.f.)
CF 2  100 2  12 2  10144
CF  10144  101 cm (3 s.f.)
[4]
(ii) Angle BCE (or ADF)
12
cos  
70
  80.1 (1 d.p.)
[4]
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AQA FM Geometry I Assessment solutions
(iii) Angle ACF
A
F
14900

10144
C
10144
14900
  34.4 (1 d.p.)
cos  
[5]
6. Circumference of circle with diameter AB is (x + 1).
Perimeter of shaded region is
( x  1)  x  ( x  1) ( x  1)

 

 ( x  1)
2
2 2
2
2
i.e. the perimeter of the shaded region is equal to the circumference of the circle with
diameter AB.
7.
sin 22.5 
2
r
2
r 
 5.226
sin 22.5
Area of circle = π x 5.2262 = 85.81 cm2
2
tan 22.5 
h
2
h
 4.828
tan 22.5
Area of octagon = 8 x 2 x 4.828 = 77.255 cm 2
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AQA FM Geometry I Assessment solutions
Percentage of area =
77.255
 100  90%
85.81
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