year 8 maths

645
4 a i E ii H
b i EH
c i ∠G
ii GH
ii ∠E
5 a i F ii I
b i FJ
c i ∠H
6 J and G; D and K; C and I
ii HI
ii ∠J
7 a 32 b 24 c 20 d 8 e 4
8 A and J; C and K; E and G
9 a (A, E ), (B, D), (C, F )
b (A, Y ), (B, X ), (C, W ), (D, Z )
c (A, W ), (B, X ), (C, Z ), (D, Y )
d (A, T ), (B, Z ), (C, X ), (D, S), (E, W )
10 a △ AMC, △ BMC
b Yes, all corresponding sides and angles will be equal.
11 a i △ ABD, △ CBD
ii Yes, all corresponding sides and angles will be equal.
b i △ ABC, △ ACD
12 yes
ii No, sides and angles will not be equal.
13 Equal radii.
14 a reflection in the y-axis then translation by the vector [
1
−2]
b rotation anticlockwise about the origin (0, 0) by 90° then
translation by the vector [
6
3]
c rotation about the origin (0, 0) by 180° then translation
by the vector [
−2
1]
d reflection in the x-axis, reflection in the y-axis and
−2
translation by the vector [ 1]
e reflection in the x-axis, reflection in the y-axis and
3
translation by the vector [ 2]
f rotation about the origin (0, 0) by 180° then translation
3
by the vector [ 2]
Exercise 8E
1 a SSS b RHS c SAS d AAS
2 a △ ABC ≡ △ EFD b △ ABC ≡ △ FED
c △ XYZ ≡ △ UST d △ ABC ≡ △ ADC
3 a SAS b SSS c RHS d AAS
e RHS f AAS
4 a x = 4, y = 1 b x = 9, a = 20
c x = 5, a = 24 d x = 5, a = 30
e x = 4, a = 95, b = 25 f x = 11, a = 50, b = 90
5 a no b yes, SAS c yes, AAS
d no e yes, SAS f yes, AAS
6 a EF = 3 m b ∠B = 30°
c AC = 6 cm d ∠C = 20°
7 a no b yes c yes d no
8 Yes, show SSS using Pythagoras’ theorem.
9 a You can draw two different triangles with SSA.
b You can draw an infinite number of triangles with the
same shape but of different size.
10 a ∠CAB = ∠CED (alternate angles)
∠ACB = ∠ECD (vertically opposite angles)
AC = EC (given equal and corresponding sides)
∴ △ ABC ≡ △ EDC (AAS)
b BD = BD (given and common equal side)
∠ADB = ∠CDB (given and equal angles)
AD = CD (given equal sides)
∴ △ ADB ≡ △ CDB (SAS)
c ∠ACB = ∠CAD (alternate angles)
∠CAB = ∠ACD (alternate angles)
AC = AC (given and common equal side)
∴ △ ABC ≡ △ CDA (AAS)
d ∠ABC = ∠ADC (given 90° angles)
AC = AC (given and common equal side)
BC = DC (given equal sides)
∴ △ ABC ≡ △ ADC (RHS)
11 a AB = AC (given equal sides)
BM = CM (given equal sides)
AM = AM (given and common equal side)
∴ △ ABM ≡ △ ACM (SSS)
∴ ∠AMB = ∠AMC
As ∠AMB + ∠AMC = 180° then
∠AMB = ∠AMC = 90°.
b ∠AEB = ∠CDB (alternate angles)
∠EAB = ∠DCB (alternate angles)
EB = BD (given equal sides)
∴ △ AEB ≡ △ CDB (AAS)
∴ AB = BC and AC = 2AB
c AD = DC (given equal sides)
AB = CB (given equal sides)
BD is a common side
∴ △ ABD ≡ △ CBD (SSS)
∴ ∠DAB = ∠DCB
d △ ACD ≡ △ ACB (SSS)
so ∠DCA = ∠BCA
Now △ DCE ≡ △ BCE (AAS)
with ∠CDE = ∠CBE (isosceles triangle)
So ∠DEC = ∠BEC
Since ∠DEC and ∠BEC are supplementary (sum to 180°)
So ∠DEC = ∠BEC = 90°
So diagonals intersect at right angles.
Exercise 8F
1 A and J; C and K; F and H; I and L
2 a i ∠D ii ∠E iii ∠F
b i AB ii BC iii CA
c i 2 ii 2 iii 2
d Yes, all side ratios are equal and all interior angles are equal.
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
Photocopying is restricted under law and this material must not be transferred to another party.