year 8 maths

647
10 a ∠A = ∠D (corresponding angles in parallel lines)
∠B = ∠B (common)
b ∠ACB = ∠ECD (vertically opposite)
∠E = ∠A (alternate angles in parallel lines)
c ∠A = ∠A (common)
∠ABC = ∠ADB (given)
d EF = BA (equal sides)
∠E = ∠B (equal interior angles in regular polygons)
DE = CB (equal sides)
11 Using Pythagoras’ theorem AC = 25.
DF
AC = 50
25 = 2
ED
AB = 14
7 = 2
∴ ΔABC ||| ΔDEF
12 a 4 b 2.4 c 16
3
Exercise 8H
1 (Other answers possible.)
a b c d
d 10
3
2 a Opposite sides of a parallelogram are equal.
b Opposite interior angles of a parallelogram are equal.
c Diagonals of a parallelogram bisect each other.
d Opposite sides of a parallelogram are parallel.
3 a All sides equal, opposite sides parallel, diagonals bisect
at right angles, opposite interior angles are equal,
diagonals bisect the interior angles through which
they pass.
b All sides equal, opposite sides parallel, diagonals
bisect at right angles, all angles equal, diagonals equal,
diagonals bisect the interior angles through which
they pass.
4 a AC b BD c DB
5 In △ ABC and △ BCD
∠ABC = ∠BCD (angles in a square are equal)
AB = BC (sides of a square are equal)
BC is common.
∴ △ ABC ≡ △ BCD (SAS)
∴ AC = BD (matching sides in congruent triangles are
equal) and the diagonals of a square are equal in length.
6 a AAS b RHS c SSS
d SAS e AAS f SSS
7 ∠ABE = ∠CDE (alternate angles in parallel lines)
∠BAE = ∠DCE (alternate angles in parallel lines)
AB = CD (given)
△ ABE ≡ △ CDE (AAS)
BE = DE and AE = CE because corresponding sides on
congruent triangles are equal.
8 ∠EFI = ∠GHI (alternate angles in parallel lines)
∠FEI = ∠HGI (alternate angles in parallel lines)
EF = GH (given)
△ EFI ≡ △ GHI (AAS)
EI = GI and FI = HI because corresponding sides on
congruent triangles are equal.
9 a equal (alternate angles in parallel lines)
b equal (alternate angles in parallel lines)
c BD d AAS e They must be equal.
10 a VU = TU, VW = TW, UW is common.
So △VWU ≡ △TWU by SSS.
b ∠VWU = ∠TWU and since they add to 180° they must
be equal and 90°.
11 a SSS (3 equal sides)
b They are equal and add to 180° so each must be 90°.
c Since △QMN is isosceles and ∠MQN is 90° then
∠QMN = 45°.
12 a AB = CB, AD = CD and BD is common.
So △ABD ≡ △CBD by SSS.
b △ABD ≡ △CBD so ∠DAB = ∠DCB
c △ABD ≡ △CBD so ∠ADB = ∠CDB
13 Let AD = BC = a and AB = CD = b.
Then show that both BD and AC are equal to √a 2 + b 2 .
14 ∠ABD = ∠CDB (alternate angles in parallel lines)
∠ADB = ∠CBD (alternate angles in parallel lines)
BD is common
So △ABD ≡ △CDB
So AB = CD and AD = BC
15 a ∠DCE b ∠CDE
c There are no pairs of equal sides.
16 a △ACD is isosceles.
b AD = CD, ∠DAE = ∠DCE and ∠ADE = ∠CDE (AAS)
c ∠AED = ∠CED and sum to 180° so they are both 90°.
17 a First show that △ABD ≡ △CDE by SSS.
So ∠ABD = ∠CDE and ∠ADB = ∠CBD and since these
are alternate angles the opposite sides must be parallel.
b First show that △ABE ≡ △CDE by SAS.
So ∠ABE = ∠CDE and ∠BAE = ∠DCE and since these
are alternate angles the opposite sides must be parallel.
c First prove that △ABD ≡ △BAC by SSS.
Now since ∠DAB = ∠CBA and they are also cointerior
angles in parallel lines then they must be 90°.
Puzzles and challenges
1 3, Reason is AAA for each pair with a right angle and a
common angle.
2 Yes, illustrates Pythagoras’ theorem using areas.
3 31
4 a (3 − r ) + (4 − r ) = 5, so r = 1
b r = 4 − 2√2
5 60
17
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.