Consolidation Exercise 11C (P.11.

9) DE EF
=
1. (a) They are similar. RP PQ
3 4
(b) They are not similar. =
r 12
(c) They are not similar. 3× 12
=r
(d) They are similar. 4
r= 9
2. A and E, B and D, C and F are three pairs of similar
figures. 7. ∵ △FGH ~ △TUS
∴ ∠H = ∠S
3. (a) ∠P and ∠Z, ∠Q and ∠Y, ∠R and ∠X are three h = 48°
pairs of corresponding angles. ∠U = ∠G
(b) PQ and ZY, QR and YX, PR and ZX are three = 42°
pairs of corresponding sides. In △STU,
48° + t + ∠U = 180°
48° + t + 42° = 180°
t = 90°
4. (a) By considering the corresponding angles and
sides of the two triangles, we have 8. ∵ △LMN ~ △ZYX
△ABC ~ △FED. ∴ LN =
LM
(b) By considering the corresponding angles and ZX ZY
m 10
sides of the two triangles, we have =
△PQR ~ △UST (or △PQR ~ △SUT). 4 5
10 × 4
m=
∵ △ABC ~ △LMN
5
5. =8
∴ ∠N = ∠C MN LM
=
x = 50° YX ZY
∠M = ∠B 14 10
=
z 5
= 60°
In △LMN, 14 × 5
10
=z

x + y + ∠M = 180° z=7
50° + y + 60° = 180°
y = 70° 9. ∵ △BAC ~ △PRQ
∴ ∠Q = ∠C
6. ∵ △DEF ~ △RPQ q = 30°
∴ DF =
RQ PQ
EF AC BC
=
RQ PQ
2 4 b 14
= =
p 12 12 10.5
2 × 12 14 × 12
=p b=
4 10.5
p=6 = 16
10. ∵ △JKL ~ △TSR 6×5
= 3k
∴ KL =
SR TR
JL 10
3 = 3k
x 10 k=1
=
9 18
x=
10 × 9
13. ∵ △PQR ~ △ZYX
=5
18
∴ ∠Q = ∠Y
∠R = ∠L 2r + 5° = 3r − 40°

r = 60° 5° + 40° = 3r − 2r

∠S = ∠K r = 45°

= 90° ∠X = ∠R

In △TSR, =r
= 45°
△XYZ,
r + t + ∠S = 180°
In
60° + t + 90° = 180°
t = 30° ∠X + ∠Y + z = 180°
45° + [3(45°) − 40°] + z = 180°
z = 40°

∵ △EFD ~ △XYZ
∵ △PQR ~ △SRP
11.
∴ ∠Z = ∠D 14.

5x = 75° ∴ QR =
RP PS
RP

x = 15° p 6
ED EF =
= 6 9
XZ XY 6×6
10 15 p=
= 9
y + 4 12 =4
10 × 12
=y+4 ∠PSR = ∠RPQ
15
x = 25°
8=y+4
y=4
15. (a) ∵ △PQR ~ △LMN
12. ∵ △ABC ~ △TSU ∴ ∠R = ∠N
∴ SU
= 40°
∵ △PQR ~ △LMN
BC AB
= (b)
TS
h − 5 10
=
∴ ∠L = ∠P
7 5 = 80°
h−5=
10 × 7
5
(c) In △LMN,
∠L + ∠M + ∠N = 180°
h − 5 = 14
80° + ∠M + 40° = 180°
h = 19
AC AB ∠M = 60°
=
TU TS
6 10 16. (a) ∵ △ABC ~ △PRQ
∴ ∠R = ∠B
=
3k 5
 = 65° 18. ∵ △ABE ~ △ACD
(b) Let RQ = x cm. ∴ CD
BE AB
∵ △ABC ~ △PRQ
=
AC

∴ RQ
7 5
BC BA =
= k 5+2
RP
7×7
12 16 =k
= 5
x 10
k = 9.8
12 × 10
=x
∵ △PQR ~ △PST
16
x = 7.5 19.
∴ RQ = 7.5 cm ∴ PT
PR QR
=
ST
5
=
15
1
=
3
∴ PT = 3PR
PR + RT = 3PR
RT = 2PR
i.e. It is true that RT = 2PR.
17. (a) ∵ △DFE ~ △YZX 20. ∵ △MNO ~ △SPO
∴ ∠F = ∠Z ∴ NO =
PO SO
MO

= 100°
(b) ∵ △DFE ~ △YZX NO 20
9
=
15
∴ ∠E = ∠X NO =
20 × 9
= 35° 15
In △DFE, = 12

∠D + ∠F + ∠E = 180° ∵ △MNO ~ △RQO

∠D + 100° + 35° = 180° ∴ MO =
RO QO
NO

∠D = 45°
20 12
(c) Let EF = x cm. =
SR + 15 9 + 6
4YZ = 3DF 20 × 15
= SR + 15
YZ 3 12
=
DF 4 SR + 15 = 25
∵ △ △YZX
DFE ~ SR = 10
∴ EF
XZ YZ
=
DF
24 3
=
x 4
24 × 4
=x
3
x = 32
∴ EF = 32 cm