Page 66 - Steady Mate Tg1 - Book A

Matematik Tingkatan 1 Bab 9 Poligon Asas

Contoh 3. 4.
x
63°

50° x 120° 75°
x
43°
x = 43° + 63° x = 50° + 50° 120° = x + 75°
= 106°
= 100° 120° – 75° = x
x = 45°

D Selesaikan masalah berikut. TP 4 ARAS S
Solve the following problems.

1. Diberi AEHB dan CEFD ialah garis lurus dan 2. Diberi ABC, AFE dan DFG ialah garis lurus, cari nilai
FE = FH. Cari nilai x dan y. w, x, y dan z.
Given that AEHB and CEFD are straight lines and FE = FH. It is given that ABC, AFE and DFG are straight lines, find
Find the values of x and y. the values of w, x, y and z.
C
D
B 60°
130° F y
A w D
B 48°
y
H 85° 85°
F
60° 105° 48°
G x E z x
A G E
C
Dalam segi tiga ADE / In triangle ADE,
130° = /GEF + 60° w + 90° + 48° = 180°
130° – 60° = /GEF w + 138° = 180°
/GEF = 70° w = 180° – 138°
w = 42°
/FEH = 180° – 105°
= 75° x = 180° – 48° atau / or x = 90° + 42°
= 132° = 132°
x = 180° – 70° – 75°
= 35° y + y = 60° atau / or /ABD = 180° – 60°
2y = 60° = 120°
y = 180° – 75°
= 105° y = 30° y = 180° – 120°
= 30° 2
atau / or
/AED dan /BEC ialah sudut
x + 70° = 105° bertentang bucu. Dalam segi tiga AFG / In triangle AFG
/AED and /BEC are
x = 105° – 70° vertically opposite angles. z + 85° + 48° = 180°
= 35° z + 133° = 180°
z = 180° – 133°
y = /FEA Oleh sebab / Since /FHE = /FEH
= 105° Maka/ Hence /FHB = /FEA = 47°

© Penerbitan Pelangi Sdn. Bhd. 64