Embed Size (px)
Citation preview
10.3 The Area Of A Triangle
Prepared By,
Asnita Mapasisi
Nurul HanisahHalbani
Sharmiza AdlinaJunaidi
10.3.1 AREA OF A TRIANGLE
Area of ∆ABC = ½ bc sin A
Consider a triangle ABC on the right. By droping a perpendicularline, h from C to meet AB at D, sin A = h/b , h = b sin A
C Area of triangle = ½ x base x height
= ½ x c x h
= ½ c x b sin A
b h a = ½ bc sin A
A D B
c
10.3.1 AREA OF A TRIANGLE
Similarity,area of ∆ABC = ½ ac sin B or ½ ab sin C. Hence,to find
the area of a triangle ABC given two sides and the included angle is
as follows :
Area of ABC = ½ bc sin A
= ½ ac sin B
= ½ ab sin C
10.3.1 AREA OF A TRIANGLE
Example 5
Find the areas of the following triangles.
a )
3cm
A 5 cm B
b ) C
8 cm
A B
32̊
C
55 ˚
10 cm
10.3.1 AREA OF A TRIANGLE
Solution :
a ) Area of ∆ ABC
= ½ bc sin A
= ½ (3)(5) sin 32
= 3.97 cm²
b ) By using the sine rule
b = c b = 40˚ 57’
sin b sin c = 180˚ - 55˚- 40˚57’
8 = 10 = 84˚3’
sin b sin 55˚
sin b = 8 ( sin 55˚ ) ABC = ½ bc sin A
10 = ½ (8)(10) sin 84˚3’
= 39.78 cm²
10.3.2 PROBLEMS INVOLVING THREE
DIMENSIONS
Example 6The diagram on the right shows a pyramid of height 9 cm and
stands on a square base of 12 cm.The vertex V is vertically above
the point D and U is the midpoint of BC.Calculate,
A ) UAV
B ) the area of ∆UAV V
D
C
AB
U12 cm
9 cm
10.3.2 PROBLEMS INVOLVING THREE
DIMENSIONS
Solution :
U V
A B D U
6 cm
12 cm
9 cm
13.416 cm
AU ² = 6² + 12²
= 180
AU = 13.416 cm
DU = AU
= 13.416
UV ² = 9² + 13.416²
= 260.989
UV = 16.155 cm
AV ² = 9² + 12²
= 225
AV = 15 cm
V
9 cm
DA 12 cm
10.3.2 PROBLEMS INVOLVING THREE DIMENSIONS
b ) Area ∆ UAV = ½ (13.416) (15) sin 69˚2’
= 93.958 cm²
V
U
A
16.155 cm
13.416 cm
15 cm
By using the cosine rule,
cos UAV = 13.416² + 15² - 16.155²
2(13.416)(15)
UAV = 69º2’
RUMUS SOH CAH TOA
Sin = T
H
Kos = S
H
Tan = T
S Tentu Tidak Susah
Saya Telah Hafal
Kalau Saya Hafal
PREPARED BY,
Nurul Hanisah Halbani
Asnita Mapasisi
Sharmiza Adlina Junaidi
