Transcript Area of Triangle - Tak Sun Secondary School Personal Web Server

HKDSE MATHEMATICS
Ronald Hui
Tak Sun Secondary School
MISSING HOMEWORK
RE2
Kelvin
SHW2-R1
Kelvin, Charles
SHW2-P1
Daniel, Kelvin, Sam L, Charles
22 October 2015
Ronald HUI
MISSING HOMEWORK
SHW3-01
Kelvin, Charles
SHW3-A1
Charles
SHW3-B1
Charles
SHW3-C1
Daniel, Kelvin(RD), Sam L, Charles
22 October 2015
Ronald HUI
MISSING HOMEWORK
SHW3-E1
Kelvin, Charles
SHW3-R1
Daniel, Charles
SHW3-P1
Kelvin, Charles, Isaac, Macro S (RD)
22 October 2015
Ronald HUI
MISSING HOMEWORK
Chapter 4 HW
SHW4-01
SHW4-A1
SHW4-B1
SHW4-C1
SHW4-D1
SHW4-R1
SHW4-P1
Deadline: 26 Nov (Thursday)
22 October 2015
Ronald HUI
Book 5A Chapter 5
Area of a Triangle
In trigonometry, we usually represent the angles and
sides of a triangle, say △ABC, as follows:
b
A
C
c
a
B
(a) The interior angles are denoted by capital letters
A, B and C respectively.
(b) The lengths of the sides opposite to angles A, B
and C are denoted by the corresponding small
letters a, b and c respectively.
In trigonometry, we usually represent the angles and
sides of a triangle, say △ABC, as follows:
b
A
C
c
a
B
Example:
A
62 5 cm
B
7 cm
C
A=
62
a=
7 cm
b=
5 cm
There is a formula to calculate
the area of a triangle when
only two of its sides and their
included angle are known.
For example:
7 cm
140
40
9 cm
9 cm
7 cm
Given two sides and their included angle
Suppose only two sides a and b of △ABC and their included
angle C are given. Consider the following two cases.
Case 1: C is an acute angle. Case 2: C is an obtuse angle.
A
b
h
B
C
D
a
Consider △ADC.
h
sin C 
b
h  b sin C
A
To find the area of
△ABC, we have to
find its height first.
B
a
b
h
C
D
Consider △ADC.
h
◄ ACD = 180 
sin (180  C ) 
b C
h  b sin (180  C )
h  b sin C ◄ sin (180  C)
= sin C
Case 1: C is an acute angle. Case 2: C is an obtuse angle.
A
A
b
h
B
C
D
B
a
b
h
C
D
a
h  b sin C
In both cases, the height of
the triangle with base a
is equal to b sin C.
h  b sin C
Case 1: C is an acute angle. Case 2: C is an obtuse angle.
A
A
b
h
B
B
C
D
a
b
h
C
D
a
h  b sin C
h  b sin C
Thus, we have:
1
Area of △ABC   base  height
2
1
  a  b sinC
2
1
 ab sin C
2
Similarly, for any △ABC, we have:
1
ab sin C
2
1
= bc sin A
2
1
= ac sin B
2
the included angle
of a and b.
Area of △ABC =
A
b
c
B
a
C
the included angle
the included angle
of b valid
and cfor
. right-angled triangles.
Note: The above formula is also
of a and c.
When C = 90,
area of △ABC 
1
1
1
ab sin C  ab sin 90  ab
2
2
2
A
10 cm
B
105
Find the area of △ABC
correct to 1 decimal
place.
6 cm
C
b = 6 cm,
c = 10 cm,
A = 105
Area of △ABC
1
  6  10  sin105 cm2
2
 29.0 cm2 (cor. to 1 d. p.)
Follow-up question
The figure shows △ABC with area 10 cm2.
If AC = 7 cm and BAC = 75, find the
y cm
value of y correct to 3 significant figures.
B
A 75
7 cm
area =
10 cm2
C
1
Area of △ABC   AB  AC  sin A
2
1
10   y  7  sin 75
2
y  2.96 (cor. to 3 sig. fig.)
When only the lengths of the
three sides of a triangle are
known, could we find the
area of the triangle?
Yes, we can use Heron’s formula
to find the area of the triangle.
Heron’s formula
Area of △ABC  s(s  a)(s  b)(s  c) ,
where s 
abc
.
2
s is called the semi-perimeter.
B
A
b
c
a
C
Let me show you how to find
the area of △ABC correct
to 3 significant figures.
A
abc
Let s 
.
2
875
∴
s
cm  10 cm
2
5 cm
7 cm
B
8 cm
Area of △ABC  s(s  a )(s  b )(s  c )
 10  (10 – 8)  (10 – 7)  (10 – 5) cm2
 17.3 cm2 (cor. to 3 sig. fig.)
C
Follow-up question
A
9 cm
There figure shows △ABC, where
B
AB = 9 cm, BC = 12 cm and AC = 13 cm.
(a) Find the area of △ABC.
12 cm
(b) Find the length of BD.
(Give your answer correct to 3 significant figures.)
abc
.
(a) Let s 
2
12  13  9
s
cm  17 cm
∴
2
D
13 cm
C
Area of △ABC  s(s  a)(s  b)(s  c )
 17  (17  12)  (17  13)  (17  9) cm2
 2720 cm2
 52.2 cm2 (cor. to 3 sig. fig.)
Follow-up question
A
9 cm
There figure shows △ABC, where
B
AB = 9 cm, BC = 12 cm and AC = 13 cm.
(a) Find the area of △ABC.
12 cm
(b) Find the length of BD.
(Give your answer correct to 3 significant figures.)
(b) ∵
∴
1
Area of △ABC   AC  BD
2
1
 AC  BD  2720 cm2
2
1
 13  BD  2720 cm
2
BD  8.02 cm
(cor. to 3 sig. fig.)
D
13 cm
C