MT_F4_7.3 Area of Polygons_1

7.3 Area of Polygons T triangle
quadrilateral

Area of triangle Modulus

1 x1 x2 x3 x1
2 y1 y2 y3 y1

 1 x1y2  x2 y3  x3 y1  x2 y1  x3 y2  x1y3 
2

Area of quadrilateral

1 x1 x2 x3 x4 x1
2 y1 y2 y3 y4 y1

 1  x1 y2  x2 y3  x3 y4  x4 y1   x2 y1  x3 y2  x4 y3  x1y4
2

Example:
1. Calculate the area of triangle ABC with the vertices A (1, 3), B (5, 1) and C (6, 7).

If the coordinates are arranged anticlockwise, y C (6, 7)

Area of ABC  1 1 5 6 1 A (1, 3)
2 317 3

 1 1 35 18 15  6  7 B (5, 1)

2 x

 1 54  28 要排好每一个 coordinates
2
A (1, 3), B (5, 1) , C (6, 7)
 1 26
2 (x1, y1) (x2, y2) (x3, y3)
要重复第一个 coordinates (x1, y1)
 13 units2
• 先向下乘 相减（积数）
• 再向上乘

If the coordinates are arranged clockwise,

Area of ABC  1 1 6 5 1 y
2 3713
A (1, 3) C (6, 7)
 1 7  6 15 18  35 1
B (5, 1)
2
 1 28  54 x

2

 1  26 Take absolute value
2

 1 26 •顺序排每一个 coordinates

2 A (1, 3), C (6, 7) , B (5, 1)
 13 units2 (x1, y1) (x2, y2) (x3, y3)

• 记得要重复第一个 coordinates (x1, y1)

• Area = 1  26 ，就要写成 positive value

2

(absolute value)

Absolute value describes the distance of a number on the number line from
0 without considering which direction from zero the number lies. The
absolute value of a number is never negative.

 The absolute value of 5 is 5.

distance from 0: 5 units
 The absolute value of –5 is 5.

distance from 0: 5 units

 |6| = 6 means the absolute value of 6 is 6.
 |–6| = 6 means the absolute value of –6 is 6.

Therefore, |x| = 6 means : x = 6 or x = –6
|2x – 3| = 5 means : 2x – 3 = 5 or 2x – 3 = –5

2. Calculate the area of quadrilateral PQRS with the vertices P (–2, 13), Q (10, 12),
R (2, 3) and S (–10, 4).

Area of quadrilateral PQRS  1  2 10 2 10  2
2
13 12 3 4 13

 1  24  30  8 130 130  24  30  8

2
 1 116 116

2
 1  232

2

 1 232

2
 116 units2

3. The vertices of a triangle are X (–2, 3), Y (0, p) and Z (–4, –1). Given that the area of
the triangle is 10 units2, find the values of p.

Area of XYZ  10

1 2 0  4  2  10
2 3 p 1 3

1  2 p  0  12 0   4 p 2  10

2

 2 p 12  4 p  2  10 2

2 p 14  20

2 p 14  20 or 2 p 14  20
2 p  34 or 2 p  6
p  17 or p  3

Page 195 (a) 24 units2
Page 196 (b) 12 units2
(c) 28.5 units2

(a) k  11 or k  3
3

(b) k  1 or k  21

(a) 52 units2
(b) 88.5 units2
(c) 19 units2

k  4