Oblique Triangles Heron’s

1. Area of Oblique Triangles
Lecture

2. The Area of a Triangle
•In ∆𝐴𝐵𝐶, let 𝐵𝐶 be the
base of the triangle and ℎ
be the altitude to this
base.
•If 𝐾 is the area of the
triangle, then 𝐾 =
1
2
𝑐ℎ.
𝛾
𝐴
𝐶
𝐵
𝛼 𝛽
𝑏
𝑎
𝑐
ℎ

3. The Area of a Triangle
•Since ℎ = 𝑏 sin 𝛼, we have
•𝐾 =
1
2
𝑐ℎ =
1
2
𝑐 𝑏 sin 𝛼 or 𝐾 =
1
2
𝑏𝑐 sin 𝛼
•If ℎ = 𝑎 sin 𝛽, then 𝐾 =
1
2
𝑎𝑐 sin 𝛽.
•If the base is 𝐴𝐶, then
•𝐾 =
1
2
𝑎𝑏 sin 𝛾
𝛾
𝐴
𝐶
𝐵
𝛼 𝛽
𝑏
𝑎
𝑐
ℎ

4. Theorem 1
•The measure of the area of a triangle is one-half the
product of the measures of two sides and the sine
of the angle included between the two sides.
•𝐾 =
1
2
𝑏𝑐 sin 𝛼 or
•𝐾 =
1
2
𝑎𝑐 sin 𝛽 or
•𝐾 =
1
2
𝑎𝑏 sin 𝛾
𝛾
𝐴
𝐶
𝐵
𝛼 𝛽
𝑏
𝑎
𝑐

5. Illustration
•In ∆𝐴𝐵𝐶 if 𝛼 = 30°, 𝑏 = 24, 𝑐 = 26, then
•𝐾 =
1
2
26 24 sin 30°
•= 13 24 ∙
1
2
= 156
•Thus area of ∆𝐴𝐵𝐶 is equal
to 156 square units.
𝛾
𝐴
𝐶
𝐵
𝛼 = 30° 𝛽
𝑏 = 24
𝑎
𝑐 = 26

6. Example 1
•Find the area of the triangle whose two sides
measure 56 meters and 93 meters and whose
included angle measures 75°.
•Answer: 2515.3 square meters

7. Solution to Example 1
•Let the 𝑏 and 𝑐 be the measures of the sides and
𝛼 be the measure of the included angle.
• If 𝐾 is the area of the triangle, then
•𝐾 =
1
2
𝑏𝑐 sin 𝛼
•𝐾 =
1
2
56 93 sin 75°
•𝐾 = 2515.3
𝛾
𝐴
𝐶
𝐵
𝛼 = 75.3° 𝛽
𝑏 = 56
𝑎
𝑐 = 93

8. Example 2
•A parallelogram has sides of lengths 16 and 24,
and the angle between two consecutive sides is
45°. Determine the area of the parallelogram.
•Answer: 192 2 square units

9. Solution to Example 2
•Let ℎ be the length of the altitude of the
parallelogram with base of length 𝑏.
•The area, 𝐴, of a parallelogram is equal to the
product of its base and altitude.
•𝐴 = 𝑏ℎ ℎ = 16 sin 45°
•𝐴 = 24 16 sin 45°
•𝐴 = 24 16 ∙
2
2
= 192 2 45°
16
𝑏 = 24
ℎ

10. Theorem 2: Heron’s Area Formula
• The area of a triangle whose sides have lengths 𝑎, 𝑏, and 𝑐 and with
semi-perimeter 𝑠 =
𝑎+𝑏+𝑐
2
has area 𝐾 = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐 .
𝐴 𝐵
𝐶
𝑐
𝑎
𝑏

11. Heron’s Formula, Derived
• From the Law of Cosines,
• 𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐 cos 𝛼
• 2𝑏𝑐 cos 𝛼 = 𝑏2 + 𝑐2 − 𝑎2
• Note that 𝐾 =
1
2
𝑐ℎ and ℎ = 𝑏 sin 𝛼
• 𝐾 =
1
2
𝑐 𝑏 sin 𝛼
• 𝐾2
=
1
4
𝑏2
𝑐2
𝑠𝑖𝑛2
𝛼
• 𝐾2 =
1
4
𝑏2𝑐2 1 − 𝑐𝑜𝑠2𝛼
• 𝐾2 =
1
16
2𝑏𝑐 2𝑏𝑐 1 − cos 𝛼 1 + cos 𝛼
• 𝐾2 =
1
16
2𝑏𝑐 − 2𝑏𝑐 cos 𝛼 2𝑏𝑐 + 2𝑏𝑐 cos 𝛼 𝐴 𝐵
𝐶
𝑐
𝑎
𝑏
ℎ
𝛼

12. Heron’s Formula, Derived
• 𝐾2
=
1
16
2𝑏𝑐 − 2𝑏𝑐 cos 𝛼 2𝑏𝑐 + 2𝑏𝑐 cos 𝛼
• 𝐾2
=
1
16
2𝑏𝑐 − 𝑏2
− 𝑐2
+ 𝑎2
2𝑏𝑐 + 𝑏2
+ 𝑐2
− 𝑎2
• 𝐾2
=
1
16
𝑎2
− 𝑏 − 𝑐 2
𝑏 + 𝑐 2
− 𝑎2
• 𝐾2 =
1
16
𝑎 + 𝑏 − 𝑐 𝑎 − 𝑏 + 𝑐 𝑏 + 𝑐 + 𝑎 𝑏 + 𝑐 − 𝑎
• 𝐾2 =
𝑎+𝑏+𝑐
2
𝑎+𝑏+𝑐
2
− 𝑎
𝑎+𝑏+𝑐
2
− 𝑏
𝑎+𝑏+𝑐
2
− 𝑐
• 𝐾2 = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐
• 𝐾 = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐 𝐴 𝐵
𝐶
𝑐
𝑎
𝑏
ℎ
𝛼

13. Example 3
• Find the area of a triangle with sides of lengths 31, 42, and 53.
• Solution:
• 𝐾 = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐
• 𝑠 =
𝑎+𝑏+𝑐
2
=
31+42+53
2
= 63
• 𝐾 = 63 63 − 31 63 − 42 63 − 53
• 𝐾 = 63 32 21 10
• 𝐾 = 7 ∙ 32 24 3 ∙ 7 2 ∙ 5
• 𝐾 = 25 ∙ 33 ∙ 5 ∙ 72
• 𝐾 = 22 ∙ 3 ∙ 7 2 ∙ 3 ∙ 5 = 84 30 ≈ 460

14. Practice Exercise A
•Find the area of triangle given the following. Do
not use calculators.
1) 𝑎 = 15, 𝑏 = 30, 𝛾 = 60°
2) 𝑏 = 25, 𝑐 = 39, 𝛼 = 45°
3) 𝑎 = 14, 𝑏 = 21, 𝛾 = 120°
4) 𝑎 = 1.5, 𝑐 = 2.4, 𝛽 = 135°

15. Practice Exercise B
•Find the area of triangle given triangle
1) 𝑏 = 34, 𝑐 = 28, 𝛼 = 68°
2) 𝑎 = 4.3, 𝑐 = 3.2, 𝛽 = 56°
3) 𝑎 = 114, 𝑏 = 153, 𝛾 = 112.5°
4) 𝑎 = 40.5, 𝑏 = 45.3, 𝛾 = 75.2°

16. Practice Exercise C
1) Find the area of triangle whose sides have lengths 5.9, 6.7, and
10.3.
2) Find the area of a parallelogram whose two sides have lengths 50
cm and 200 cm, and one diagonal with length 177 cm.