Find the area (in square units) of each triangle described.
1. a = 8, c = 16, B = 60°
2. b = 6, c = 4 square root of 3, A = 30°
3. a = 1, b = 1, c = 1
4. a = 49, b = 33, c = 18
First of all, what is the formula to get the area of a triangle?
area= height times base over two
For #1 and #2, you can use the fact that
area = 1/2 bc sinA (use two sides and the angle between them)
#3 is an equilateral triangle, which I'm sure you know how to do
For #4, use the law of cosines to find an angle, say A, then solve as with #1,2
To find the area of each triangle, we can use different formulas depending on the given information. Let's go through each question one by one.
1. In triangle ABC, we are given that side a is 8 units, side c is 16 units, and angle B is 60°.
To find the area, we can use the formula:
Area = (1/2) * a * c * sin(B), where sin(B) is the sine of angle B.
Since we are given all the required values, we can substitute them into the formula:
Area = (1/2) * 8 * 16 * sin(60°)
Using the value of sin(60°) = √3/2, we can calculate the area:
Area = (1/2) * 8 * 16 * (√3/2)
Area = 64 * (√3/2)
Area = 32√3 square units
Therefore, the area of the triangle is 32√3 square units.
2. In triangle ABC, we are given that side b is 6 units, side c is 4√3 units, and angle A is 30°.
To find the area, we can use the formula:
Area = (1/2) * b * c * sin(A), where sin(A) is the sine of angle A.
Since we are given all the required values, we can substitute them into the formula:
Area = (1/2) * 6 * 4√3 * sin(30°)
Using the value of sin(30°) = 1/2, we can calculate the area:
Area = (1/2) * 6 * 4√3 * (1/2)
Area = 12√3 square units
Therefore, the area of the triangle is 12√3 square units.
3. In triangle ABC, we are given that all sides have a length of 1 unit.
To find the area, we can use Heron's formula, which is based on the lengths of the sides:
Area = √(s * (s - a) * (s - b) * (s - c)), where s is the semiperimeter of the triangle and is calculated as (a + b + c)/2.
Since all sides are equal to 1, we have s = (1 + 1 + 1)/2 = 3/2.
Substituting the values into the formula, we get:
Area = √((3/2) * ((3/2) - 1) * ((3/2) - 1) * ((3/2) - 1))
Simplifying the expression inside the square root, we have:
Area = √((3/2) * (1/2) * (1/2) * (1/2))
Area = √(3/16)
Area = √(3)/4
Therefore, the area of the triangle is √(3)/4 square units.
4. In triangle ABC, we are given that side a is 49 units, side b is 33 units, and side c is 18 units.
To find the area, we can again use Heron's formula:
Area = √(s * (s - a) * (s - b) * (s - c)), where s is the semiperimeter of the triangle.
The semiperimeter s is calculated as (a + b + c)/2.
Substituting the given values, we get:
s = (49 + 33 + 18)/2 = 100/2 = 50.
Now we can substitute the values into the formula:
Area = √(50 * (50 - 49) * (50 - 33) * (50 - 18))
Simplifying the expression inside the square root, we have:
Area = √(50 * 1 * 17 * 32)
Area = √(50 * 17 * 32)
Area = √(27200)
Area ≈ 164.97 square units (rounded to two decimal places)
Therefore, the area of the triangle is approximately 164.97 square units.