Q1) Find area of the triangle ABC where AB=c, AC=b and BC=a. When a=b=c=12m
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Area of tiangle = 1/2 base*altitude
= 1/2 *12*12
Area of tiangle = 1/2 base*altitude
= 1/2 *12*12
= 72sq. unit
1. Label all the sides as a, b and c. Where A is 15m, B is 10m & C is 12m (as these are corresponding to their sides.
2. Write down the formula for the cosine rule (because were looking for the angle, CosA will be the subject rather than a^2):
CosA=b^2+c^2-a^2/2bc
2.Substitute your values:
CosA=10^2+12^2-15^2/2*10*12
4. Now, because you will be looking for the angle, you need to enter SHIFT + Cos on your calculator in order for it find the angle.
5. Your final answer should be 85.5 degrees (Rounded to 1 d.p.)
To find the area of triangle ABC, we can use Heron's formula. Heron's formula states that for a triangle with side lengths a, b, and c, the area (A) can be calculated as:
A = sqrt(s * (s - a) * (s - b) * (s - c))
where s is the semiperimeter of the triangle, given by:
s = (a + b + c) / 2
In this case, we have a triangle ABC with sides AB = c, AC = b, and BC = a, where a = b = c = 12m. Substituting these values, we have:
s = (12m + 12m + 12m) / 2
= 18m
Now we can use Heron's formula to find the area:
A = sqrt(18m * (18m - 12m) * (18m - 12m) * (18m - 12m))
= sqrt(18m * 6m * 6m * 6m)
= sqrt(648m^4)
= 3 * 6m^2
= 18m^2
Therefore, the area of triangle ABC is 18m^2.