It is sqrt{s*(s-a)*(s-b)*(s-c)} where the lengths of the three sides are a, b and c units and s = (a+b+c)/2.
32 Chess Men of the Board
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Suppose the sides are a, b and c units. Calculate s= (a+b+c)/2 Then Area = sqrt[s*(s-a)*(s-b)*(s-c)] square units
Use the Hero's formula: Let s = (a + b + c)/2. Then the area of the triangle equals√[s(s - a)(s - b)(s - c)], where a, b, and c denote the sides of the triangle.
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
The information depends on what information is available. For example, for a triangle in which all three sides are known (a, b and c) then calculate s = (a+b+c)/2 And the area = sqrt[s*(s-a)*(s-b)*(s-c)] If a, b and the angle between then, C, is known then area = 0.5*a*b*sin(C). There are other formulae for other circumstances.
Take a map S from set A to set B, denote S: A ---> B We call A to be our domain, B our codomain. We call, with an small abuse of notation, S(A) our range, that is the set of all maps of elements of A. Or, we call set C the range of S if C = {c | c = S(a) for all a from A} Remark, C is a subset of B. Just for further knowledge, if for all a in A, S(a) is different, or S(a) != S(b) => a != b for all a, b from A (S(a) and S(b) from B), then we say S is one-to-one. It can happen when the "size" of A is smaller or equal than that of B. if the range of S is the same as the codomain. Or for all elements c from B, c = S(a) for some a from A, then we call S to be onto. It can happen when "size" of A is larger or equal to that of B. Further, if S is one-to-one AND onto, it is invertible. I will leave the proof as an exercise. Just two more note: 1. S is linear if it's a map between vector space A, B over field F which also satisfies S(a + b) = S(a) (+) S(b) and S(kb) = k.S(b) where + and x are addition and scalar multiplication from A while (+) and . are for B. 2. S is not necessarily a function.
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120 sq metres. To see how you get this answer, read on: If the sides are a, b and c, then calculate s = 0.5*(a+b+c) Then the area is sqrt[s*(s-a)*(s-b)*(s-c)]
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