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.
c
The other 2 angles are c 62 and 90 degrees
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.
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
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
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)]
The answer depends on the information that you do have. Suppose you know all the edge lengths: the three sides of the triangle are a, b and c and the length of the prism is d. Let s = (a + b + c)/2 Then the area of the triangular cross section is sqrt[s*(s-a)*(s-b)*(s-c)] square units. So, surface area = 2*sqrt[s*(s-a)*(s-b)*(s-c)] + d*(a+b+c) square units. Volume = sqrt[s*(s-a)*(s-b)*(s-c)]*d cubic units.
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.
If A is the area of a triangle with side length a, b and c and s is (a+b+c)/2 ( known as the semiperimeter of the triangle) Then, Heron's formula tells us that A=square root of [s(s-a)(s-b)(s-c)].
90
One possible form is sqrt{s*(s-a)*(s-b)*(s-c)} square units where the lengths of the three sides of the triangle are a, b and c units and s = (a+b+c)/2.