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Two sides to a coin.
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What is the formula for the area of a triangle given only the lengths of the sides?
L=sqr((1/2 a+b+c) * (s-a) * (s-b) * (s-c))
If the perimeter of a square is 36 what is the circumference of a circle?
If you mean a circle inscribed in the square: C = circumference, π = pi, r = radius, s = side, P = perimeter C = 2πr r = s/2 C = πs s = P/4 C = πP/4 So for this problem, the circumference is 9π, or about 28.3 If you mean a square inscribed in the circle, the computations are practically the same, except: r = sqrt(2)s/2 C = sqrt(2)πs C = sqrt(2)πP/4 So for this problem, the circumference is sqrt(2)9π, or about 40.0
C program to find area and perimeter of triangle?
include #include#includevoid main(){int a,b,c:float s,A,x=1/2;clrscr();printf(" three sides of a triangle\n");scanf("%d %d %d",&a,&b,&c);s=(a+b+c)/2;A=pow((s(s-a)(s-b))(s-c)),x);printf("AREA OF TRIANGLE:%d\n"'A);gech();}
How do you find the surface and lateral area of a triangular prism?
If the bases have sides of length a, b and c units and the length of the prism is d units thenlateral area = (a+b+c)*darea of base = sqrt{s*(s-a)*(s-b*(s-c)} where s = (a+b+c)/2Then total surface area = lateral area + 2*area of base.
Area of triangle 12Cm 7Cm 10Cm?
90 centimeters* * * * *That is so wrong!Let the three sides be a, b and c and let s = (a+b+c)/2then area = sqrt[s*(s-a)*(s-b)*(s-c)]So here s = 14.5 cm and so area = sqrt(14.5*2.5*7.5*4.5) = sqrt(1223.4375)= 34.98 sq cm (to 2 dp)
What r the dimensions of mutual inductance?
Obviously there is more than one way to do this. VL = Ldi/dt Volts has units of Joules/Coulomb: J/C i has units of Coulombs/second: C/s So di/di is C/s^2 L has units of J/C / C/s^2 = Js^2/C^2 Ic = CdV/dt => Ic/dV/dt = C/s / J/C-s = C/s * C-s/J = C^2/J C has units of C^2/J OR you could just type Q = CV => C = Q/V = C/J/C = C^2/J same answer R = V/I => J/C / C/s = J-s/C^2
What is the area formula for an triangle?
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.
2 S on a c?
2 sides on a coin.
What is the area of a 2 by 8 by 8 and a half cm triangle?
Assuming 2, 8, and 8.5 are the lengths of the sides, the area is 7.929 cm^2. Using Heron's formula, A = sqrt(s * (s-a) * (s-b) * (s-c)) where a, b, and c are the lengths of the sides and s = (a + b + c)/2
Area of scalene triangle?
The answer depends on what information you do have.Suppose you know only the lengths of the sides (a, b and c), then let s = (a + b + c)/2.Then area = sqrt[s*(s - a)*(s - b)*(s - c)]If 2 sides and the included angle, then area = 1/2*a*b*sin(C).There are other formulae.
How is the area of a triangle affected if the height is doubled but the length of the base remains the same?
The area is doubled. a,b - cathetus; c - hypotenuse; h - height; S - area. S = (a*b)/2 = (c*h)/2 obviously if k is the doubled height. and A is the new area. A = (c*k)/2 = (c*2h)/2 = c*h and A = S*2
2 equals S of a C?
2 Sides of a Coin
2 S T E C?
?What?
What is the surface area triangular prism?
If the sides of the triangular base are a, b and c and the height is h, the area is 2sh + 2 sqrt (s(s-a)(s-b)(s-c)) where s = (a +b +c)/2 [I. e. s is the semiperimeter of the base] and sqrt means"the square root of "
What is the area of a triangle when given are the length and the sides?
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
What is the area of oblique triangles?
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.
How do you find the area of an oblique triangle which 3 sides are given?
Let the sides be a, b, c Area = sq rt [s(s-a)(s-b)(s-c)] where s= 1/2 (a+b+c)
