I suspect it is:
220 squares on a Chess board, counting the 2*2, 3*3.... squares as well, though the answer should be 204, and I guess the question setter added the 16 5*5 squares twice!
no, b/c the 0 at the end means it is a product of 10
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.
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)
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)].
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.
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.
#include<stdio.h> #include<math.h> main() { float A,a,b,c,s; printf("entr the value of a,b and c"); scanf("%f %f %f",&a,&b,&c); A=sqrt(s*s-a*s-b*s-c) printf("area of the triangle %f\n"A); getch(); }