determine the area of triangle abc when side ab= square root of x, side bc= square root of x, and side ca= the square root of the square root of x.
So how do I solve it. How do I add the square root of x + the square root of x + the square root of the square root of x?
what is the area actually. How would I solve for that?
1.)4(z-2)+z= -13
area in terms of the three sides of the triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c) where s is the semiperimeter of the triangle, that is, s = (a + b + c)/2.
s= (3/2 sqrt x)
area = sqrt [s(s – a)(s – b)(s – c) ]
= sqrt [ (3/2 sqrt x) ( (3/2 sqrt x) – sqrtx)( (3/2 sqrt x) – sqrtx)( (3/2 sqrt x) – (3/2 sqrt x) ) ]
= sqrt ((3/2 sqrt x) (1/2 sqrt x)^3 )
= sqrt ( 3/2 * 1/8)(sqrt x)^4
= 1 x sqrt (3/16)= x/4 * sqrt 3
check that. lots of square roots to mess up.
To find the area of triangle ABC, you can use Heron's formula, which states that the area (A) of a triangle with side lengths a, b, and c can be found using the semiperimeter (s) of the triangle:
s = (a + b + c)/2
Then, the area (A) can be calculated using the formula:
A = √(s(s - a)(s - b)(s - c))
In this case, side AB = √x, side BC = √x, and side CA = √√x.
Let's substitute these values into the formula:
s = (√x + √x + √√x)/2
= (2√x + √√x)/2
= (√x + 1/2 √√x)
Now we can calculate the area (A):
A = √(s(s - √x)(s - √x)(s - √√x))
= √((√x + 1/2 √√x)((√x + 1/2 √√x) - √x)((√x + 1/2 √√x) - √x)((√x + 1/2 √√x) - √√x))
Simplifying the expression inside the square root:
A = √((√x + 1/2 √√x)(1/2 √√x)(1/2 √√x)(1/2 √√x))
= √((√x + 1/2 √√x)(1/2 √√x)^3)
= √(√x * (1/2 √√x)^3)
= √(1/16 * x^2)
= (1/4) * √3 * x
So the area of triangle ABC is equal to (1/4) * √3 * x.