Can any expert me find the area of an triangle for brainlest and points

Step-by-step explanation:

2s=a+b+c

area=√(s(s-a)(s-b)(s-c))

where a,b,c are the sides of triangle.

2s=28.8+18+12=58.8

s=29.4

area=√((29.4)(29.4-28.8)(29.4-18)(29.4-12))

=√((29.4)(0.6)(11.4)(17.4))

≈59.15

≈59.2 sq.inch

58.8 in^2

Step-by-step explanation:

There is a way of finding the area of a triangle by knowing all three side lengths. It is called Heron's formula. You can look it up and use it by just plugging in the lengths you are given.

I'm going to use the more common formula, A = (1/2)bh.

In order to use the more common formula, we need a base and a height. Any side of a triangle can be used as a base. Then you need the height drawn from the opposite vertex to that side.

Let's use the side that has length 18 inches as the base. Prolong the 18-inch side to the left until it is below the left upper vertex. Now drop a perpendicular from the left upper vertex to the extension of the 18-inch side. That perpendicular segment is the height we need. We can use trigonometry to find its length.

Now you have a right triangle on the left. Its hypotenuse is 12 inches. The vertical side is the height we need. The given angle measures 147 degrees. The adjacent angle measures 33 degrees. The height is the opposite leg to the 33-deg angle.

sin x = opp/hyp

sin (33 deg) = h/(12 in.)

h = 12 in. * sin (33 deg)

h = 6.54 in.

Now we have the height and the base of the triangle, we can find the area.

A = (1/2)bh = (1/2)(18 in.)(6.54 in.)

A = 58.8 in^2

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: