<h2>
The area of a triangle is =54 square units</h2><h2>
The perpendicular distance from B to AC is = 
</h2>
Step-by-step explanation:
Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

The area of a triangle is= ![\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bx_1%28y_2-y_3%29%20%2Bx_2%20%28y_3-%20y_1%29%2Bx_3%28y_1-y_2%29%5D)
=![|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|](https://tex.z-dn.net/?f=%7C%5Cfrac%7B1%7D%7B2%7D%20%5B2%282-8%2B12%288-1%29%2B12%281-2%29%5D%7C)
=
= 54 square units
The length of AC = 
= 
=
units
Let the perpendicular distance from B to AC be = x
According To Problem

⇔
units
Therefore the perpendicular distance from B to AC is = 
Hello,
As 4²+(5/2)²=22.25=89/4 (Pytagorean theorem)
Perimeter of the triangle =2*√89 /2 +5 =5+√89
Lateral area: 18*(5+√89)
Total area= 90+18√89+ 4*5/2 *2 =110+18√89 ≈279.811..(ft²)
The answer you are looking for is -5x^5-5x^4+6x^3-x^2-7x+20
-I hope this is the answer you are looking for, please feel free to post your future questions on Brainly.com.
Plug in 2 for x. f(20=2-51 = -49