So, to find the solution to this problem, we will we using pretty much the same method we used in your previous question. First, let's find the area of the rectangle. The area of a rectangle is length x width. The length in this problem is 16 and the width is 3, and after multiplying these together, we have found 48 in^2 to be the area of the square. Next, we can find the area of the trapezoid. The area of a trapezoid is ((a+b)/2)h where a is the first base, b is the second base, and h is the height. In this problem, a=16, b=5, and h=10. So, all we have to do is plug these values into the area formula. ((16+5)/2)10 = (21/2)10 = 105. So, the area of the trapezoid is 105 in^2. Now after adding the two areas together (48in^2 and 105in^2), we have found the solution to be 153in^2. I hope this helped! :)
<h3>Hey .... </h3>
here's your answer in the given attachment
The irrational in between should be 7.8
You can use substitution and solve y-x=2 for y which is y= x+ 2 and plug it into the y of the other equasion and you get x=5 y=7
Y=17.8 Please let me know if it is correct :)
