The area of ​a figure defined by the set of solutions of a system of inequalities I have been trying to solve this problem over a week since i got it but no luck. I have translated problem form my native language into English, hopefully i have made problem clear. Thanks in advance!
For each positive value of the parameter ''a'', consider a figure defined by the set of solutions of the system of inequalities in the oxy rectangular coordinate plane. Find the value of ''a'' for which the area of ​​this figure will be equal to 24 units
y<=5-|x| and |y-1|<=a
 A: 
Firstly, the first inequality isn’t an isosceles triangle, rather an infinite region. The area bounded by the region $y \leq 5—|x|$ and the x axis is an isosceles triangle. The shaded region represents the first inequality.
CASE 1. If a $\geq$ 4.
The upper violet line either passes through (5,0) or does not touch the shaded region of the first inequality. Also, the lower violet line must stay below the x-axis because $-a+1 \leq -3$. So the required area is the part of the turquoise area between the two violet lines, which will obviously be greater than the area of the isosceles triangle bounded by the x-axis and the region $y \leq 5—|x|$. The area of the triangle is 25 sq. units which is greater than 24, so a cannot be greater than or equal to 4.
CASE 2. If $1 \lt a \lt 4$
 The lower limit on a ensures that the lower violet line stays below the x-axis.
 The area of the final trapezium formed is $$(\frac{5-(—a+1)}{5})^2 25 — (\frac{5-a-1}{5})^2 25 = 16a $$ because the ratio of the areas of similar triangles is equal to the square of the ratio of their heights. What I did was take the smaller triangle (bounded by the region $y \leq 5—|x|$ and the upper violet line) and subtract its area from the larger triangle (bounded by the region $y \leq 5—|x|$ and the lower violet line).
Thus, $16a = 24$, which implies that $a= \frac{3}{2}$ .
CASE 3. If a $\leq 1$
Both violet lines will be above the x-axis, so the area will be (computed in the same way as the second case) 16a = 24. This gives $a= \frac{3}{2}$, which clearly does not satisfy $a\leq 1$. Thus this case has no solutions which means that the only solution is $a= \frac{3}{2}$.
