The area (in sq. units) bounded by $x=4, y=-4$ and $y=x$ is
- $48$
- $32$
- $24$
- $16$
Solution

The required bounded area is $\triangle A B C$. Here $A C=8 \& B C=8$ $\therefore$ Required area $=\frac{1}{2} \times A C \times B C=\frac{1}{2} \times 8 \times 8=32$
Asked in: AP EAMCET 2023 (17 May Shift 1)