How do I find the are of a convex quadrialater with vertices (-1,0),(0,1),(3,0) and (0,-5)?
How do I find the are of a convex quadrialater with vertices P1(-1,0),P2(0,1),P3(3,0) and P4(0,-5)?
Hello Guest!
\(A=\dfrac{(y_2-y_4)(x_3-x_1)}{2}\\ A=\dfrac{(1-(-5))(3-(-1))}{2}\\ A= \dfrac{6\cdot 4}{2}\)
\(A=12\)
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