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and we have:

base is the segment PR = 5 - ( -4 ) = 5 + 4 = 9 unitss

height is: 3 units

then area is

[tex]A=\frac{bh}{2}=\frac{9\cdot3}{2}=\frac{27}{2}=13.5\text{ }[/tex]

for the perimeter, we use the distance between two points to find the required sides:

for segment QR

[tex]\begin{gathered} QR=\sqrt[]{(x2-x1)^2+(y2-y1)^2} \\ QR=\sqrt[]{(5-(-1))^2+(-4-(-1))^2} \\ QR=\sqrt[]{(5+1)^2+(-4+1)^2} \\ QR=\sqrt[]{6^2+(-3)^2} \\ QR=\sqrt[]{36+9} \\ QR=\sqrt[]{45}=3\sqrt[]{5} \end{gathered}[/tex]

For segment PQ:

[tex]\begin{gathered} PQ=\sqrt[]{(-1-(-4))^2+(-1-(-4))^2} \\ PQ=\sqrt[]{(-1+4)^2+(-1+4)^2} \\ PQ=\sqrt[]{3^2+3^2} \\ PQ=\sqrt[]{9+9} \\ PQ=\sqrt[]{18}=3\sqrt[]{2} \end{gathered}[/tex]

therefore the perimeter is:

[tex]P=9+3\sqrt[]{5}+3\sqrt[]{2}=19.95[/tex]

answer: a = 13.5 sq units and p = 19.95 units

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