what is thr exact perimeter and are of the figure in the coordinate plane?
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Answer:
Step-by-step explanation:
in order to calculate the lengths of side AB,AC and BC we need to use the pythagorean theorem for triangles ADB,AEC and BFC.
for ΔADB:
[tex](AB)^{2}=(AD)^{2}+(DB)^2\\\\(AB)^{2}=4^2+8^2=16+64=80\\\\AB=4\sqrt{5}[/tex]
similarly for ΔAEC
[tex](AC)^{2}=(AE)^{2}+(EC)^2\\\\(AC)^{2}=4^2+4^2=16+16=32\\\\AB=4\sqrt{2}[/tex]
similarly for ΔBFC
[tex](BC)^{2}=(BF)^{2}+(FC)^2\\\\(BC)^{2}=8^2+4^2=16+64=80\\\\AB=4\sqrt{5}[/tex]
Hence, perimeter of ΔABC= AB+AC+BC=[tex]4\sqrt{5}+4\sqrt{2}+4\sqrt{5}[/tex]
=[tex]8\sqrt{5}+4\sqrt{2}units[/tex]
area of square DBFE= (side)^2
since the length of the side of square DBFE=8 units.
Hence, Area of square DBFE= (8)^2=64 square units.
Area of triangle is given as: [tex]\dfrac{1}{2}bh[/tex]
where b denotes the base of the triangle and h denotes the height of triangle and for right angled triangle it is equal to the perpendicular side.
for ΔADB; h=4 units and b=8 units
Hence, area ΔADB=[tex]\dfrac{1}{2}\times 4\times 8=16 square units[/tex]
for ΔAEC ; h=4 units ,b=4 units
Hence, area ΔAEC = [tex]\dfrac{1}{2}\times4\times4=8 squareunits[/tex]
for ΔBFC ; h=8 units and b=4 units
Hence area of ΔBFC=[tex]\dfrac{1}{2}\times 4\times 8=16 square units[/tex]
Hence area ΔABC=64-(16+16+8)=24 square units.