Respuesta :

Answer:

  • perimeter ΔABC=[tex]8\sqrt{5}+4\sqrt{2}units[/tex]
  •  area ΔABC=64-(16+16+8)=24 square units.

Step-by-step explanation:

  • The perimeter of triangle is the length of all the sides of a triangle.

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]

  • Now area of ΔABC=area of square DBFE-( area ΔADB+ area ΔAEC+ area ΔBFC)

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.

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