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Trapezoid area calculations

Trapezoid Area Calculations class=

Sagot :

#1

  • 1/2(Sum of parallel sides)Height=Area

Area

  • 1/2(6+14)(5.5)
  • 1/2(20)(5.5)
  • 10(5.5)
  • 55in²

#2

Apply Pythagorean theorem

  • B²=14²-12²=196-144=52
  • B\approx 7

Area

  • 1/2(52+8)(12)
  • 6(60)
  • 360yd²

#3

Area

  • 1/2(Diagonals)
  • 1/2(6+6)(9+3)
  • 1/2(12)12)
  • 6(12)
  • 72ft²
semsee45

Answer:

Formulas Used

[tex]\textsf{Area of a Trapezoid}=\dfrac{1}{2}(a+b)h[/tex]

where:

  • a and b are the bases (parallel sides)
  • h is the height (perpendicular to the parallel sides)

[tex]\textsf{Pythagoras' Theorem}: \quad a^2+b^2=c^2[/tex]

where:

  • a and b are the legs of the right triangle
  • c is the hypotenuse (longest side, opposite the right angle)

[tex]\textsf{Area of a Kite}=\dfrac{1}{2}pq[/tex]

where:

  • p and q are the diagonals

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Question g (Trapezoid)

[tex]\textsf{Formula}: \quad A=\dfrac{1}{2}(a+b)h[/tex]

[tex]\textsf{Substitution}: \quad A=\dfrac{1}{2}(6+14)5.5[/tex]

[tex]\textsf{Answer}: \quad A=55\:\: \sf in^2[/tex]

Question h (Trapezoid)

[tex]\textsf{Formula}: \quad A=\dfrac{1}{2}(a+b)h[/tex]

Find the missing side length of the right triangle using Pythagoras' Theorem:

[tex]\implies a=\sqrt{14^2-12^2}=\sqrt{52}[/tex]

Therefore, top edge of trapezoid = √52 + 8

[tex]\textsf{Substitution}: \quad A=\dfrac{1}{2}(\sqrt{52}+8+8)12[/tex]

[tex]\textsf{Answer}: \quad A=96+12\sqrt{13}=139.3\:\: \sf yd^2\:(nearest\:tenth)[/tex]

Question i (Kite)

[tex]\textsf{Formula}: \quad A=\dfrac{1}{2}pq[/tex]

[tex]\textsf{Substitution}: \quad A=\dfrac{1}{2}(6+6)(3+9)[/tex]

[tex]\textsf{Answer}: \quad A=72\:\: \sf ft^2[/tex]

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