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]
