6.5: The Area of a Trapezoid
- Page ID
- 34150
In Figure \(\PageIndex{1}\), \(b_1\) and \(b_2\) are the bases of trapezoid \(ABCD\) and \(h\) is the height or altitude. The formula for the area is given in the following theorem:
The area of a trapezoid is equal to one-half the product of its height and the sum of its bases.
\[A = \dfrac{1}{2} h(b_1 + b_2)\]
- Proof
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In Figure \(\PageIndex{1}\) draw \(BD\) (see Figure \(\PageIndex{2}\)). Note that \(CD = b_2\) is the base and \(BF = h\) is the height of \(\triangle BCD\). Area of trapezoid \(ABCD =\) Area of \(\triangle ABD\) + Area of \(\triangle BCD\) = \(\dfrac{1}{2} b_1h + \dfrac{1}{2} b_2 h = \dfrac{1}{2} h(b_1 + b_2)\).
Find the area:
Solution
\(A = \dfrac{1}{2} h(b_1 + b_2)= \dfrac{1}{2} (6)(28 + 16) = \dfrac{1}{2} (6)(44) = 132\).
Answer: \(A = 132\).
Find the area and perimeter:
Solution
Draw heights \(DE\) and \(CF\) (Figure \(\PageIndex{3}\)). \(\triangle ADE\) is \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle. So \(AE\) = short leg = \(\dfrac{1}{2}\) hypotenuse = \(\dfrac{1}{2}(10) = 5\), and \(DE\) = long leg = (short leg)(\(\sqrt{3}\)) = \(5\sqrt{3}\). \(CDEF\) is a rectangle so \(EF = CD = 10\). Therefore \(BF = AB - EF = 22 -10 - 5 = 7\). Let \(x = BC\).
\[\begin{array} {rcl} {\text{CF}^2 + \text{BF}^2} & = & {\text{BC}^2} \\ {(5\sqrt{3})^2 + 7^2} & = & {x^2} \\ {75 + 49} & = & {x^2} \\ {124} & = & {x^2} \\ {x} & = & {\sqrt{124} = \sqrt{4} \sqrt{31} = 2\sqrt{31}} \end{array}\]
Area = \(\dfrac{1}{2} h(b_1 + b_2) = \dfrac{1}{2} (5\sqrt{3}) (22 + 10) = \dfrac{1}{2} (5\sqrt{3})(32) = 80\sqrt{3}\).
Perimeter = \(22 + 10 + 10 + 2\sqrt{31} = 42 + 2\sqrt{31}\).
Answer: \(A = 80\sqrt{3}\), \(P = 42 + 2\sqrt{31}\).
Problems
1 - 2. Find the area of \(ABCD\):
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3 - 12. Find the area and perimeter of \(ABCD\):
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13 - 14. Find the area and perimeter to the nearest tenth of \(ABCD\):
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15. Find \(x\) if the area of \(ABCD\) is 50:
16. Find \(x\) if the area of \(ABCD\) is 30: