gorden wrote:
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In the figure, AB = AE = 8, BC = CD = 13, and DE = 2. What is the area of region ABCDE ?
A. 76
B. 84
C. 92
D. 100
E. 108
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areaofabcde.png [ 38.37 KiB | Viewed 1152 times ]
Area of figure ABCDE =
area of trapezoid ABDE +
area of ∆ BCD.
This problem contains two special right triangles with Pythagorean triplets, which makes lengths for area of ∆ BCD easy to find.
1.
Connect B to D and find area of trapezoid ABDE
Area ABDE = \(\frac{b_1 + b_2}{2} * h\)
Given: AB = 8, DE = 2, and AE = 8
Area of trapezoid ABDE = \(\frac{8 + 2}{2} * 8\) =
402. Find area of ∆ BCD -- we need the base and height.
3. Base of ∆ BCD
To find base of ∆ BCD,
draw a line from D to AB (in the diagram,
DX) that is parallel to AE in order to create a right triangle BDX where
DX is one leg of the right triangle
BX is the other leg, and
BD, the base we need, is the hypotenuse
DX = 8
BX = 6 (AX = DE; they are parallel sides of the new rectangle created by the pink line. BX = AB - AX.
BX = 8 - 2 =
6)
This is a 3-4-5 right triangle: 6: 8:
10. Base BD =
104. Height of ∆ BCD
Because ∆ BCD is an isosceles triangle, its altitude will create two congruent right triangles (see second small figure in diagram). Draw an
altitude from C to BD
BD is bisected. Each right triangle has
short leg of length
5Hypotenuse CD = 13. This is a 5-12-13 triangle.
Height of ∆ BCD is
125. Area of ∆ BCD =\(\frac{10*12}{2}\)=
60Add the areas of the trapezoid and the triangle: 40 + 60 = 100
Answer D
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