【Haichuan Teahouse】Brain-teasing Series: Feels completely clueless
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As shown in the figure, quadrilateral ABCD is a right trapezoid with upper base AD = 7 and lower base BC = 11. A equilateral triangle is constructed using CD as one of its sides, and its vertex E lies exactly on side AB. Find the area of △CDE.
I’ve been too busy these past few days and don’t have time to think about it; I’ll deal with it once I’m done with this workload;P
Draw a perpendicular line from E to CD; the point of intersection is F. Then, considering that the new triangles DEF and DEA are similar, the problem is solved: the area of an equilateral triangle with side length 14 is 7*7*3^0.5/2=42.435.
This post was last edited by zqwqiwu123 on 2021-8-11 10:20. Triangles DEF and DEA are similar; how can this be proven?
With E as the center and EA as the radius, does it pass through point F? In other words, DA and DC are tangents to circle EA.
There seems to be a problem; according to these results, angle BEC should be 90 degrees
This post was last edited by pzhmotor on 2021-8-13 at 11:29. Geometry problems for middle school students; I can’t do the calculations. :'( 31√3 = 53.6936 is correct! Using SolidWorks constraints, it’s exactly the same as this value.
With EG⊥CD and GF⊥AB, it follows that GF=9, and AF=FB. That is, AE-BE=2*EF; let CD=2*X, and X can be found by solving the equation. The computational load is significant; the trick lies in using an appropriate format for the variables, rather than simply using X, which can simplify the calculations.