Isosceles TrapezoidDate: 6/23/96 at 1:21:41 From: Anonymous Subject: Isosceles Trapezoid The bases AB and CD of an isosceles trapezoid ABCD are 12 units apart. AB = 10 and CD = 8. Point R is on the axis of symmetry of the trapezoid so that the angle CRB measures 90 degrees. Find the possible distances from R to AB. I don't even know where to begin. Please help! Date: 6/23/96 at 8:12:30 From: Doctor Anthony Subject: Re: Isosceles Trapezoid Draw the following diagram and refer to it as I go through the work. Draw the isosceles trapezium ABCD and let E be midpoint of AB and F the midpoint of CD. Draw in the lines CR and BR to meet at right angles on the line EF. We now let RE = x so that RF = 12-x and we consider the two right- angled triangles BRE and RCF. These triangles are similar because they are both right-angled and angle BRE = angle RCF (both are complementary to angle CRF). We can equate ratios of corresponding sides, so BE/RE = RF/CF. Filling in the lengths we get: 5/x = (12-x)/4 20 = x(12-x) 20 = 12x - x^2 x^2 - 12x + 20 = 0 (x-2)(x-10) = 0 so x = 2 or x = 10 So point R can be 2 units or 10 units from point E. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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