1: Triangle WXY has side lengths XY = 14 and WX = 8. The tangent to the circumcircle of triangle WXY at X is drawn, and the line through W that is parallel to this tangent intersects line XY at Z. Find YZ. I have been directed to https://web2.0calc.com/questions/please-help_21594#r2 multiple times, but it is incorrect.  Here is the picture:






2: This is one I need all work shown for.  Let ABCD be an isosceles trapezoid, with bases AB and CD. A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base AB is 2x, and the length of base CD is 2y. Prove that the radius of the inscribed circle is the square root of xy.

 Apr 22, 2020

1)  If all we know about the triangle is that two of its sides measure 14 and 8 and, from this information, we can calculate the answer, then that answer must be true for however we draw the figure. (I find this hard to believe; is there something else about the triangle that we know?)


Assuming that we do not need to know anything else, I'll choose to draw a right triangle.

I'm going to place point X at the origin of the coordinate plane; point W at point (0,8); and point Y t point (14,0).

Then, the center of the circumcircle (call it "C") will be the midpoint of the hypotenuse (7, 4).

The slope of the line from X to C is 4/7; this line is a radius of the circle.

The tangent drawn through point X will be perpendicular to line CX, so its slope is -7/4.

Line WZ is parallel to this tangent and, therefore, also has slope = -7/4.

The equation of the line through C with this slope is:  y - 7  =  (-7/4)(x - 4)

     --->     4y - 28  =  -7x + 28     --->     7x + 4y  =  56

To find where this line intersects XY (the x-axis), substitute 0 for y:  7x + 4(0) = 56     --->  x = 8.

So, the distance ZY = 6.

 Apr 22, 2020

2)  Call the center of the circle "Q". 

     Since CD || AB, angle(DCB) + angle(ABC)  =  180o.

     QC bisects angle(DCB) and QB bisects angle(ABC).

     Therefore angle(QBC) + angle(QCB) = 90o, making angle(CQB) a right angle.

     The line segment from Q to the point of tangency on line BC is perpendicular to BC and has length r.

     Being an altitude of a right triangle, it is the mean proportional between the segments of BC.

     Therefore, r is the mean proportional between x and y, making r = sqrt(x·y).

 Apr 22, 2020

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