Dragan

Quadrilateral ABCD is 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 the base AB is 2x, and the length of the base CD is 2y. Prove that the radius of the inscribed circle is sqrt(xy).

r = sqrt( x*y )    or     r² = x*y                   x < r < y

r = 2         x = 1.25          y = r² / x       y = 3.2

Important:     AD = BC = x + y

Proof:        sqrt [( y-x )² + ( 2r )² ] = x + y

                                 4.45 = 4.45  

This is one way to prove it.

Find y if the point (-6, y) is on the line that passes through (-1, 7) and (3, -2).

https://web2.0calc.com/questions/some-more-questions#r1

What is the length of AB?

2.5/5.8 = 2.7/x

Let us have a triangle ABC and a point D on BC such that BD = DC = DA. If angle ABC = 50 degrees, then how many degrees are in angle ACB?

Given: EG - diameter m arc FG =41°, m arc EH =53° Find: m∠E, m∠GPF

Angle E = 20.5º

Angle GPF = 47º

It is D!!!

If x = -1

sqrt( x +3 ) / sqrt( 1 - x ) = sqrt( -1 + 3 ) / sqrt[ 1 - (-1)] = 1  

18-gon interior angles are 120º

Heptagon interior angles are 128.5714286º

Side lengths of both polygons are 1 unit.

Coordinates of Y, X, W, and Z

Y ( 0,0 )

X ( 1,0 )

W (1.93969262, 0.342020143)

Z (1.623489802, -0.781831482) 

Angle YZW = 80º  

x = [ sin(60º) * 8 ] / sin(45º)

If this diagram is not drawn to scale, then 'x' could be 45 degrees.

The solid is a couple of cones that share the same base.

My diagram may help you in solving this geometric problem.