jugoslav

I'll use your graph, if I may?! 

Note that the y-axis is parallel to a chord AB, and the distance between them is 3 cm.

The coordinates of the points P and O are:     P (0, 2)       O (0, 0)        MD = ND ==>  r

PO = 2       PM = 3          ON = 5

PO² + (PM + r)² = (ON - r)²

2² + (3 + r)² = (5 - r)²    ==>     r = 0.75

A = (3/4)²pi cm²

DX / DY = DY / DZ

DX = DY² / DZ        Everything else is easy.

The sum of the areas of the black regions is 2pi square units because the sum of interior angles of a hexagon is 2 full circles or 720 degrees.

Nope, it's not 7.

The distance between the centers is a hypotenuse of a triangle, and the legs are 1 and 7 units.

Distance is 5√2 units.

What is the radius of the circle inscribed in triangle ABC if AB = 5, AC=7, BC=8? Express your answer in the simplest radical form.

r = √3        area of ΔABC = 10r         area of a circle is  3pi   

The area of the triangle AOD is 1/4 of [ABCD].

To find a radius of a circle we ought to multiply [AOD] by 2 and then divide the product by 25.

r = 300 / 25  

A = [3*(3 / tan60º + 3 / tan30º)] / 2 = 6√3

The chord of a circle divides the circle into two parts.  A square of area 16 is inscribed in one part, and a square of area 144 is inscribed in the other part.  The radius of the circle is?

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Note that only 2 vertices of each square are on the circle!

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∠CBR = ∠PQO = tan^-1(CR / BR) ==> 14.03624347º

P and Q are midpoints on TS and BC

CQ = 1/2 * sqrt(CR² + BR²) = √68

PQ = 1/4(AB + CD) = 4

OQ = PQ / cos∠PQO = √17

CO = r = sqrt(OQ² + CQ²)   ==>     r = √85

Here's a link:    http://mathforum.org/library/drmath/view/54103.html

A circle is centered at O  and has an area of 48pi. Let  Q and R  be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO  and triangle PQR  is equilateral, then find the area of triangle PQR.

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OQ = OR = r = 4√3

∠QOR = 30º         ∠QOP = 15º           ∠QPR = 60º

QR = 2 (OQ * sin∠QOP) ==>         QR = 3.586301889 

Height of ΔPQR      h = sqrt[PQ² - (QR/2)²] = 3.105828542

[PQR] = 1/2(h * QR) = 5.569219383 u²