# Inscribed kissing circles in an isosceles trapezoid

5 equal circles in an isosceles trapezoid. Radius of circle is 4. Find black colored area.

I don't have any ideas, could you give me a hand? Thanks.

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What is the question? Evaluate the black area? Dimensions of the trapezoid? –  Ross Millikan Jun 6 '11 at 19:34
@Ross find area, en.wikipedia.org/wiki/Area (that one) –  Templar Jun 6 '11 at 19:37

Clearly the area of the colored region is 3 times the area of one of the curved triangles.

Draw a regular hexagon around one of the circles, in such a way that all of its sides are tangent to the circle (i.e., the circle is inscribed in the hexagon). There are 6 regions of the hexagon that are not contained in the circle, one at each vertex of the hexagon:

Note that any one of the curved triangles in your picture consists of 3 of these hexagon corners.

Let $A$ be the area of a hexagon with inradius 4. Let $B$ be the area of a circle with radius 4. The total area of 6 hexagon corner regions is $A-B$. The region we want to find is made of 3 curved triangles, each of which is made of 3 hexagon corners, for a total of 9 hexagon corners. Thus, the area of the colored region is $\frac{3}{2}(A-B)$.

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+1, Note to OP: in particular, the fact that it was a trapezoid didn't matter at all. –  Eric Naslund Jun 6 '11 at 19:52
OMG thanks, I would never think of it myself –  Templar Jun 6 '11 at 20:00

It's pretty easy to figure out the area of a hexagon and easier to look it up, but I don't know it off the top of my head and I think the problem can be done without it. The area formulas for trapezoids and circles would be enough considering this figure:

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In reference to stuart's drawing: Note that the height of the (inner) trapezoid, the top of it and the two non parallel sides are all equal to the diameter of the a circle (2r), and the longest side is two times the diameter of a circle. The area comprised of sectors of circles inside the smaller trapezoid total 3/2 time the area of one circle: top and bottom left sectors are supplementary (total 1/2 circle), same with top and bottom circles, and 1/2 of the middle bottom circle is contained in the trapezoid. Area trapezpoid - 3/2(area circle) = desired area. –  amWhy Jun 6 '11 at 21:04
The height of the inner trapezoid isn't the diameter of the circles, but it's still easy to figure out. –  stuart Jun 6 '11 at 21:19
Ahhh...dah...you're right about the height; it's still easy to figure out given all the other info; one could construct a right triangle by drawing the segment extending from the center of upper left circle, intersecting lower side perpendicularly: that gives hypotenuse = 2d, short leg = 1/2 d...the rest is quite obvious...This was precisely the approach I took when solving the problem...by the time I had drawn the picture, you had already posted yours! :D –  amWhy Jun 6 '11 at 21:27

The hexagonal solution is lovely, and the trapezoid picture is pretty nice. But there are cruder ways that perhaps require less insight.

Let's find the area of one of the three black chunks. To do this, join the centers of the circles it fits between. We get an equilateral triangle of side $8$. From the area of this, you must subtract three $60^\circ$ sectors, which together have half the area of a circle of radius $4$, that is, $8\pi$.

With basic trigonometry, or otherwise, one finds that an equilateral triangle of side $a$ has area $a^2\sqrt{3}/4$. Here $a=8$, so the area is $16\sqrt{3}$. Thus one little black chunk has area $16\sqrt{3}-8\pi$, and all three have combined area $$48\sqrt{3}-24\pi.$$

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