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JH
Numerade Educator

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Problem 92 Hard Difficulty

In the figure at the right there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.

Answer

$\frac{11 \pi}{96}$

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Video Transcript

figure out the right there infinitely many circles approaching the furnace ese of the equilateral triangle. And we see each circle is touching other circles, or and also the size of the triangle were given that the triangle has side likes one, and now we'd like to find all the areas inside the circles. So let's observe that since the side length is one, this tells us that the medium length sounds wrong, that I'll drown the medium if you forgot what this was, so it's going to draw the median and blue in the picture that we really have. So for this people lateral triangle, we just go to one of the verses and we just draw the straight line down perpendicular to the office inside again. This trunk GoldenEye, true is not perfect at all. Should be symmetric, equal animals. Rangel. So it's not the best drawing here, but that's the media, that blue line. So that length Well, if we have side length one, then this sign over here. It's just one hat so we could use photography. They're going to find this opposite side length room three over, too. That will be the medium length so since this is true, the radius of the circle. So she's called the first circle. So that's this guy right here. A one circle one area one is given my r equals X over three. And let me explain why. So this is coming from geometry that this sensor of the circle is the central. So this is a geometry packed of the triangle. So of course, by the triangle, they mean that as in the picture we have the triangle circumscribe is the large circle, so the radius is median over three. So this tells us that a equals pi and then x over three square radius over three because our was X over three. Excuse me. That's why this is my art median over three. And then we square that so has gone to the next page. So the total sum before we go to the total sum notice that the radio of the next three identical circles No, no, we'll just be and sober nine. So the radius of each progressing circle is just a third of the previous. So therefore, we're basically done here. We just have one computation to do. So the sun of the areas is let's write this some this way. Let's call it a And then I was separate a one from the others because the other one's came in and triples and came in threes. So a ones the area of the large circle with a note that each one and then here this will be my radius of the smaller circles. So pi x pi r square for my area. So I'm writing the three year because those come in threes, the smaller circles and then we start dividing by We've already taken out the case where we divided by three. So this is where this is coming from a one and then each time we go, we progress. We increased and buy one, so we should be adding And there So that's the radios for the end progression. So if you want this guy right here, this will be a n plus one. Now, let's go ahead and simplify this So I'll need more room here. So let me go to the next page show. That's a one. So that's pie And then we have from three this home getting sloppy here Room three over two. But then we're also divided by three. So that's four times three square down there. So it's Pirates twelve. And then for the other song we can rewrite us by pulling us in terms. Come on, one. So this is Shia measuring. And not only that, we could go on. It simplifies in terms over here. So that's a three. That's a three that'LL cancel with that down there. So we just get pi over four in the front and then using the formula for geometric series the first term over one minus what we multiplying by each time? One overnight. Esther the R No. So now I'm gonna come back down here in power for and then we have one over nine divided by eight over nine. That's one over eight. So Ohio or thirty two. So adding this back today one are total area A will be pi over twelve plus time or thirty two, and we cannot and simplify that to be eleven pie over ninety six. And that's our final answer