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The School of Athens
Euclid, as imagined by Raphael in this detail from The School of Athens.
Jamshed Al-Kashi
Al-Kashi was one of the greatest mathematicians of the 15th century, calculating the sin(1*) correct to 16 decimal places.
Al-Beruni
The 11th century scientist and philosopher managed to estimate the radius of the earth using a very creative method that employed trigonometry.
16 comments:
For the last three problems on p. 278/279, I have found the surface area, but have not found the volume because I can't find the altitude. I have tried looking at different right triangles, but none of the numbers fit. Any suggestions?
Victoria, if you look at the diagram on 279, then for say, 15 the base is an equilateral triangle and the lateral faces are isosceles triangles. You have AB, you can find the height of an equilateral triangle CK. Since AV, CV and BV equal 10. So you now have CV and CK. Next think about where N has to be located on CK. This will help you find CN. Then it is just Pythagorean theorem from there and then the volume formula.
Also, remember that for every regular pyramid, the vertex V is always equidistant from the vertices of the base.
Finding where N is located is what gets me. I know VN intersects at right angles, but I can't think of anything else. Suggestions?
Do you agree that N is the circumcenter of the base? In an equilateral triangle, where do the altitudes meet and where do the angle bisectors meet and where do the medians meet? Are they all the same point? What do you know about any of those points and can you find a way to use other triangles formed within the base to get CN?
I'm having some trouble getting started on page 274 #18. Can someone give me a push in the right direction?
I am having trouble getting started on 7 on p. 273. Can anyone help me?
Michael, for #18 on page 274, I am sure you have figured that the ratio of APQ to ACB is 1/9. Now I know that it is hard to find anything about BPQ, but consider the auxiliary line CQ. Compare the areas of BPQ, CPQ, CPB and CQB. Remember that PQBC is a trapezoid. See where you get from such comparisons.
For #7 on page 273, it is important to note that the areas of ABD and CDB are equal since they have the same base and height, i.e. the height of the parallelogram. So try and determine the ratio of the area of BPA to ABD or APD.
This may not lead you directly to the answer but is a starting point.
Okay, so for #7, it may be a good idea to draw a line through P that is parallel to AB and DC and meets AD and BC at R and Q, respectively. Now you'll note that BC, BD and AD act as transversals. BPQ and BDC are similar and their ratio of similitude is 1/3. This way you can find what fraction of the area of BDC is BPQ. Then you can note that PQB and PRD are similar.
Let me know if you are following so far.
In my last comment, ignore the PQB, PRD thing at the end. Instead, once you find PQB's ratio to DCB, you'll also know what fraction PQB's area is to that of the whole parallelogram since BDC is half the area of the parallelogram.
Next, construct another line through P that is parallel to AD and meets DC and AB at S and T respectively. You will then notice that ABCD is divided into four parallelograms. Which two parallelograms are similar inside the ABCD?
What is the relationship between the areas of BQPT and TPRA? (I am assuming you have drawn diagrams much to my specs.)
They both have the same height but one's base is twice the other since PQ:QR is 1:3, which means that PQ:PR is 1:2. So the area of BQPT = PQ * h and the area of ATPR = PR * h = 2PQ * h.
In the same waythe ratio of ABQR's area can be compared with the ratio of ABCD if you make BQ and BC the bases. Once you have this you can notice that BPA's area is exactly half the area of ABQR. Let me know if you understand this.
Michael, how are you doing on #18?
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