umer007 Posted February 24, 2005 Posted February 24, 2005 There is a semi-circle with diameter XY. Rectange PQRS is inscribed in the semi-circle with PQ=12 and QR=28. Square STUV has T on RS, U on the semi-circle and V on XY. The area of STUV is closest to.... Answer is 16. I just want to know steps to come up to this answer. I made a drawing on paint, it dznt look xactly like the one I have but it shd give u guyz and idea of the question. If someone can show me steps to get to this answer it wd b helpful alot. Thx.
albertlee Posted February 25, 2005 Posted February 25, 2005 Not closest to 16, but it is exactly 16!!! The solution is: 1) Call the center of XY, O 2) We know that QP = RS, hence the angle, QOP = ROS, therefore, PO = OS 3) from 2), we know the radius is QO or RO, and we also know that PS = 28, and RS = 12, so according to Pytha. theorem, RO = root(144+196) = root(340) = radius 4) now parameterize TU = x, which equals to SV, we know that OS = 14, ie, half of PS, so OV = 14 + x 5) we know the radius is root(340), using pytha. theorem again, that 340 = x2 + (x+14)2 6) now rearange 5), we get the quadratic, x2 + 14x - 72, and its factorized form will be (x+18)(x-4) 7) x could either be -18 or 4, -18 is not possible, because we cant have a side of square being negative, ie, it is not a vector quantity, so 4 is the answer for x 8) Finally....... 4*4 = 16!!!!! Does that help??? Albert
timo Posted February 25, 2005 Posted February 25, 2005 Well, I can tell you the steps I´d take to get the answer: a) To deternime the diameter of the circle you use the height-sentence (or whatever it´s called in english): (PQ)² = XP * PY = XP * (PS + SY) = XP * (QR + XP). Solving for XP gives you the position of all points except T, U and S b) To determine VU and thus the position of the other points you can use the same method again: (UV)² = (XS + SV) * (XY - XS - SV) Note that UV = SV because SVUT is a square. Solve and be happy EDIT: Hehe, 12 hours without response and then two people answering within 10 minutes .
umer007 Posted February 25, 2005 Author Posted February 25, 2005 oic, thx alot. I understand how to get it from the first response but the 2nd response confuses me Maybe if u go a bit slower, but thats alrite atleast I know one way now. Thanks again both of u. Well another answer another question:P A solid cube of side length 4 cm is cut into two peices by a plane that passed through the midpoints of six edges, as shown. To the nearest centrimetre, the surface area of each half cube created is? I have learned the answer to be 69....is this right bcuz i havent learnd how to prove it:P personally i get confused by just looking at the picture, where are these half cubes??? are they supposed to be infront and behind the hexagon? Doc1.doc
albertlee Posted February 26, 2005 Posted February 26, 2005 Ok, for that question, first of all, find the area of the hexagon.... first find the length of 1 side of hexagon, which is root(8) or 2*root(2) the area of the hexagon will be [root(2)*root(6)]*6, which is = root(12)*6 or 2*root(3)*6 which is 12*root(3) Can you follow the above? Then, Think about how many sides there will be, 7, right??? the cube has 6 sides, plus a side of a plane cutting through, you can "visualize" in my picture now you have already known the area of hexagon, so 6 sides left Let's start with, what's the area of 1 red triangle???? area of 1 red triangle is 2 and how many red triangles are there??? 3, so the area of 3 of them is 6 now we have 3 sides left, the other 3 sides are all equaled, which 1 of them has an area of square subtracted by the blue triangle, which is same area of the red triangle, so, the side of the cube is 4*4=16, 3 of them is 48, subtracting 3 triangles so is 48-6 = 42 now the total surface area is the total of numbers in bold, ie, the answer is: [12*root(3)]+6+42, which should be approximately 69, I guess. Does that help?? Albert
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