1) let r be radius of cone, R be radius of quarter circle
From quarter circle, L = r*theta, 2pi*r = R*pi/2, R = 4r
From cone, sintheta = r/R, hence sintheta = 1/4
There are three questions I'm having trouble solving and would appreciate if you guys could solve and explain why (some of these are from the Year 11 and 12 section of the AMC test).
1) Screenshot_2.jpg
Answer is 1/4
2) Screenshot_3.jpg
Answer is between 7/16 and 1/2
3) Screenshot_1.jpg
Answer is 25pi - 25
1) let r be radius of cone, R be radius of quarter circle
From quarter circle, L = r*theta, 2pi*r = R*pi/2, R = 4r
From cone, sintheta = r/R, hence sintheta = 1/4
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I'm gonna label the corners of the outer square with A,B,C and D to make it easier, with A being the top left corner. And the corners of the inner rectangle with E, F, G, H with E being the upper most corner. [My apologies, all these letters may seem confusing to memorise so I suggest writing them down on a diagram somewhere so the solution is a little easier to follow]
Let angle AED = theta. From the larger unshaded triangle, we can tell that tantheta = 2.
Since AED, HEF, and BEF are supplementary, they add to 180, and HEF equals 90 whilst AED is theta. So doing the algebra (I trust that you can do this part since this question is one where if you can see it, the maths is pretty simple), and using angle sum of triangle, EFB equals theta. Using the same proof (supplementary angles, angle sum of triangle) angle FGC is also theta. Using the tantheta identity in each of the unshaded white triangles, you can get tantheta = something. (For the upper triangle, 1/2 is already given to you as a side length.) Let the other adjacent side be x, and let
1/2 over x = 2. (From the tantheta=2 identity you proved earlier on) And solve to get 1/4. Now since its a square the side length must be 1, so 1-1/4 gives 3/4 as the opposite angle in tantheta=2 for the bottom one.
(Oh, just a quick reminder that tantheta = opposite/adjacent, so the fraction itself can equal 2. Don't put tan(fraction)=2!)
Repeat the process for the bottom one, and you should get 3/8. Then, using pythagoreas theorem in each of the triangles, you should get respectively, rt(5)/4 and 3rt(5)/8. (Remember to square root the squares! It's a common silly error :P)
Then multiply them. I got 15/32 which is between 7/16(7/16=14/32) and 1/2. (1/2=16/32)
Ah, wow I did it differently and never thought of something as fast and simple as that.
But anyway, the much longer way I thought of for the question was:
I drew a 10x10cm square around the circle, and drew circles of 5cm radii as shown.
From there I just subtracted the green area ([square - circle]/2)
Then subtracted the 2 quarter circles (the navy) and got an answer of 50 HAHAHAHAHAHAHAHAH way more trek though
Last edited by 30june2016; 19 Jul 2017 at 6:37 PM.
UTS BIT 2018
Yep, the answer is 50.
25pi/4+25pi/4+(25-25pi/4)+(25-25pi/4)
Yep my bad, the answer was 50.
Thank you all for your help.
The radius of the circle is 5, so the area is 25pi
Area of shape = Area of Quarter circle + Area of Quarter circle + (25 - Area of Quarter circle) + (25 - Area of Quarter circle)
25 is the area of a 5*5 square
The top left and bottom right form quarter circles
the top right and bottom left areas are given by 25 - Area of Quarter circle
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