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### FAQ

How do you work out the area of this triangle?
This can be solved using the Law of Sines, the Law of Cosines and the general formula for the area of a triangle: Area = a * b * sin(C) / 2. This general law can be easily derived from other formulas for the area of a triangle. All three of these can be found in Wikapedia. I have a Perl script that runs an evaluator loop (in a commnd line) which is convenient for this sort of thing since it gives a compete trace. Here is the solution (you can round it yourself). Also note, that I work entirely in radians and so convert back and forth when degrees are given. I have interspersed some comments. $AC = 4.94.9$AD = 3.83.8 $A = 80 * pi / 1801.39626340159546== Use Law of Cosides to get length of CD$CD = sqrt($AC**2 +$AD**2 - 2 * $AC *$AD * cos($A))5.65538167267988== Use Law of Sines to get unknown angles$D1 = asin($AC * sin($A) / $CD)1.02222099951733$C1 = asin($AD * sin($A) / $CD)0.723108252476999 ($A + $C1 +$D1) * 180 / pi180== At this point the bottom left triangle ACD is completely known. $C2 = 25 * pi / 1800.436332312998582$D2 = pi - $D12.11937165407246$B = pi - $C2 -$D20.585888686518749 ($C2 +$D2 + $B) * 180 / pi180== Use Law of Lines to get DB and CB$DB = $CD * sin($C2) / sin($B)4.32247125729336$BC = $CD * sin($D2) / sin($B)8.72709037836453== At this point the top right triangle BCD is completely known.== For completeness get unknowns for large triangle.$C = $C1 +$C21.15944056547558 ($A +$B + $C) * 180 / pi180$AB = $AD +$DB8.12247125729336== Use formula for area to areas for all three triangles. $AreaACD =$AC * $AD * sin($A) / 29.16856018054366 $AreaBCD =$CD * $DB * sin($D2) / 210.4291678555696 $AreaABC =$AC * $AB * sin($A) / 219.5977280361132
There is curfew in my area and Internet service is blocked, how can I fill my exam form as today is the last day to fill it out?
Spend less time using your blocked Internet to ask questions on Quora, andTravel back in time to when there was no curfew and you were playing Super Mario Kart, and instead, fill out your exam form.
Is there a formula/algorithm to find out the radius of n circles needed to fill a square area?
Major edit:OK, I'm not good at formal mathematics. I had an initial hunch, which ended up with the following reasoning, valid only for quadratic n x n matrices of circles within squares:Let n be the number of circles of radius r along each side of the square with sides 1, and you get:$r_{n} = \frac{1}{2n} \sqrt{2}$Now, solving the problem only for very easy cases isn't much fun, so I tried to make a model and draw some of the other alternatives. Maybe someone who's better in formal mathematics can use them to explain this rigorously, but to me it seems that the quadratic stacking of circles is the most efficient solution.The drawings that follow were constructed in MATLAB assuming that the most efficient distribution of circles would have them intersect in the center of the square, A, and for n circles place their centers (B,C, etc.) on the n lines drawn from A in such a manner that connecting B,C etc. forms a equilateral n-angle.Here, there are three circles with radius $r = \frac{2}{5} \sqrt{2}$. I don't know if the exactly circumscribe the square, but it's pretty close. With four, which clearly gives the same result as 8, you get this:Where the black circles obviously play no role. Moving the square around within the circumscribed area doesn't change this, but it does for the 5-circle variant, although here the circles need a slightly larger radius, about 0.52 x sqrt(2), and the intersection of all of the circles needs to be slightly offset from the center of the square in order to get the smallest circle size, about 0.04 units here:The situation is similar with 6 and 7 circles. The diameter of the circles needs to be at least 0.5 (for 6) and somewhere between 0.5 and 0.54 with a slight offset (for 7, the circles here are 0.5, 0.52 and 0.54 times sqrt(2)):I already showed the picture with 8 circles, so for the last one, let's skip to 9. Here, it's obvious that the square arrangement is more efficient, and using this picture, I think it's possible to see the logic: If you remove the central circle (center at A) and expand all of the rest, while moving their centers toward A, the edges of the circles on the diamust also intersect at A, and their centers must be placed halfway between A and the corners. If this condition isn't met, either all the circles won't be the same size, or the entire square won't be circumscribed.For the last approach I'm able to think of, you could take the central circle as your starting point, remove some of the outer circles and expand and move the remaining circles around laterally to achieve total circumscription. But I haven't been able to make great gains that way either. For 6 circles, using 0.48 x sqrt(2) as the radius (admittedly, less than 0.5 x sqrt(2)), spacing the centers from A by 0.78 units to form an equilateral pentagon and offsetting A by 0.05 units, I get this:I got a similar arrangement with 7 circles down to 0.41 x sqrt(2) and with 8 to about 0.395. So now I'm interested, as well: Is there any general solution to this?
How much cement, sand and aggregate are required to fill one cubic meter area?
Since you have not mentioned the grade of concrete, I’m assuming grade of concrete as M25.Volume of concrete = 1 m³Grade of concrete M25 (1:1:2)Wet volume of concrete = 1 m³Here, wet volume is nothing but Volume of the concrete after mixing with water.∵ Dry volume = Wet volume x 1.54 (You can also take it as 1.50 to 1.55, But it became standard practice to considering 1.54 as multiplication factor).Dry volume of concrete = 1 x 1.54 =1.54 m³Quantity of Cement:∵ Quantity of cement = (Dry volume of concrete x cement ratio)/(sum of the ratio)Cement = (1.54 x 1)/(1+1+2)=0.385 m³∵ Density of Cement = 1440 kg/m³Weight of Cement = 1440 x 0.385 = 554.4 Kg∵ 1 bag of cement contains 50 kg of cementNumber of bags = 554.4 Kg/50 kg = 11.088 ≅ 11 No’sQuantity of Sand:∵ Cement : Sand :: 1:1Quantity of sand =Quantity of cement x 1∴ Quantity of Sand = 0.385 x 1 = 0.385 m³∵ 1m³=35.3147 Cubic Feet (CFT)Quantity of sand = 0.385 x 35.3147 = 13.596 CFT∵ Weight of sand = volume x density, Density of Sand = 1600 kg/m^3∴ Weight of sand= 0.385 x 1600= 616 kg (0.616 tonnes)Quantity of Coarse Aggregates:∵ Cement : Coarse Aggregates :: 1:2Quantity of Coarse Aggregates =Quantity of cement x 2∴ Quantity of Coarse Aggregates = 0.385 x 2 = 0.77 m³Quantity of Coarse Aggregates = 0.77 x 35.3147 = 27.217 CFT∵ Weight of Coarse Aggregates = volume x density, Density of Coarse Aggregates = 1520 kg/m^3∴ Weight of Coarse Aggregates= 0.77 x 1520 = 1170.4 kg (1.170 tonnes)Source:- How to calculate Quantity of Cement, Sand, Aggregate & Water in M25 concrete?Hope it Helps
If an alien spacecraft landed on Earth (and its crew died later for some reason), how quickly could we learn from its spaceship how to build our own spaceship that could match its travelling ability?