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A-Level Maths
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Not that long ago, there was a thread on AB about an O-level maths question, with the general agreement that it was a very easy question to answer.
Well here is a question that is supposedly an A-level maths question that I would judge to be at the level to be expected of O-level. The solution is quite easy, but I would elevate it to O-level, in that it involves a modicum of thought in how to divide up the area under consideration to get to the solution.
If students entering university are unable to answer this question, either degree courses have been significantly dumbed-down, or they will seriously struggle with the syllabus.
https:/
Well here is a question that is supposedly an A-level maths question that I would judge to be at the level to be expected of O-level. The solution is quite easy, but I would elevate it to O-level, in that it involves a modicum of thought in how to divide up the area under consideration to get to the solution.
If students entering university are unable to answer this question, either degree courses have been significantly dumbed-down, or they will seriously struggle with the syllabus.
https:/
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There is a square and 4 triangles to add up. The 4 triangles have an area of 2. The sides of the square are 2+sqrt(2). So as it says in the picture it's (2+√2)² - now that's the method but the final execution of that, as CTG points out is what is sometimes wrong but we know you get most points for the method so even if the final result of the arithmetic is wrong it's not as important.... is so here I go......
(2+2**0.5)**2 = 11.66 to 2dps. How did I do?
(2+2**0.5)**2 = 11.66 to 2dps. How did I do?
Maybe I've missed something, but isn't that p easy ?
You can divide the star up any way you like but perhaps easiest is to see the square that has sides of 2+root2. Then use your ½ base × height formula to work out the other 4 triangles and add them in.
No doubt I must've missed why that doesn't work.
You can divide the star up any way you like but perhaps easiest is to see the square that has sides of 2+root2. Then use your ½ base × height formula to work out the other 4 triangles and add them in.
No doubt I must've missed why that doesn't work.
As, zebu says, the key to unlocking the solution is to verify the 4 triangles are right angled.
Everybody seems to have assumed they are by just looking at the sketch. However, those marking the paper may want to see proof because without it, using Pythagoras to calculate the length of the unknown side may not be considered valid.
So, the sum of the internal angles of a polygon is (2n-4) * 90 (where n is the number of sides). For an octagon this becomes (16-4) * 90 = 1080 degrees. So each internal angle is 1080/8 = 135 degrees.
At the point where two octagons meet a triangle there are two angles of 135 degrees, so the remaining angle (that of the triangle) must be 360-270 = 90.
Everybody seems to have assumed they are by just looking at the sketch. However, those marking the paper may want to see proof because without it, using Pythagoras to calculate the length of the unknown side may not be considered valid.
So, the sum of the internal angles of a polygon is (2n-4) * 90 (where n is the number of sides). For an octagon this becomes (16-4) * 90 = 1080 degrees. So each internal angle is 1080/8 = 135 degrees.
At the point where two octagons meet a triangle there are two angles of 135 degrees, so the remaining angle (that of the triangle) must be 360-270 = 90.
