500 million points for anyone who can solve this maths problem

[da_beast]

I think the answer is 4

lol i need working so i can check it. I want to know how its actually done. Ive been trying for 2 days now

[Toyman]

Yes just the top figure. the +0.25 is the limit of error and is for another question i have already answered. It isnt relevant to this question.

Im completely stumped.

The base of the triangle i drew is 2.098. Thats half of the width (4.195) so that triangle run from the right hand side to the centre of the diagram.

The 1.775 side is half of the height (3.55).

Again im not even sure if my triangle is correct

OK,

but can you assume the dimensions of the triangle are half the width/height? I could see half the height being possible, but the triangle doesnt seem to be half the width of the figure.

[DoktorSleepless]

I took calculus years ago and I really don't remember much of it.

Your triangle though, wouldn't it have to be an isosceles triangle?

[Toyman]

I took calculus years ago and I really don't remember much of it.

Your triangle though, wouldn't it have to be an isosceles triangle?

If it is isosceles, then i think he can calculate the area of the four corners as long as he knows the length of the 2 equal sides, but it doent say anywhere to asume its isosceles

[DoktorSleepless]

You know what? I'm not sure what I'm supposed to be looking at in the second picture of diagram 2. Can you explain what's going on there? It is part of the first picture somehow?

[tesla_power]

What subject is this? That will proved a clue. Without calculus, this would be a bit hard.

But if i interpret it correctly, that is a square inside a circle.
Thus, calculate the area of the circle using 4.195 as diameter. Then calculate area of the square with 3.55. The difference you divide by 2( as two outer segments are chopped off). Then, just add this to the area of the square.

This is just on top of my head.

[miniq]

The answer is 12.983 mm^2

The rectanguler section = 9.017 mm^2
The 'circle section' = 1.983 mm^2


Took me way too long...I've given the units of measurement as mm as none were given...

[Toyman]

Actually, i think i see the OP's problem now.

the "triangles" at the side can't just be simply cut off with a trick, since they aren't isosceles. the shape of the two protruding round sides isnt actually a circle if you put it together, it's more like an ellipse.

so in order to get the closest estimate, i think you need to draw an imaginary rectangle encompassing each half of the ellipse. The remaining area outside each "semi-circle" will have to be approximated with triangles, base of (3.55/2) and height (4.195/4). That would enable you to estimate the area of each triangle, but the question is asking you to round the angles...which im guessing means they want you to use the sin, cosin rules along with a TI calculator or somethin.


edit: just tried it again;

area of entire figure would be 4.195 * 3.55 = 14.89

then each of those small "triangles" would have base (4.195/4) and height (3.55/2) with area 0.93. since theres 4 triangles, they comprise in total about 3.72. so if you take that away from 14.89, you get 11.17. thats my best guess as of now, lol

[tesla_power]

Actually, i think i see the OP's problem now.

the "triangles" at the side can't just be simply cut off with a trick, since they aren't isosceles. the shape of the two protruding round sides isnt actually a circle if you put it together, it's more like an ellipse.

so in order to get the closest estimate, i think you need to draw an imaginary rectangle encompassing each half of the ellipse. The remaining area outside each "semi-circle" will have to be approximated with triangles, base of (3.55/2) and height (4.195/2). That would enable you to estimate the area of each triangle, but the question is asking you to round the angles...which im guessing means they want you to use the sin, cosin rules along with a TI calculator or somethin.

I think this should be treated as a square inside a problem, the clue is the statement "sensor surface been a complete circle".

[da_beast]

OK,

but can you assume the dimensions of the triangle are half the width/height? I could see half the height being possible, but the triangle doesnt seem to be half the width of the figure.

For the triangle i used this

You know what? I'm not sure what I'm supposed to be looking at in the second picture of diagram 2. Can you explain what's going on there? It is part of the first picture somehow?

This is a past paper on mathematics for com*****g. Its supposed to be a projector thats activated by the sensor shown.

The bottom part of the picture is like a birdseye view of the sensor from above. The only relevant figure i can see in that diagram is the 2.54 which im gives you the width of the rectangle in the middle of the sensor.

Using that info i can use the general 4.195 with and subtract the width of the rectangle (2.54) this gives me 1.655.

This 1.655 is the width of both arcs combined so 0.8275 sis the width of each circle

Using this i can calculate the area of each circle but it doesnt show me the working or the formula.

The rectangle with 1 arc included is 17 squared. -9.02 = 7.8 for the area of one circle

Grrrrr. that cant be right as there is no way its only 1.4 smaller than the rectangle.

I cant do this lol