Beyond the course?

[Richard Feynman 101]

Just wondering if VCAA went beyond on the SD on this question

[VeryJuicyLemon]

No, why?

Use cosine rule to find the vertical length of the triangle, divide by 2 then use Pythagoras to find the horizontal length of the triangle. Add that with the radius and the length of the square = the length of pathway.

Sunrise is beyond the course 

[Richard Feynman 101]

Do you know what the sagitta is?

[VeryJuicyLemon]

Do you know what the sagitta is?

yes but you can combine rules learnt in further maths to solve it, just like majority of past geometry question ever questioned at the last page

[Richard Feynman 101]

yes but you can combine rules learnt in further maths to solve it, just like majority of past geometry question ever questioned at the last page

Mmmm. Very interesting

[AngelWings]

No, why?

Use cosine rule to find the vertical length of the triangle, divide by 2 then use Pythagoras to find the horizontal length of the triangle. Add that with the radius and the length of the square = the length of pathway.

An alternative solution:
1. the diameter of the circle is 100m, add this with the 65m from the square
2. find the overlap between the two shapes and subtract this from the length found in Step 1:
   a. use Pythagoras' theorem using 50m and (65/2)m to find the distance between the centre of the circle to where it meets the square (You can do this because parallel lines from square and corresponding angles gives right-angled triangle.)
   b. radius distance minus distance between the centre of the circle to where it meets the square

This sort of thing uses a little out of the box thinking, but isn't unusual, especially given that they need to separate people at the top end. The use of out of the box thinking usually happens at the end of Methods Exam 1, but has on rare occasions appeared in Further Exam 2 to intentionally stump people. Good news is now you're prepared for this type of question.