S.o.s yr12 cambridge qst! In the hood prank (gone wrong) (1 Viewer)

[Wonderlay]

Please see the attached pic, I'm referring to pt (b)
This question is from 12Cambridge2U and is under the Financial Series/Sequence chapter,
tho the question (b) is more "thinking outside the box"

I'm slumped, there was a post from 2006 about the same question but no answer was posted
Would be greatly appreciated if you can solve this!

 

[1729]

Please see the attached pic, I'm referring to pt (b)
This question is from 12Cambridge2U and is under the Financial Series/Sequence chapter,
tho the question (b) is more "thinking outside the box"

I'm slumped, there was a post from 2006 about the same question but no answer was posted
Would be greatly appreciated if you can solve this!

View attachment 34095

 

[1729]

Although perhaps there is a much more elegant method.

eg. (alternatively) \alpha and \beta can be used to find the lengths of the legs of the triangle made by two adjacent sides of the rectangle and the row adjacent to the diagonal. In which Pythagoras' Theorem can then be applied to find the desired length (which would be the triangle's hypotenuse).

Also note: although there are two rows adjacent to the diagonal, they are equal in length, by symmetry.

 

[Wonderlay]

Thank you!
The method was genius

 

[seanieg89]

If you want to avoid using angles, an alternate method is as follows:

Let d be the length of the diagonal, and let p be the length of the perpendicular from a vertex to the diagonal that does not pass through it.

By considering the area of the triangles the diagonal bisects the rectangle into, we have pd=7500.

Now by similar triangles, the length of the row adjacent to the diagonal is (p-3)d/p=d-3d/p=d-3d^2/7500=d-d^2/2500.

Just chuck in the d=125 you get from Pythagoras to complete.

 

[Wonderlay]

Thank you! That works brilliantly