Can you solve these math problems?

[akubi]

Originally posted by: chuckywang
2) The answer is 103. Man, I had to go through a lot of work to get that, so I don't think my way was the best way. What's the shortcut in this problem, Vertimus?

nah it's 110.

it's easy:

let BC be x, CD by y.

now the triangles ABC and BCD are similar so you have the proportion:
sqrt(11)/x = x/y

so y = x^2/sqrt(11)

also on the other side of the trapezoid... drop down A to point E such that ABCE is a rectangle. (assume AB longer than CD, if not flip).

then... AE^2 + DE^2 = x^2 + DE^2 = AD^2 = 1001 (pyth)
solve for DE^2 = sqrt(1001 - x^2)

so..... sqrt(1001-x^2) + y = sqrt(11). plug in for y.
sqrt(1001-x^2) + x^2/sqrt(11) = sqrt(11).
move over the x^2 term to other side and square both sides. get fourth degree polynomial but it factors nicely to (x^2 - 110) (x^2 + 99) = 0.

obviously you want positive answer so x^2 = 110

 

[chuckywang]

Originally posted by: akubi

Originally posted by: chuckywang
2) The answer is 103. Man, I had to go through a lot of work to get that, so I don't think my way was the best way. What's the shortcut in this problem, Vertimus?

nah it's 110.

it's easy:

let BC be x, CD by y.

now the triangles ABC and BCD are similar so you have the proportion:
sqrt(11)/x = x/y

so y = x^2/sqrt(11)

also on the other side of the trapezoid... drop down A to point E such that ABCE is a rectangle. (assume AB longer than CD, if not flip).

then... AE^2 + DE^2 = x^2 + DE^2 = AD^2 = 1001 (pyth)
solve for DE^2 = sqrt(1001 - x^2)

so..... sqrt(1001-x^2) + y = sqrt(11). plug in for y.
sqrt(1001-x^2) + x^2/sqrt(11) = sqrt(11).
move over the x^2 term to other side and square both sides. get fourth degree polynomial but it factors nicely to (x^2 - 110) (x^2 + 99) = 0.

obviously you want positive answer so x^2 = 110

You're right....I made a stupid arithmetic mistake when I, for some reason, had 81^2 as 898 instead of 891, hence the difference of 7 between your correct answer and mine.