Question

Explanation please? (will medal and fan)

What are the values of x and y?
Pic below....

Answer 1

Except for the fact it is a right triangle, ignore the interior. It has hypotenuse length 12 + 16 = 28. So Pythagorus says the sum of the squares of the other two sides must equal 28 x 28 = 784. Try the given combo's to see which x and y match. My money's on the last choice. :-)

Answer 2

Hmm... I'm not too sure if I understand. Could you possibly elaborate a little more? (I'm sorry, I'm just trying to /fully/ understand the concept and really appreciate your help!)

Answer 3

Forget the distracting stuff inside the triangle
\[28^{2} = 784 = 336 + 448 =\]\[ (16 \times 21) + (64 \times7) = (4 \sqrt{21})^{2} + (8 \sqrt{7})^{2} \]

Answer 4

So you got 16 x 21 and 64 x 7 by just guessing?

Answer 5

like, inputting numbers until they equaled 336 and 448?

Answer 6

No, I tried the x and y combo's as given forming \[x^{2}+y^{2}\]until 784 turned up. :-)

Answer 7

Say the first choice, \[x = 2 \sqrt{3}\]and\[ y = 4\]gives \[(2 \sqrt{3})^{2}+(4)^2= (4 \times 3) + 16 = 28\]but that's not what we want. We want ( because of Pythagorus' Thereom ) 28 x 28 = 784

Answer 8

Would the same thing apply to a problem like this? (asking for 2 different variables again)

Answer 9

Yes indeed. But they are putting extraneous information in, so part of solving the problem is working out what to concentrate on. For this second question you want to focus on only part of the diagram : hence you want to solve\[6^{2} + a^{2} = b^{2}\]or\[b^{2} - a^{2} = 36\]thus again by trial and error, you find that the first choice works\[(15/2)^{2}-(9/2)^{2}=225/4 - 81/4 = 144/4 = 36\]

Answer 10

Oh! I see what they're trying to do there.. I thought the other information was relavent- guess not!
Thanks for all your help :-)