How can I solve for X?

What I did was solve for C, and then took (6/6+2)*C. It seems to work but I am not completely sure. Is there a better way?

enter image description here


closed as off-topic by Apep, Glorfindel, Ankoganit, APrough, Mithrandir Nov 30 '17 at 16:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Apep, Glorfindel, Ankoganit, APrough, Mithrandir
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What is it that leaves you unsatisfied with what you've done so far? $\endgroup$ – Gareth McCaughan Nov 30 '17 at 14:26
  • $\begingroup$ And: why (given that you are "not completely sure") did you do that particular calculation rather than some other? $\endgroup$ – Gareth McCaughan Nov 30 '17 at 14:28
  • $\begingroup$ I guess it works but I am not capable of formulating a proof. Was also wondering if there were any alternative formulas to use. $\endgroup$ – amikic Nov 30 '17 at 14:29
  • $\begingroup$ It's all I could think of....I'm not a math genius, hence why I am here. $\endgroup$ – amikic Nov 30 '17 at 14:31
  • 1
    $\begingroup$ I think you shouldn't be thinking in terms of "what formula can I find that will solve this?" but in terms of "what do I know about triangles in general, and this particular triangle, that will help?". So first of all you found C (you didn't say how, but there's an obvious way which I guess you used). Very sensible. But then why -- I'm not looking for a formalized proof but for any kind of explanation -- the particular calculation you did next? $\endgroup$ – Gareth McCaughan Nov 30 '17 at 14:33

Well ,

Both outer and inner triangles are similar (AAA similarity) Therefore , $\frac{6}{6+2} = \frac{x}{7}$ . i.e., x = $\frac{21}{4}$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.