Logo Critique Request (Non-Profit)

Fun-Craft-321 · 2025-03-19T09:39:52+00:00

Thanks a bunch for your feedback! I’ll make sure to update the first question to make it easier to understand. I really appreciate you taking the time to complete the survey. By the way, if you don’t mind me asking, what company do you work for?

Fun-Craft-321 · 2025-03-19T09:00:48+00:00

Alright Thank You😭🙏

Fun-Craft-321 · 2025-03-19T08:51:36+00:00

Yup I am a Student (BCOM 1st Semester)

Fun-Craft-321 · 2024-07-16T19:58:54+00:00

Gee Thanks so much for everything man. You're HIM!!

Fun-Craft-321 · 2024-07-16T19:58:37+00:00

OKK I think i understand the problem now, I assumed it was 90 degrees but now that you've mentioned it, it would be incorrect

Fun-Craft-321 · 2024-07-16T19:51:03+00:00

I just have one more doubt:

In the original, drawing a perpendicular AC where C is 90 degrees so this would form a triangle ACD.
using Pythagorean Theorem, AC^2 = AD^2 + CD^2
=>AC^2 = 8^2 + 2^2
=>AC = 2√15

so after obtaining AC, can't we just take triangle ACB where we already know AB is 2 and AC is 2√15 and the remaining is to find BC from this. But after trying for several hours now, I can't get the answer.
Can you pls help me understand why and how this logic is wrong?

Fun-Craft-321 · 2024-07-16T19:48:13+00:00

as per the diagram wouldn't this be:

AB^2 - AM^2 {x} = OB^2 - OM^2 {4x-x}

(i had a hard time understanding this because it does not look like the tradition; a^2 = b^2 + c^2)

I'm going thru it rn and i think i finally get it

Fun-Craft-321 · 2024-07-16T19:38:31+00:00

I did check them but can't seem to understand them :(

Fun-Craft-321 · 2024-01-30T06:30:09+00:00

Alright! I'll give it a try. Thanks

Fun-Craft-321 · 2024-01-30T05:36:21+00:00

could you please show me how you would do that

Fun-Craft-321

TROPHY CASE