Cyniez@lemmy.world to Asklemmy@lemmy.ml · 17 hours agoCan anyone solve this math for me?lemmy.worldimagemessage-square38fedilinkarrow-up1120arrow-down117
arrow-up1103arrow-down1imageCan anyone solve this math for me?lemmy.worldCyniez@lemmy.world to Asklemmy@lemmy.ml · 17 hours agomessage-square38fedilink
minus-squareTransient Punk@sh.itjust.workslinkfedilinkEnglisharrow-up46arrow-down1·edit-217 hours agoX=15
minus-squareRentlar@lemmy.calinkfedilinkarrow-up61·edit-216 hours agoNote that the problem states that the outer shape is a quarter circle, information not provided in OP’s question. Knowing it is a quarter circle is important because it allows us to validate that the bottom-right angle is 90 degrees.
minus-squareflicker@lemmy.dbzer0.comlinkfedilinkEnglisharrow-up3·8 hours agoI actually came to the comments to see if we had this information! Thanks.
minus-squaregezginorman@lemmy.mllinkfedilinkEnglisharrow-up2·14 hours agobut does it have to be a given, or can we actually prove that it has to be
minus-squarebstix@feddit.dklinkfedilinkarrow-up16·13 hours agoIt has to be given, otherwise there would be infinitely many solutions. You would need some other information to link the line segment X to the rest of the figure.
minus-squareocean@lemmy.selfhostcat.comlinkfedilinkEnglisharrow-up1·9 hours agoTats so cool! Did you just do that or find it?
minus-squareolosta@lemmy.worldlinkfedilinkarrow-up5arrow-down1·16 hours agoThe explanation don’t explain why AE must be a diameter of the circle. What makes that obvious?
minus-squarekambusha@sh.itjust.workslinkfedilinkarrow-up4·14 hours agoThanks. I had the same Q: https://en.m.wikipedia.org/wiki/Thales’s_theorem
minus-squaretehn00bi@lemmy.worldlinkfedilinkarrow-up3·14 hours agoHow have I had years of math and not seen this? I mean, it’s not super useful for me, but I would have thought I would have seen something like this in pre calculus at least.
minus-squaremaxprime@lemmy.mllinkfedilinkarrow-up1·8 hours agoI teach this to my grade 9 class in Canada. It’s on the curriculum.
minus-squaretehn00bi@lemmy.worldlinkfedilinkarrow-up1·3 hours agoNice. I have no recollection of seeing this before.
minus-squaremaxprime@lemmy.mllinkfedilinkarrow-up1·2 hours agoTbf most Canadian grade 9 teachers skip it.
X=15
Note that the problem states that the outer shape is a quarter circle, information not provided in OP’s question.
Knowing it is a quarter circle is important because it allows us to validate that the bottom-right angle is 90 degrees.
I actually came to the comments to see if we had this information! Thanks.
but does it have to be a given, or can we actually prove that it has to be
It has to be given, otherwise there would be infinitely many solutions.
You would need some other information to link the line segment X to the rest of the figure.
Tats so cool! Did you just do that or find it?
The explanation don’t explain why AE must be a diameter of the circle. What makes that obvious?
Thales’ Theorem
Thanks. I had the same Q: https://en.m.wikipedia.org/wiki/Thales’s_theorem
How have I had years of math and not seen this? I mean, it’s not super useful for me, but I would have thought I would have seen something like this in pre calculus at least.
I teach this to my grade 9 class in Canada. It’s on the curriculum.
Nice. I have no recollection of seeing this before.
Tbf most Canadian grade 9 teachers skip it.
Geometry, class six or seven.
Wow, that’s cool
Ooh clever