Cyniez@lemmy.world to Asklemmy@lemmy.ml · 2 months agoCan anyone solve this math for me?lemmy.worldimagemessage-square43fedilinkarrow-up1138arrow-down119
arrow-up1119arrow-down1imageCan anyone solve this math for me?lemmy.worldCyniez@lemmy.world to Asklemmy@lemmy.ml · 2 months agomessage-square43fedilink
minus-squareTransient Punk@sh.itjust.workslinkfedilinkEnglisharrow-up49arrow-down1·edit-22 months agoX=15
minus-squareRentlar@lemmy.calinkfedilinkarrow-up67·edit-22 months 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-up3·1 month agoI actually came to the comments to see if we had this information! Thanks.
minus-squaregezginorman@lemmy.mllinkfedilinkEnglisharrow-up2·2 months agobut does it have to be a given, or can we actually prove that it has to be
minus-squarebstix@feddit.dklinkfedilinkarrow-up17·2 months 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-squareolosta@lemmy.worldlinkfedilinkarrow-up5arrow-down1·2 months agoThe explanation don’t explain why AE must be a diameter of the circle. What makes that obvious?
minus-squarekambusha@sh.itjust.workslinkfedilinkarrow-up4·2 months agoThanks. I had the same Q: https://en.m.wikipedia.org/wiki/Thales’s_theorem
minus-squaretehn00bi@lemmy.worldlinkfedilinkarrow-up3·2 months 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-up1·1 month agoI teach this to my grade 9 class in Canada. It’s on the curriculum.
minus-squaretehn00bi@lemmy.worldlinkfedilinkarrow-up1·1 month agoNice. I have no recollection of seeing this before.
minus-squaremaxprime@lemmy.mllinkfedilinkarrow-up1·1 month agoTbf most Canadian grade 9 teachers skip it.
minus-squareocean@lemmy.selfhostcat.comlinkfedilinkEnglisharrow-up1·2 months agoTats so cool! Did you just do that or find 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.
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.
Geometry, class six or seven.
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.
Wow, that’s cool
Ooh clever
Tats so cool! Did you just do that or find it?