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#1
September 6th 16, 02:43 AM
posted to sci.astro.amateur
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Math!
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#2
September 6th 16, 03:43 AM
posted to sci.astro.amateur
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Math!
On Monday, September 5, 2016 at 6:43:28 PM UTC-7, StarDust wrote:
https://qph.ec.quoracdn.net/main-qim...to_webp=t rue This can be 'roughly' calculated using basic math. OK, the large square has an area of 50 cm^2, so the length of its side is the square root of 50. Since 7 squared is 49, the length of that side must be a little more than 7. The area of the smaller square is 18 cm^2, so the length of its side is the square root of 18. 4 squared is 16, so the length of that side is a little more than 4. Subtracting a little more than 4 from a little more than 7 tells us that the answer is somewhere near 3. Which of those available answers is about 3? This should be very apparent. Hint: the square root of 2 is about 1.414... \Paul A
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#3
September 6th 16, 10:22 AM
posted to sci.astro.amateur
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Math!
On 06/09/2016 03:43, palsing wrote:
On Monday, September 5, 2016 at 6:43:28 PM UTC-7, StarDust wrote: https://qph.ec.quoracdn.net/main-qim...to_webp=t rue This can be 'roughly' calculated using basic math. OK, the large square has an area of 50 cm^2, so the length of its side is the square root of 50. Since 7 squared is 49, the length of that side must be a little more than 7. The area of the smaller square is 18 cm^2, so the length of its side is the square root of 18. 4 squared is 16, so the length of that side is a little more than 4. Subtracting a little more than 4 from a little more than 7 tells us that the answer is somewhere near 3. Which of those available answers is about 3? This should be very apparent. Hint: the square root of 2 is about 1.414... But it can be *exactly* calculated by noting that 50 and 18 are suspiciously both multiples of exact squares 25=5^2, 9=3^2. Like most relativity questions these sorts of puzzles usually involve either a 3,4,5 triangle or a 5,12,13 one. -- Regards, Martin Brown
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#4
September 6th 16, 11:52 AM
posted to sci.astro.amateur
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Math!
On Tuesday, September 6, 2016 at 2:22:41 AM UTC-7, Martin Brown wrote:
On 06/09/2016 03:43, palsing wrote: On Monday, September 5, 2016 at 6:43:28 PM UTC-7, StarDust wrote: https://qph.ec.quoracdn.net/main-qim...to_webp=t rue This can be 'roughly' calculated using basic math. OK, the large square has an area of 50 cm^2, so the length of its side is the square root of 50. Since 7 squared is 49, the length of that side must be a little more than 7. The area of the smaller square is 18 cm^2, so the length of its side is the square root of 18. 4 squared is 16, so the length of that side is a little more than 4. Subtracting a little more than 4 from a little more than 7 tells us that the answer is somewhere near 3. Which of those available answers is about 3? This should be very apparent. Hint: the square root of 2 is about 1.414... But it can be *exactly* calculated by noting that 50 and 18 are suspiciously both multiples of exact squares 25=5^2, 9=3^2. Like most relativity questions these sorts of puzzles usually involve either a 3,4,5 triangle or a 5,12,13 one. -- Regards, Martin Brown Yes! Solution is - G -!
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