Laura Corin Posted September 24, 2009 Share Posted September 24, 2009 'A rectangle has a diagonal of 12cm and its long sides are twice the length of its short sides. What are the lengths of the sides of the rectangle?' Calvin gets as far as: H squared = x squared + y squared 144 = x squared + (2x)squared But he can't get to the number with algebra. He tried various values to check whether x was a whole number, and it's not. The answer is 5.37 and 10.73, but how does he get to it? My algebra is more than rusty... Thank you Laura Quote Link to comment Share on other sites More sharing options...
8filltheheart Posted September 24, 2009 Share Posted September 24, 2009 I'm not sure why he isn't getting the correct answer. (though I am not sure why he used a y) s=short side 2s=long side B/c a^2+b^2=c^2 s^2 + (2s)^2= 144 s^2+ 4s^2=144 5s^2=144 s^2=28.8 square root both sides and s= 5.37 (Only thing I can think of is that he didn't square his 2???) Quote Link to comment Share on other sites More sharing options...
LinRTX Posted September 24, 2009 Share Posted September 24, 2009 Everything you have so far is correct. 144= x squared + (2x) squared. Now (2x) squared is 4(x squared), so we now have 144 = x squared + 4 x squared 144 = 5 x squared 144/5 = x squared 12/ (square root of 5) = x I don't have my calculato, but that will be less than 12/2 so the answer of 5.37 seems reasonable. Linda Quote Link to comment Share on other sites More sharing options...
Halftime Hope Posted September 24, 2009 Share Posted September 24, 2009 (edited) He was on the right track. :) See below. 'A rectangle has a diagonal of 12cm and its long sides are twice the length of its short sides. What are the lengths of the sides of the rectangle?' Calvin gets as far as: H squared = x squared + y squared 144 = x squared + (2x)squared He's got it started correctly, just needs to continue: 144= x^2 + (2x)^2 144= x^2 + 4x^2 (or 4 x squared) 2x times 2x = 4x^2 144= 5x^2 Add x squared plus 4x squared 144 divided by 5 = x^2 5.37 (rounded) = x 10.73 = 2x = y But he can't get to the number with algebra. He tried various values to check whether x was a whole number, and it's not. The answer is 5.37 and 10.73, but how does he get to it? My algebra is more than rusty... Thank you Laura Edited September 24, 2009 by Valerie(TX) Quote Link to comment Share on other sites More sharing options...
Laura Corin Posted September 24, 2009 Author Share Posted September 24, 2009 I'm not sure why he isn't getting the correct answer. (though I am not sure why he used a y) The theorem is written out with x and y in his book, so those were the letters he used. Thanks again Laura Quote Link to comment Share on other sites More sharing options...
Laura Corin Posted September 24, 2009 Author Share Posted September 24, 2009 Where would this problem come in the US maths sequence? Algebra 1 or before that? Thanks again, Laura Quote Link to comment Share on other sites More sharing options...
Pamela H in Texas Posted September 24, 2009 Share Posted September 24, 2009 yes, Algebra I Quote Link to comment Share on other sites More sharing options...
lgm Posted September 24, 2009 Share Posted September 24, 2009 Where would this problem come in the US maths sequence? Algebra 1 or before that? Thanks again, Laura It's considered Pre-Algebra in NY and taught in 7th grade for both honors and on-grade level. http://www.glencoe.com/sites/common_assets/workbooks/math/Pre-AlgebraOK/papw.pdf or http://phschool.com/webcodes10/index.cfm?fuseaction=home.gotoWebCode&wcprefix=adk&wcsuffix=0099 Quote Link to comment Share on other sites More sharing options...
Halftime Hope Posted September 24, 2009 Share Posted September 24, 2009 Where would this problem come in the US maths sequence? Algebra 1 or before that? Thanks again, Laura It has come up in our pre-algebra text, and then in *every* level after that. :) Quote Link to comment Share on other sites More sharing options...
Laura Corin Posted September 24, 2009 Author Share Posted September 24, 2009 It has come up in our pre-algebra text, and then in *every* level after that. :) This text would be used by pupils aged (depending on ability/school) between about age 12 and 14 in the UK. Laura Quote Link to comment Share on other sites More sharing options...
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