QR - Geometry

[belmont3]

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Which of the following pairs could represent the length and width of a rectangle that has an area of 7 and a perimeter of 12?

a) (3 + [square root of 2] , 3 - [square root of 2])
b) (6 + [square root of 5] , 6 - square root of 5])
c) (3, 3)
d) [square root of 12]+ [square root of 7] , [square root of 12] - [square root of 7])
e) None of the above.

Thank you!!

 

[mel5142]

l*w=7 (area)
2l + 2w = 12
substitute: l =7/w
2(7/w) + 2w = 12
multiply both sides by w
14 + 2w^2 = 12w
so 2w^2 -12w + 14=0
divide both sides by 2
w^2 - 6w +7 =0
can't factor so use quadratic formula
=[6 + or - sq rt (36-(4*1*7)]/2= [6+ or - sq rt (36-28)]/2= [6+sq rt(8)]/2 or [6-sq rt (8)]/2
but sq rt of 8 is 2 sq rt(2) so:
[6+ 2*sqrt(2)]/2 or [6-2*sqrt(2)]/2
so 3+sqrt(2) or 3-sqrt(2) could be one of the sides

I have only figured two possible values for one of the sides.
Come on, they are not going to test the quadratic formula on the exam.

BUT, the trick to this problem is to see that l*w=7. so which of these answers for l and w multiplied times each other = 7.
Note that:
in a., you get l*w = 9-2 =7
in b., you get 36-5 = 31
in c., you get 3*3 = 9
in d. you get 12-7=5

so a. looks like a good answer but then does 2L + 2 W =12
2[3+sq rt (2)] + 2[3-sq rt (2)] where sq rt of 2 is @ 1.4 = 2(4.4)+2(1.6)= 8.8+3.2=12 , so correct answer is A.

 

[mel5142]

I failed to mention that: notice you don't even need to approximate the square root of 2. Just note that :
2[3+ sqrt(2)] + 2[3- sqrt(2)] = 6+2 sqrt(2) +6-2 sqrt(2) = 12 because the 2sqrt(2) terms cancel since one is positive and the other negative so problem just becomes 6+6 =12!!

 

[mel5142]

Also something u need to know in this problem is that sqrt (x) * sqrt (x) = just "x".

 

[belmont3]

Also something u need to know in this problem is that sqrt (x) * sqrt (x) = just "x".

Got it 🙂. Thank you so much!!!

 

[mel5142]

np