Find the dimensions of a rectangular enclosure with perimeter 40 yd and are 91 yd^2
Let the length be x. Width = Perimeter /2 - x = 20 - x. Area = length x width = (20 - x) = 20x - x^2 = 91. x^2 - 20x + 91 = 0. (x - 7) (x -13) = 0. x = 7 or 13. ANSWER: Length = 13yds Width = 7 yds.
thank you
You're welcome!! Xo!
wait so its a rectangle so perimeter/2 how do you know it gives you the length not the width or does it matter
doesn't matter
Im slightly confused still can you go through all of the steps i have a huge test tomarrow lol
ok so a rectangle has 4 sides, correct? 2 sides for width and 2 for length.
yes.
so perimeter should be equal to 2 width + 2 length
yea.
and area is equal to length times width
well yea i just dont know what to do with the numbers
ok well think about it this way…. whats half of the perimeter? L+W=20 right?
so if you try to solve for width you'll get W = 20-L
so say perimeter equals 34ft and area equals 60ft^2 width=perimeter/2-x again?
so length is = L and width is = 20-L if we fill this into our area formula we get…. area = L(20-L) we know area is 91 yrds^2 91 =L(20-L)
yes, you're correct… the reason why we divide it is so we don't have to work with 2 sides, as long as you remember we have used half the perimeter, its fine.
so 60=17l-l^2 for the problem i just said
yes
and then 60/17 leaves me with 3.5=l-l^2?
no, you can't do that… if you're dividing you have to be able to take it from all terms on each side…. bring your 60 over to the other side like this… 0 = -60+17L-L^2
then factor to get (12+L)(7+L)??
right
close.. what adds to give you -17 and multiplies to give you +60?
12 and 5 i mean
they would have to be negative… remember the - in front of L^2
but you are correct (L-5)(L-12)
okay thank you so much i think i understand now
great! good luck on your test. Practice really makes perfect with this stuff.
one more question if it is not factorable can i use the quadratic formula
yes
Join our real-time social learning platform and learn together with your friends!