SOLUTION: Find the dimensions of a rectangle whose area is 216 cm2 and whose perimeter is 60 cm

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Question 752687: Find the dimensions of a rectangle whose area is 216 cm2 and whose perimeter is 60 cm
Found 2 solutions by MathLover1, tommyt3rd:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
area is A=216cm%5E2 and
perimeter is P=60cm
if the length is L and the width is W, then
L%2AW=216cm%5E2......eq.1 and
2%28L%2BW%29=60cm.....eq.2
__________________________
start with 2%28L%2BW%29=60cm.....eq.2 and solve for L
L%2BW=60cm%2F2
L%2BW=30cm
L=30cm-W.....plug it in eq. 1 and solve for W
%2830cm-W%29b=216cm%5E2......eq.1
30%2AWcm-W%5E2=216cm%5E2
0=W%5E2-30%2AWcm%2B216cm%5E2 or
W%5E2-30%2AWcm%2B216cm%5E2=0 ....use quadratic formula

W+=+%28-%28-30%29+%2B-+sqrt%28+%28-30%29%5E2-4%2A1%2A216+%29%29%2F%282%2A1%29+

W+=+%2830+%2B-+sqrt%28900-864+%29%29%2F2+

W+=+%2830+%2B-+sqrt%2836+%29%29%2F2+
W+=+%2830+%2B-+6+%29%2F2+
solutions;
W+=+%2830+%2B6+%29%2F2+
W+=+36%2F2+
W+=+18cm+
or
W+=+%2830+-6+%29%2F2+
W+=+24%2F2+
W+=+12cm+

now find L
L=30cm-W
if W+=+18+cm => L=30cm-18cm => L=12cm
if W+=+12cm+ => L=30cm-12cm => L=18cm
since the length is usually greater we will say that L=18cm and W+=+12cm+



Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
l%2Aw=216



2%28l%2Bw%29=60 therefore l%2Bw=30

l%2830-l%29=216

l%5E2-30%2Al%2B216=0

l=12 or l=18

so l=12 and w=18


:)