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Find the dimensions of a rectangle whose perimeter is 26 meters and whose area is 40 square meters.

Sagot :

luana
[tex]x,y-\ dimensions\ of\ rectangle\\\\\ Perimeter\\\\ P=2x+2y\\ P=26\\\\ 26=2x+2y\ |:2\\ 13=x+y\\ x=13-y\\\\\ Area:\\\\ A=x*y\\ A=40\\ 40=x*y\\\\ Substituting\ x=13-y\\\\ 40=(13-y)*y\\ 40=13y-y^2\\y^2-13y+40=0\\ \Delta=13^2-4*1*40=169-160=9\\ \sqrt{\Delta}=\sqrt9=3\\\\ y=\frac{13-3}{2}=\frac{10}{2}=5\ \ or y=\frac{13+3}{2}=\frac{16}{2}=8\\x=13-y\\x=13-5=8\ or\ x=13-8=5\\\\Dimensions\ are\ 5m\ and\ 8m. [/tex]
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