A rectangle has an area of 120cm^2 and a diagonal that is 17cm long. Let the length and breath of the rectangle be x and y respectively. What is the length and breath of the rectangle? Find it using Simultaneous Equation.

1 Answer
Jim G.
Jan 18, 2018

#15" and "8" cm"#

Explanation:

#"the diagonal 'splits' the rectangle into 2 right triangles"#

#"using "color(blue)"Pythagoras' theorem "#

#x^2+y^2=17^2#

#"area of rectangle "=xy=120#

#"thus we have 2 equations"#

#x^2+y^2=289to(1)#

#xy=120to(2)#

#"from equation "(2)toy=120/xto(3)#

#"substitute "y=120/x" into equation "(1)#

#rArrx^2+(120/x)^2=289#

#rArrx^2+14400/x^2=289#

#"multiply through by "x^2#

#rArrx^4+14400=289x^2#

#rArrx^4-289x^2+14400=0#

#"let "u=x^2#

#rArru^2-289u+14400=0larrcolor(blue)"in standard form"#

#"with "a=1,b=-289" and "c=14400#

#"solve for u using the "color(blue)"quadratic formula"#

#u=(289+-sqrt(83521-57600))/2=(289+-161)/2#

#rArru=(289-161)/2" or "u=(289+161)/2#

#rArru=64" or "u=225#

#u=x^2rArrx=sqrtu#

#rArrx=sqrt64=8" or "x=sqrt225=15#

#"substitute these values into equation "(3)#

#x=8toy=120/8=15#

#x=15toy=120/15=8#

#"length can be "15" or "8#

#"breadth can be " 8" or "15#

#rArr15xx8" or "8xx15" are the dimensions"#

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