the length and width of a rectangle are (x+4) cm and x cm respectively. A square has the same perimeter as the rectangle. If you add the areas of the rectangle and the square the results is 94 cm^2. solve for x.
what do u mean by adding the areas of the rectangle and the square
hmm what do you think is the perimeter of the RECTANGLE? keep in mind that, length = x+4 and width = x
hmm.... well.. is not 4... lemme poke a bit
what do u mean by adding the areas of the rectangle and the square
The problem says if you add it you get 94 cm^2
\(\bf \textit{perimeter of the rectangle}=2l+2w=2(x+4)+2(x)\\ \quad \\ 2x+8+2x\implies 4x+8 = \textit{perimeter of the square}\)
so now we know that the perimeter of the rectangle, which equals to the perimeter of the square is 4x +8 now recall that a square has 4 EQUAL SIDES, that means, if we divide the perimeter by 4 we should get 1 side by itself
so.... what would you get for 1 side? in terms of "x" that is
\(\bf \cfrac{4x+8 }{4} \implies \square ?\)
3
\(\bf \cfrac{4x+8 }{4} \implies 3?\)
yes?
well... ahemm we dunno what "x" is for one, and the "x" wen MIA in your answer
went MIA == missing in action =)
x=3 ?
is it? well... we really dunno what "x" is, if we did... then we can just plug it in the original equation and use that, but ... we dunno what "x" is
\(\bf \cfrac{4x+8 }{4} \implies \cfrac{4x}{4}+\cfrac{8}{4}\implies \square ?\)
is not a number, just an expression in terms of "x"
this is what u do...
the perimeter is the same as that of the square. that gives us the following eqn; 2(x+4)+2x=4l where l=length of side of square
and the sum of the squares equals 94 leaves us with a 2nd eqn, which is; l^2+x(x+4)=94
the perimeter of the rectangle is the same as the one for the square you use the perimeter of the rectangle, which is the square's divide it by 4, you get 1 side for the square, all 4 sides are equal in a square once you have 1 side for the square, side * side is the area of the square, and add that to the rectangle's area, which is (x+4) * x
simplifying the 2 eqns, we are left with the following; 4x-4l=-8 x^2+4x+l^2=94
so it would stay like that?^^
or it has to be solved
making x the subject in the 1st eqn, we are left with x=-2+l
substitute that into the 2nd eqn and get the ff; l=7
now that u have l, substitute l=7 into the 1st eqn. u should have the following 4x-4(7)=-8 4x-28=-8 4x=28-8 4x=20 x=5
are u ok with it now
yes so the x is 5
yes
I got it now Thanks !
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