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alliemeade1
17.12.2020 •
Mathematics
PLEASE HELP QUICK WILL MARK BRAINLIEST
4) There is a square and a rectangle. The length of the rectangle is two centimeters more
than the sides of the square. The width of the rectangle is one centimeter less than the
square's sides. The sum of the areas of the two figures is thirty-four square centimeters,
Find the dimensions of each figure.
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Ответ:
given/equations formed from text:
I: l=w+3
II: A=l*w
III: new area B=(l-2)*w
IV: B=A-30
substitute l in II with I to remove l:
II': A=l*w=
(w+3)*w=
w^2+3w
substitute l in III with I to remove l:
III': B=(l-2)*w=
(w+3-2)*w=
(w+1)*w=
w^2+w
substitute A and B from II' and III' into IV:
B=A-30
w^2+w=w^2+3w-30
30=2w
w=15
insert w=15 into I:
l=15+3=18
->
A) dimensions of full size rectangle are width 15 and length 18
dimensions of reduced size rectangle are width 15 and length 16
B) full size: 15*18=10*18+5*18=180+90=270cm^2
reduced size: 15*16=10*16+5*16=160+80=240cm^2
THANKS
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