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Perimeter and area of a rectangle enter number for lengthl 4 enter enter number for widthw 5 enter 2 x l 2 x w 18 l x w 20 the input is a rectangle?
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∙ 15y ago
Perimeter: 18 Area: 10 sq units
Wiki User
∙ 15y ago
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The length of a rectangle is 4 cm greater than its width The perimeter is 32 cm. Find the length of the rectangle?
w= widthw+4=length2(w+l)= perimeter2(w+w+4)=32 substitution2(2w+4)=324w+8=32 distributive property4w=32-8 subtraction property of equality4w=24w=6 division property of equalitylength is 6+4 or 10cm.
Rectangular room has width twice of its length When 6 is decreased from both length and width then its area is differed by 108 so find the width?
The width of the room is equal to twice the Length. Suppose Length = L, width = W, and A = areaW = 2L from the information in the questionNow we know area, A is equal to length times widthW*L=A, plug in 2L for W and we get 2L*L=A or 2L^2=ANext, we see that when 6 is subtracted from both length and width A becomes 108 less.So (2L-6)*(L-6)=A-108Multiply (2L-6)*(L-6) out and the result is (2L^2-18L+36) Set that equal to A-108(2L^2-18L+36)=A-108. We found out that A=2L^2 earlier so we can substitute the terms.(2L^2-18L+36)=2L^2-108. Now solve for LSubtract 2L^2 from both sides(-18L+36)=(-108)Subtract 36 from both sides(-18L)=(-144)Divide by (-18)L=8We know W=2L so W=16Now lets test our answer.16*8=128(16-6)*(8-6)=10*2=20128-20=108So the answer of L=8 and W=16 is correct.
