Find the original dimension of the rectangle












0












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A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle? Explain your answer.






geometry







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  • 2




    $begingroup$
    Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
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    – drhab
    Sep 13 '14 at 10:08










  • $begingroup$
    if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
    $endgroup$
    – Vikram
    Sep 13 '14 at 10:14


















0












$begingroup$


A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle? Explain your answer.






geometry







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:08










  • $begingroup$
    if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
    $endgroup$
    – Vikram
    Sep 13 '14 at 10:14
















0












0








0





$begingroup$


A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle? Explain your answer.






geometry







share|cite|improve this question









$endgroup$




A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle? Explain your answer.






geometry




geometry






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 13 '14 at 10:05









Adam LewisAdam Lewis

12




12








  • 2




    $begingroup$
    Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:08










  • $begingroup$
    if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
    $endgroup$
    – Vikram
    Sep 13 '14 at 10:14
















  • 2




    $begingroup$
    Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:08










  • $begingroup$
    if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
    $endgroup$
    – Vikram
    Sep 13 '14 at 10:14










2




2




$begingroup$
Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
$endgroup$
– drhab
Sep 13 '14 at 10:08




$begingroup$
Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Do not use expressions like: "explain your answer". We are not here to do your homework but are eager to teach you some maths.
$endgroup$
– drhab
Sep 13 '14 at 10:08












$begingroup$
if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
$endgroup$
– Vikram
Sep 13 '14 at 10:14






$begingroup$
if the "inches" is troubling you, note that all the dimensions are in inches and hence for this problem you can drop it. You must know that area of a rectangle = (length) x (width)
$endgroup$
– Vikram
Sep 13 '14 at 10:14












2 Answers
2






active

oldest

votes


















0












$begingroup$

Hint:



Denote its original width by $x$ and find an expression in $x$ for the area of the reformed rectangle. It should equalize $60$ so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:23












  • $begingroup$
    Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:30










  • $begingroup$
    Okay I think I got it w=4?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:45










  • $begingroup$
    That is correct. Well done!
    $endgroup$
    – drhab
    Sep 13 '14 at 10:50



















0












$begingroup$

$$ Width=a, Length=4a , $$
$$text {so the total original area is: } A=4a^2$$
$$text {now we know that the new area is}$$
$$ A'=(4a+4)(a-1)=4a^2-4a+4a-4=4a^2-4=60$$
$$text{so } 4a^2=A=64$$
$$ a^2=64/4=16$$
$$ a=4$$
$$text{so} Width=4text{ and } Length=16$$
$$text{(all expressed in inches)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:54










  • $begingroup$
    sorry, just edited now
    $endgroup$
    – Mosk
    Sep 13 '14 at 11:06











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Hint:



Denote its original width by $x$ and find an expression in $x$ for the area of the reformed rectangle. It should equalize $60$ so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:23












  • $begingroup$
    Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:30










  • $begingroup$
    Okay I think I got it w=4?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:45










  • $begingroup$
    That is correct. Well done!
    $endgroup$
    – drhab
    Sep 13 '14 at 10:50
















0












$begingroup$

Hint:



Denote its original width by $x$ and find an expression in $x$ for the area of the reformed rectangle. It should equalize $60$ so...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:23












  • $begingroup$
    Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:30










  • $begingroup$
    Okay I think I got it w=4?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:45










  • $begingroup$
    That is correct. Well done!
    $endgroup$
    – drhab
    Sep 13 '14 at 10:50














0












0








0





$begingroup$

Hint:



Denote its original width by $x$ and find an expression in $x$ for the area of the reformed rectangle. It should equalize $60$ so...






share|cite|improve this answer









$endgroup$



Hint:



Denote its original width by $x$ and find an expression in $x$ for the area of the reformed rectangle. It should equalize $60$ so...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 13 '14 at 10:12









drhabdrhab

103k545136




103k545136












  • $begingroup$
    so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:23












  • $begingroup$
    Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:30










  • $begingroup$
    Okay I think I got it w=4?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:45










  • $begingroup$
    That is correct. Well done!
    $endgroup$
    – drhab
    Sep 13 '14 at 10:50


















  • $begingroup$
    so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:23












  • $begingroup$
    Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:30










  • $begingroup$
    Okay I think I got it w=4?
    $endgroup$
    – Adam Lewis
    Sep 13 '14 at 10:45










  • $begingroup$
    That is correct. Well done!
    $endgroup$
    – drhab
    Sep 13 '14 at 10:50
















$begingroup$
so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
$endgroup$
– Adam Lewis
Sep 13 '14 at 10:23






$begingroup$
so it'll be like 60=lw 60=(4w+4)(w-1) cause the length is 4x the width and they added 4 to the "new length" then it's w-1 cause they deducted 1 from the "new width". Is my equation right or am I in the wrong path?
$endgroup$
– Adam Lewis
Sep 13 '14 at 10:23














$begingroup$
Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
$endgroup$
– drhab
Sep 13 '14 at 10:30




$begingroup$
Your equation is correct. The second step is to solve it. Keep in mind that $w$ cannot be negative here.
$endgroup$
– drhab
Sep 13 '14 at 10:30












$begingroup$
Okay I think I got it w=4?
$endgroup$
– Adam Lewis
Sep 13 '14 at 10:45




$begingroup$
Okay I think I got it w=4?
$endgroup$
– Adam Lewis
Sep 13 '14 at 10:45












$begingroup$
That is correct. Well done!
$endgroup$
– drhab
Sep 13 '14 at 10:50




$begingroup$
That is correct. Well done!
$endgroup$
– drhab
Sep 13 '14 at 10:50











0












$begingroup$

$$ Width=a, Length=4a , $$
$$text {so the total original area is: } A=4a^2$$
$$text {now we know that the new area is}$$
$$ A'=(4a+4)(a-1)=4a^2-4a+4a-4=4a^2-4=60$$
$$text{so } 4a^2=A=64$$
$$ a^2=64/4=16$$
$$ a=4$$
$$text{so} Width=4text{ and } Length=16$$
$$text{(all expressed in inches)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:54










  • $begingroup$
    sorry, just edited now
    $endgroup$
    – Mosk
    Sep 13 '14 at 11:06
















0












$begingroup$

$$ Width=a, Length=4a , $$
$$text {so the total original area is: } A=4a^2$$
$$text {now we know that the new area is}$$
$$ A'=(4a+4)(a-1)=4a^2-4a+4a-4=4a^2-4=60$$
$$text{so } 4a^2=A=64$$
$$ a^2=64/4=16$$
$$ a=4$$
$$text{so} Width=4text{ and } Length=16$$
$$text{(all expressed in inches)}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:54










  • $begingroup$
    sorry, just edited now
    $endgroup$
    – Mosk
    Sep 13 '14 at 11:06














0












0








0





$begingroup$

$$ Width=a, Length=4a , $$
$$text {so the total original area is: } A=4a^2$$
$$text {now we know that the new area is}$$
$$ A'=(4a+4)(a-1)=4a^2-4a+4a-4=4a^2-4=60$$
$$text{so } 4a^2=A=64$$
$$ a^2=64/4=16$$
$$ a=4$$
$$text{so} Width=4text{ and } Length=16$$
$$text{(all expressed in inches)}$$






share|cite|improve this answer











$endgroup$



$$ Width=a, Length=4a , $$
$$text {so the total original area is: } A=4a^2$$
$$text {now we know that the new area is}$$
$$ A'=(4a+4)(a-1)=4a^2-4a+4a-4=4a^2-4=60$$
$$text{so } 4a^2=A=64$$
$$ a^2=64/4=16$$
$$ a=4$$
$$text{so} Width=4text{ and } Length=16$$
$$text{(all expressed in inches)}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 13 '14 at 11:05

























answered Sep 13 '14 at 10:44









MoskMosk

1,5324912




1,5324912












  • $begingroup$
    Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:54










  • $begingroup$
    sorry, just edited now
    $endgroup$
    – Mosk
    Sep 13 '14 at 11:06


















  • $begingroup$
    Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
    $endgroup$
    – drhab
    Sep 13 '14 at 10:54










  • $begingroup$
    sorry, just edited now
    $endgroup$
    – Mosk
    Sep 13 '14 at 11:06
















$begingroup$
Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
$endgroup$
– drhab
Sep 13 '14 at 10:54




$begingroup$
Nothing wrong with it, but the OP was asked for the dimensions of the original rectangle. Not explicitly for its area.
$endgroup$
– drhab
Sep 13 '14 at 10:54












$begingroup$
sorry, just edited now
$endgroup$
– Mosk
Sep 13 '14 at 11:06




$begingroup$
sorry, just edited now
$endgroup$
– Mosk
Sep 13 '14 at 11:06


















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