Mixing
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm. Find the dimensions of the lot?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I'm in grade 11 math. I need help to solve this.
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm. Find the dimensions of the lot?
So far I did this but it doesn't make sense:
$P=2l+2w$ or $44 = 2l+2w$
$A = lw$ or $44 = 2l + 2w$
$44=2l+2w$ all divided by 2 is $22=l+w$ than I isolate the l so it is: $l = 22-w$
sub $l=22-w$ into $107=lw$
$107=(22-w)w$
$107=-w^2 + 22w$
$107=-(w^2-22w)$
$107=-(w^2-22w+121-121)$
$107=-(w^2-22w+121)+121$
but I'm lost from there. Any help would be appreciated.
Thanks!
algebra-precalculus word-problem
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
I'm in grade 11 math. I need help to solve this.
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm. Find the dimensions of the lot?
So far I did this but it doesn't make sense:
$P=2l+2w$ or $44 = 2l+2w$
$A = lw$ or $44 = 2l + 2w$
$44=2l+2w$ all divided by 2 is $22=l+w$ than I isolate the l so it is: $l = 22-w$
sub $l=22-w$ into $107=lw$
$107=(22-w)w$
$107=-w^2 + 22w$
$107=-(w^2-22w)$
$107=-(w^2-22w+121-121)$
$107=-(w^2-22w+121)+121$
but I'm lost from there. Any help would be appreciated.
Thanks!
algebra-precalculus word-problem
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm in grade 11 math. I need help to solve this.
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm. Find the dimensions of the lot?
So far I did this but it doesn't make sense:
$P=2l+2w$ or $44 = 2l+2w$
$A = lw$ or $44 = 2l + 2w$
$44=2l+2w$ all divided by 2 is $22=l+w$ than I isolate the l so it is: $l = 22-w$
sub $l=22-w$ into $107=lw$
$107=(22-w)w$
$107=-w^2 + 22w$
$107=-(w^2-22w)$
$107=-(w^2-22w+121-121)$
$107=-(w^2-22w+121)+121$
but I'm lost from there. Any help would be appreciated.
Thanks!
algebra-precalculus word-problem
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm in grade 11 math. I need help to solve this.
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm. Find the dimensions of the lot?
So far I did this but it doesn't make sense:
$P=2l+2w$ or $44 = 2l+2w$
$A = lw$ or $44 = 2l + 2w$
$44=2l+2w$ all divided by 2 is $22=l+w$ than I isolate the l so it is: $l = 22-w$
sub $l=22-w$ into $107=lw$
$107=(22-w)w$
$107=-w^2 + 22w$
$107=-(w^2-22w)$
$107=-(w^2-22w+121-121)$
$107=-(w^2-22w+121)+121$
but I'm lost from there. Any help would be appreciated.
Thanks!
algebra-precalculus word-problem
algebra-precalculus word-problem
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 39 mins ago
Zebert
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 39 mins ago
Zebert
344
asked 39 mins ago
Zebert
344
344
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Zebert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago
add a comment |Â
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm." would turn into $lw=107$.
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=frac107w$$ $$implies 2cdotfrac107w+2w=44$$ $$implies frac214w+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$implies 214+2w^2=44w$$ $$implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $frac-bpmsqrtb^2-4ac2a$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac-(-44)pmsqrt(-44)^2-(4cdot 2cdot 214)2cdot 2$$ $$impliesfrac44pmsqrt2244$$ $$impliesfrac11pm 4sqrt144$$
We now divide by four getting
$$w=11pmsqrt14$$
Now let's find $l$. $107div (11pmsqrt14)=11mpsqrt14$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt14$, then the width is $11+sqrt14$. When the length of the lot is $11+sqrt14$, then the width is $11-sqrt14$.
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
If you choose to solve this by completing the square, observe that $$beginalign107 & =-underbrace(w^2-22w+121)_(w-11)^2+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tagmove 121 over\ implies 14 &= (w-11)^2 tagdivide both sides by $-1$ \implies pm sqrt14 &= w - 11,endalign$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
answered 23 mins ago
Decaf-Math
2,865822
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
add a comment |Â
up vote
0
down vote
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt 14 = w - 11$
$11 pm sqrt14 = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt14) = 11 mp sqrt14$.
So $2l + 2w = 2(11mp sqrt14) + 2(11 pm sqrt14) = 44$. So far so good.
$lw = (11 + sqrt14)(11 - sqrt14 = 11^2 + 11sqrt14 -11sqrt14 - sqrt14^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$. So the Average of $l$ and $w$ is $11$. Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide. It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = sqrt14$ (well, it could be $-sqrt14$ but we assumed $d ge 0$ when we assumed it was longer than it was wide. It doesn't make a difference.)
So $l = 11 + sqrt 14$ and $w = 11 - sqrt14$.
Just a trick.
answered 3 mins ago
fleablood
62.1k22678
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm." would turn into $lw=107$.
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=frac107w$$ $$implies 2cdotfrac107w+2w=44$$ $$implies frac214w+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$implies 214+2w^2=44w$$ $$implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $frac-bpmsqrtb^2-4ac2a$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac-(-44)pmsqrt(-44)^2-(4cdot 2cdot 214)2cdot 2$$ $$impliesfrac44pmsqrt2244$$ $$impliesfrac11pm 4sqrt144$$
We now divide by four getting
$$w=11pmsqrt14$$
Now let's find $l$. $107div (11pmsqrt14)=11mpsqrt14$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt14$, then the width is $11+sqrt14$. When the length of the lot is $11+sqrt14$, then the width is $11-sqrt14$.
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
accepted
On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm." would turn into $lw=107$.
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=frac107w$$ $$implies 2cdotfrac107w+2w=44$$ $$implies frac214w+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$implies 214+2w^2=44w$$ $$implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $frac-bpmsqrtb^2-4ac2a$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac-(-44)pmsqrt(-44)^2-(4cdot 2cdot 214)2cdot 2$$ $$impliesfrac44pmsqrt2244$$ $$impliesfrac11pm 4sqrt144$$
We now divide by four getting
$$w=11pmsqrt14$$
Now let's find $l$. $107div (11pmsqrt14)=11mpsqrt14$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt14$, then the width is $11+sqrt14$. When the length of the lot is $11+sqrt14$, then the width is $11-sqrt14$.
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm." would turn into $lw=107$.
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=frac107w$$ $$implies 2cdotfrac107w+2w=44$$ $$implies frac214w+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$implies 214+2w^2=44w$$ $$implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $frac-bpmsqrtb^2-4ac2a$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac-(-44)pmsqrt(-44)^2-(4cdot 2cdot 214)2cdot 2$$ $$impliesfrac44pmsqrt2244$$ $$impliesfrac11pm 4sqrt144$$
We now divide by four getting
$$w=11pmsqrt14$$
Now let's find $l$. $107div (11pmsqrt14)=11mpsqrt14$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt14$, then the width is $11+sqrt14$. When the length of the lot is $11+sqrt14$, then the width is $11-sqrt14$.
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm." would turn into $lw=107$.
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=frac107w$$ $$implies 2cdotfrac107w+2w=44$$ $$implies frac214w+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$implies 214+2w^2=44w$$ $$implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $frac-bpmsqrtb^2-4ac2a$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac-(-44)pmsqrt(-44)^2-(4cdot 2cdot 214)2cdot 2$$ $$impliesfrac44pmsqrt2244$$ $$impliesfrac11pm 4sqrt144$$
We now divide by four getting
$$w=11pmsqrt14$$
Now let's find $l$. $107div (11pmsqrt14)=11mpsqrt14$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt14$, then the width is $11+sqrt14$. When the length of the lot is $11+sqrt14$, then the width is $11-sqrt14$.
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 mins ago
answered 27 mins ago
Max0815
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 27 mins ago
Max0815
36712
answered 27 mins ago
Max0815
36712
36712
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Max0815 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
If you choose to solve this by completing the square, observe that $$beginalign107 & =-underbrace(w^2-22w+121)_(w-11)^2+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tagmove 121 over\ implies 14 &= (w-11)^2 tagdivide both sides by $-1$ \implies pm sqrt14 &= w - 11,endalign$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
answered 23 mins ago
Decaf-Math
2,865822
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
add a comment |Â
up vote
2
down vote
If you choose to solve this by completing the square, observe that $$beginalign107 & =-underbrace(w^2-22w+121)_(w-11)^2+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tagmove 121 over\ implies 14 &= (w-11)^2 tagdivide both sides by $-1$ \implies pm sqrt14 &= w - 11,endalign$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
answered 23 mins ago
Decaf-Math
2,865822
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you choose to solve this by completing the square, observe that $$beginalign107 & =-underbrace(w^2-22w+121)_(w-11)^2+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tagmove 121 over\ implies 14 &= (w-11)^2 tagdivide both sides by $-1$ \implies pm sqrt14 &= w - 11,endalign$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
answered 23 mins ago
Decaf-Math
2,865822
If you choose to solve this by completing the square, observe that $$beginalign107 & =-underbrace(w^2-22w+121)_(w-11)^2+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tagmove 121 over\ implies 14 &= (w-11)^2 tagdivide both sides by $-1$ \implies pm sqrt14 &= w - 11,endalign$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
answered 23 mins ago
Decaf-Math
2,865822
answered 23 mins ago
Decaf-Math
2,865822
answered 23 mins ago
Decaf-Math
2,865822
answered 23 mins ago
Decaf-Math
2,865822
2,865822
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
add a comment |Â
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
– Max0815
8 mins ago
add a comment |Â
up vote
0
down vote
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt 14 = w - 11$
$11 pm sqrt14 = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt14) = 11 mp sqrt14$.
So $2l + 2w = 2(11mp sqrt14) + 2(11 pm sqrt14) = 44$. So far so good.
$lw = (11 + sqrt14)(11 - sqrt14 = 11^2 + 11sqrt14 -11sqrt14 - sqrt14^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$. So the Average of $l$ and $w$ is $11$. Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide. It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = sqrt14$ (well, it could be $-sqrt14$ but we assumed $d ge 0$ when we assumed it was longer than it was wide. It doesn't make a difference.)
So $l = 11 + sqrt 14$ and $w = 11 - sqrt14$.
Just a trick.
answered 3 mins ago
fleablood
62.1k22678
add a comment |Â
up vote
0
down vote
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt 14 = w - 11$
$11 pm sqrt14 = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt14) = 11 mp sqrt14$.
So $2l + 2w = 2(11mp sqrt14) + 2(11 pm sqrt14) = 44$. So far so good.
$lw = (11 + sqrt14)(11 - sqrt14 = 11^2 + 11sqrt14 -11sqrt14 - sqrt14^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$. So the Average of $l$ and $w$ is $11$. Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide. It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = sqrt14$ (well, it could be $-sqrt14$ but we assumed $d ge 0$ when we assumed it was longer than it was wide. It doesn't make a difference.)
So $l = 11 + sqrt 14$ and $w = 11 - sqrt14$.
Just a trick.
answered 3 mins ago
fleablood
62.1k22678
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt 14 = w - 11$
$11 pm sqrt14 = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt14) = 11 mp sqrt14$.
So $2l + 2w = 2(11mp sqrt14) + 2(11 pm sqrt14) = 44$. So far so good.
$lw = (11 + sqrt14)(11 - sqrt14 = 11^2 + 11sqrt14 -11sqrt14 - sqrt14^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$. So the Average of $l$ and $w$ is $11$. Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide. It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = sqrt14$ (well, it could be $-sqrt14$ but we assumed $d ge 0$ when we assumed it was longer than it was wide. It doesn't make a difference.)
So $l = 11 + sqrt 14$ and $w = 11 - sqrt14$.
Just a trick.
answered 3 mins ago
fleablood
62.1k22678
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt 14 = w - 11$
$11 pm sqrt14 = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt14) = 11 mp sqrt14$.
So $2l + 2w = 2(11mp sqrt14) + 2(11 pm sqrt14) = 44$. So far so good.
$lw = (11 + sqrt14)(11 - sqrt14 = 11^2 + 11sqrt14 -11sqrt14 - sqrt14^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$. So the Average of $l$ and $w$ is $11$. Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide. It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = sqrt14$ (well, it could be $-sqrt14$ but we assumed $d ge 0$ when we assumed it was longer than it was wide. It doesn't make a difference.)
So $l = 11 + sqrt 14$ and $w = 11 - sqrt14$.
Just a trick.
answered 3 mins ago
fleablood
62.1k22678
answered 3 mins ago
fleablood
62.1k22678
answered 3 mins ago
fleablood
62.1k22678
answered 3 mins ago
fleablood
62.1k22678
62.1k22678
add a comment |Â
add a comment |Â
Zebert is a new contributor. Be nice, and check out our Code of Conduct.
Zebert is a new contributor. Be nice, and check out our Code of Conduct.
Zebert is a new contributor. Be nice, and check out our Code of Conduct.
Zebert is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2930942%2fa-rectangular-lot-has-a-perimeter-of-44cm-the-area-of-the-lot-is-107cm-find-th%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
– Andrei
30 mins ago
From the step $107 = -w^2 + 22w$, you can simply rearrange it to yield $w^2 - 22w + 107 = 0$, and then use the quadratic formula.
– Decaf-Math
30 mins ago