Mixing
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm$^2$. Find the dimensions of the lot?
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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$^2$. What are 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
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
$endgroup$
add a comment |
$begingroup$
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$^2$. What are 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
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
$endgroup$
$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
1
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42
add a comment |
$begingroup$
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$^2$. What are 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
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
$endgroup$
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$^2$. What are 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
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
edited Sep 26 '18 at 3:07
Bob Jarvis
1356
1356
asked Sep 25 '18 at 23:56
ZebertZebert
364
asked Sep 25 '18 at 23:56
ZebertZebert
364
asked Sep 25 '18 at 23:56
ZebertZebert
364
364
$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
1
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42
add a comment |
$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
1
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42
$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
1
1
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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$^2$." 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 $$l=frac{107}{w}$$ $$implies 2cdotfrac{107}{w}+2w=44$$ $$implies frac{214}{w}+2w=44$$ 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 $frac{-bpmsqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac{-(-44)pmsqrt{(-44)^2-(4cdot 2cdot 214)}}{2cdot 2}$$ $$impliesfrac{44pmsqrt{224}}{4}$$ $$impliesfrac{44pm 4sqrt{14}}{4}$$
We now divide by four getting
$$w=11pmsqrt{14}$$
Now let's find $l$. $107div (11pmsqrt{14})=11mpsqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt{14}$, then the width is $11+sqrt{14}$. When the length of the lot is $11+sqrt{14}$, then the width is $11-sqrt{14}$.
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
If you choose to solve this by completing the square, observe that $$begin{align}107 & =-underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tag{move 121 over}\ implies 14 &= (w-11)^2 tag{divide both sides by $-1$} \implies pm sqrt{14} &= w - 11,end{align}$$
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 Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
add a comment |
$begingroup$
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt {14} = w - 11$
$11 pm sqrt{14} = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt{14}) = 11 mp sqrt{14}$.
So $2l + 2w = 2(11mp sqrt{14}) + 2(11 pm sqrt{14}) = 44$. So far so good.
$lw = (11 + sqrt{14})(11 - sqrt{14} = 11^2 + 11sqrt{14} -11sqrt{14} - sqrt{14}^2 = 121 - 14 = 107$.
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 $d = sqrt{14}$ (well, it could be $-sqrt{14}$ 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 $w = 11 - sqrt{14}$.
Just a trick.
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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$^2$." 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 $$l=frac{107}{w}$$ $$implies 2cdotfrac{107}{w}+2w=44$$ $$implies frac{214}{w}+2w=44$$ 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 $frac{-bpmsqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac{-(-44)pmsqrt{(-44)^2-(4cdot 2cdot 214)}}{2cdot 2}$$ $$impliesfrac{44pmsqrt{224}}{4}$$ $$impliesfrac{44pm 4sqrt{14}}{4}$$
We now divide by four getting
$$w=11pmsqrt{14}$$
Now let's find $l$. $107div (11pmsqrt{14})=11mpsqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt{14}$, then the width is $11+sqrt{14}$. When the length of the lot is $11+sqrt{14}$, then the width is $11-sqrt{14}$.
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
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$^2$." 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 $$l=frac{107}{w}$$ $$implies 2cdotfrac{107}{w}+2w=44$$ $$implies frac{214}{w}+2w=44$$ 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 $frac{-bpmsqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac{-(-44)pmsqrt{(-44)^2-(4cdot 2cdot 214)}}{2cdot 2}$$ $$impliesfrac{44pmsqrt{224}}{4}$$ $$impliesfrac{44pm 4sqrt{14}}{4}$$
We now divide by four getting
$$w=11pmsqrt{14}$$
Now let's find $l$. $107div (11pmsqrt{14})=11mpsqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt{14}$, then the width is $11+sqrt{14}$. When the length of the lot is $11+sqrt{14}$, then the width is $11-sqrt{14}$.
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
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$^2$." 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 $$l=frac{107}{w}$$ $$implies 2cdotfrac{107}{w}+2w=44$$ $$implies frac{214}{w}+2w=44$$ 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 $frac{-bpmsqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac{-(-44)pmsqrt{(-44)^2-(4cdot 2cdot 214)}}{2cdot 2}$$ $$impliesfrac{44pmsqrt{224}}{4}$$ $$impliesfrac{44pm 4sqrt{14}}{4}$$
We now divide by four getting
$$w=11pmsqrt{14}$$
Now let's find $l$. $107div (11pmsqrt{14})=11mpsqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt{14}$, then the width is $11+sqrt{14}$. When the length of the lot is $11+sqrt{14}$, then the width is $11-sqrt{14}$.
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
$endgroup$
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$^2$." 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 $$l=frac{107}{w}$$ $$implies 2cdotfrac{107}{w}+2w=44$$ $$implies frac{214}{w}+2w=44$$ 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 $frac{-bpmsqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$frac{-(-44)pmsqrt{(-44)^2-(4cdot 2cdot 214)}}{2cdot 2}$$ $$impliesfrac{44pmsqrt{224}}{4}$$ $$impliesfrac{44pm 4sqrt{14}}{4}$$
We now divide by four getting
$$w=11pmsqrt{14}$$
Now let's find $l$. $107div (11pmsqrt{14})=11mpsqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-sqrt{14}$, then the width is $11+sqrt{14}$. When the length of the lot is $11+sqrt{14}$, then the width is $11-sqrt{14}$.
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
edited Jan 29 at 15:16
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
answered Sep 26 '18 at 0:09
Max0815Max0815
81418
81418
add a comment |
add a comment |
$begingroup$
If you choose to solve this by completing the square, observe that $$begin{align}107 & =-underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tag{move 121 over}\ implies 14 &= (w-11)^2 tag{divide both sides by $-1$} \implies pm sqrt{14} &= w - 11,end{align}$$
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 Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
add a comment |
$begingroup$
If you choose to solve this by completing the square, observe that $$begin{align}107 & =-underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tag{move 121 over}\ implies 14 &= (w-11)^2 tag{divide both sides by $-1$} \implies pm sqrt{14} &= w - 11,end{align}$$
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 Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
$endgroup$
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
add a comment |
$begingroup$
If you choose to solve this by completing the square, observe that $$begin{align}107 & =-underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tag{move 121 over}\ implies 14 &= (w-11)^2 tag{divide both sides by $-1$} \implies pm sqrt{14} &= w - 11,end{align}$$
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 Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
$endgroup$
If you choose to solve this by completing the square, observe that $$begin{align}107 & =-underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \ &= -(w-11)^2 + 121 \ implies -14 & = - (w-11)^2 tag{move 121 over}\ implies 14 &= (w-11)^2 tag{divide both sides by $-1$} \implies pm sqrt{14} &= w - 11,end{align}$$
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 Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
answered Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
answered Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
answered Sep 26 '18 at 0:13
Decaf-MathDecaf-Math
3,422926
3,422926
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
add a comment |
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
$begingroup$
Wow! An upvote for this solution!!! I never would have thought of something so slick! :P I learned something today xD.
$endgroup$
– Max0815
Sep 26 '18 at 0:27
add a comment |
$begingroup$
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt {14} = w - 11$
$11 pm sqrt{14} = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt{14}) = 11 mp sqrt{14}$.
So $2l + 2w = 2(11mp sqrt{14}) + 2(11 pm sqrt{14}) = 44$. So far so good.
$lw = (11 + sqrt{14})(11 - sqrt{14} = 11^2 + 11sqrt{14} -11sqrt{14} - sqrt{14}^2 = 121 - 14 = 107$.
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 $d = sqrt{14}$ (well, it could be $-sqrt{14}$ 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 $w = 11 - sqrt{14}$.
Just a trick.
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt {14} = w - 11$
$11 pm sqrt{14} = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt{14}) = 11 mp sqrt{14}$.
So $2l + 2w = 2(11mp sqrt{14}) + 2(11 pm sqrt{14}) = 44$. So far so good.
$lw = (11 + sqrt{14})(11 - sqrt{14} = 11^2 + 11sqrt{14} -11sqrt{14} - sqrt{14}^2 = 121 - 14 = 107$.
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 $d = sqrt{14}$ (well, it could be $-sqrt{14}$ 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 $w = 11 - sqrt{14}$.
Just a trick.
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt {14} = w - 11$
$11 pm sqrt{14} = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt{14}) = 11 mp sqrt{14}$.
So $2l + 2w = 2(11mp sqrt{14}) + 2(11 pm sqrt{14}) = 44$. So far so good.
$lw = (11 + sqrt{14})(11 - sqrt{14} = 11^2 + 11sqrt{14} -11sqrt{14} - sqrt{14}^2 = 121 - 14 = 107$.
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 $d = sqrt{14}$ (well, it could be $-sqrt{14}$ 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 $w = 11 - sqrt{14}$.
Just a trick.
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
$endgroup$
Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$pm sqrt {14} = w - 11$
$11 pm sqrt{14} = w$.
That doesn't seem right but it could be. Would the teacher give a problem with an irrational solution? Well, maybe....
$l = 22 - (11 pm sqrt{14}) = 11 mp sqrt{14}$.
So $2l + 2w = 2(11mp sqrt{14}) + 2(11 pm sqrt{14}) = 44$. So far so good.
$lw = (11 + sqrt{14})(11 - sqrt{14} = 11^2 + 11sqrt{14} -11sqrt{14} - sqrt{14}^2 = 121 - 14 = 107$.
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 $d = sqrt{14}$ (well, it could be $-sqrt{14}$ 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 $w = 11 - sqrt{14}$.
Just a trick.
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
answered Sep 26 '18 at 0:32
fleabloodfleablood
73.8k22891
73.8k22891
add a comment |
add a comment |
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$begingroup$
Do you know how to solve a quadratic equation? $w^2-22w+107=0$
$endgroup$
– Andrei
Sep 26 '18 at 0:06
$begingroup$
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.
$endgroup$
– Decaf-Math
Sep 26 '18 at 0:06
1
$begingroup$
Area is not measured in $text{cm}$! Some of your answerers seem to simply quote the error without comment.
$endgroup$
– Glen_b
Sep 26 '18 at 0:41
$begingroup$
:P I will edit it to fix xD
$endgroup$
– Max0815
Sep 26 '18 at 0:42