Struggling with a geometry puzzle - area of garden path

[musicgold]

Homework Statement

I am trying to find the area of the garden path described in the attached puzzle.

Relevant Equations

While I figured that the triangular piece at the left end can be moved to the right end to create a rectangle of 1 yard width, I am not able to determine two sides of the triangle.

Could you please help me understand why the answer picture says the length of the two sides of the triangle are 1.33 yard each?

Thanks

 

[Hill]

Homework Statement: I am trying to find the area of the garden path described in the attached puzzle.
Relevant Equations: While I figured that the triangular piece at the left end can be moved to the right end to create a rectangle of 1 yard width, I am not able to determine two sides of the triangle.

picture says the length of the two sides of the triangle are 1.33 yard each?

They are not. The side is $1 \frac 1 3$ and the hypothenuse is $1 \frac 2 3$.

 

[Steve4Physics]

Homework Statement: I am trying to find the area of the garden path described in the attached puzzle.
Relevant Equations: While I figured that the triangular piece at the left end can be moved to the right end to create a rectangle of 1 yard width, I am not able to determine two sides of the triangle.

Could you please help me understand why the answer picture says the length of the two sides of the triangle are 1.33 yard each?

Thanks

There are some issues in addition to the one noted by @Hill.

You haven’t posted the original question – only the official answer.

The official answer states that “The area of the garden path is 55 x 40 = 2200”. Presumably that’s a mistake and it is the whole garden (including path) that has an area of 2200 yd².

The official answer gives a formula for $x$. But the formula is dimensionally incorrect: under the square root sign, ‘BCL’ has dimension [length]³ but the other terms have dimension [length]⁴.

The explanation given in the official answer is (to me anyway) very confused.

Maybe your best bet is to post the original question and some attempt.

 

[haruspex]

The equation for x should read
$x=\frac{B^2\sqrt{B^2-C^2+L^2}-LBC}{B^2-C^2}$.
This gives $66\frac 23$ as desired.


Defining z as the length of L that borders the path, labelled $1\frac 23$, and y as the length labelled $1\frac 13$, we have similar triangles:
$\frac Bx=\frac Cz$ and $\frac{L-z}x=\frac yz$, whence $LBC=B^2y+C^2x$.
And by Pythagoras, $B^2+L^2=C^2+(x+y)^2$.
Eliminating y produces the desired equation.


The magic behind the rational solution is that the triangles left when the path is removed are 3:4:5, as are the two tiny triangles at the ends of the path.

 

[musicgold]

There are some issues in addition to the one noted by @Hill.

You haven’t posted the original question – only the official answer.

The official answer states that “The area of the garden path is 55 x 40 = 2200”. Presumably that’s a mistake and it is the whole garden (including path) that has an area of 2200 yd².

The official answer gives a formula for $x$. But the formula is dimensionally incorrect: under the square root sign, ‘BCL’ has dimension [length]³ but the other terms have dimension [length]⁴.

The explanation given in the official answer is (to me anyway) very confused.

Maybe your best bet is to post the original question and some attempt.

My apologies! I am adding the correct picture here.
I need to find the area of the garden path that runs diagonally across the garden.

My attempt: I assumed that the sides of the triangular piece at the bottom left of the path are: 1, 1, and sqrt 2 yards.
Area of the bottom right triangle = 40 x (55 - sqrt 2) / 2 = 1071.7 sq yds
Area of the top left triangle = 40 x (55 - sqrt 2) / 2 = 1071.7 sq yds
Area of the whole rectangle = 40 x 55 = 2200 sq yds
Area of the path = 2200 - 1071.7 - 1071.7 = 56.6 sq yards.

The official answer is 66.7 sq yards.

Also, I don't find my approach very elegant given the author's claim that the puzzle is easier than it looks.

 

[Hill]

I assumed that the sides of the triangular piece at the bottom left of the path are: 1, 1, and sqrt 2 yards.

This assumption is incorrect.

 

[Steve4Physics]

I assumed that the sides of the triangular piece at the bottom left of the path are: 1, 1, and sqrt 2 yards.

As already noted by @Hill, that's wrong and without justification. Note that the small triangles are not isosceles - this should be evident from the absence of $45^o$ angles.

@haruspex outlined a method in Post #4. Here's a step-by-step approach you might want to try if you are still stuck.

All values in yards. We are told. PR =55, RT=40 and SQ=1.

Let $x=$QT, $y=$PS and $z=$PQ.

1) Apply Pythagoras to triangle QRT. Note that the 3 sides have lengths $55-z, 40$ and $x$.
You now have an equation with 2 unknowns, $x$ and $z$.

2) Prove that triangles PQS and QRT are similar.
(In fact these 2 triangles are both 3:4:5 triangles, but you can’t tell that at this stage.)

3) Since these triangles are similar, confirm that $\frac z1 = \frac x{40}$.

4) You now have 2 equations for the two unknowns, $x$ and $z$. Solve these. If you’ve done it correctly you should get $x=66\frac23$ and $z=1\frac 23$.

5) Use Pythagoras, find $y$.

6) You can now find the area of the path.

If you get stuck, post your working up to the sticking point.

 

[musicgold]

As already noted by @Hill, that's wrong and without justification. Note that the small triangles are not isosceles - this should be evident from the absence of $45^o$ angles.

@haruspex outlined a method in Post #4. Here's a step-by-step approach you might want to try if you are still stuck.
View attachment 337266
All values in yards. We are told. PR =55, RT=40 and SQ=1.

Let $x=$QT, $y=$PS and $z=$PQ.

1) Apply Pythagoras to triangle QRT. Note that the 3 sides have lengths $55-z, 40$ and $x$.
You now have an equation with 2 unknowns, $x$ and $z$.

2) Prove that triangles PQS and QRT are similar.
(In fact these 2 triangles are both 3:4:5 triangles, but you can’t tell that at this stage.)

3) Since these triangles are similar, confirm that $\frac z1 = \frac x{40}$.

4) You now have 2 equations for the two unknowns, $x$ and $z$. Solve these. If you’ve done it correctly you should get $x=66\frac23$ and $z=1\frac 23$.

5) Use Pythagoras, find $y$.

6) You can now find the area of the path.

If you get stuck, post your working up to the sticking point.

Thanks!

I have reached up to the following equations. Not sure how to get the neat $z=1\frac 23$ from here.

$1599z^2 = 4625 -110z$
$z^2 +0.07z-2.89=0$

 

[PeroK]

Thanks!

I have reached up to the following equations. Not sure how to get the neat $z=1\frac 23$ from here.

$1599z^2 = 4625 -110z$
$z^2 +0.07z-2.89=0$

It looks like $z = \frac 5 3$ solves that equation.

This problem is trickier than it looks in my opinion. The complexity comes in trying to express the length of the path as a function of the given width of the path and the dimensions of the garden. That's the equation given in the text, as corrected by @haruspex.

The area of the path is the length times the width. Hence:
$$Area = Cx = C\frac{B^2\sqrt{B^2-C^2+L^2}-LBC}{B^2-C^2}$$There is no need to calculate the other intermediate quantities. You just plug $B, C, L$ into that equation.

 

[Steve4Physics]

$1599z^2 = 4625 -110z$
$z^2 +0.07z-2.89=0$

First equation is good. But note that dividing by $1599$ gives $z^2 +0.068793z -2.8924 =0$ (approximately).

In your second equation you rounded excessively. In particular, you rounded $0.068793$ to $0.07$. You can’t expect to get a precise value for $z$ if you round so much in the middle of a calculation.

In any case, all rounding must be avoided if you want a fractional (exact) answer.

Your first equation for $z$ can be written in standard quadratic form: $1599z^2 +110z - 4625=0$. You can then use $z = \frac {-b \pm \sqrt{b^2 – 4ac}}{2a}$ with $a=1599$ etc.

There’s some messy arithmetic, but try it!

 

[mathwonk]

i don't see how to simplify that horrible quadratic equation, but getting there, it is
"clear by inspection": that 40z = x = area of the path, using the height times base formula for a parallelogram, since either x or z can be considered as the base, and then the height is either 1 or 40. hence the entire solution is given in steve4physics step 4, post 7. i.e. area = x = 66 2/3.

 

[musicgold]

Thank you all for helping me!