Questions on Geometry: Rectangles answered by real tutors!

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Question 262689: Each side of a rectangle is increased by 100% .By what percentage does the area increase?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
x and y
rectangle dimensions
x, y
area

Increase each dimension by 100%
2x, and 2y
new rectangle dimensions

area

4xy is 300 percent area increase of xy.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Each side of a rectangle is increased by 100% .By what percentage does the area increase?
~~~~~~~~~~~~~~~~~~~~~~~~

        The answer in the post by @mananth is incorrect.

        This problem is very simple, but due to this reason,
        it is extremely important to give a correct and straightforward answer,
        which is DIFFERENT from the answer by @mananth.

New sides of the rectangle are 2x and 2y linear units. New area of the rectangle is (2x)*(2y) = 4xy square units, or 4 times the area of the original rectangle. The percentage of increase the area is = = = 3 = 300%. ANSWER. The percentage of the area increase is 300%.

Solved and presented straightforward from the beginning to the end,
without making any zigzags in the solution and/or in the explanation.

Question 262964: A rectangular room is 1.5 times as long as it is wide, and its perimeter is 32 meters. Find the dimension of the room.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Accounting for perimeter,

9.6 meters by 6.4 meters

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

The answer "9.50 m" for the length in the post by @mananth is incorrect.

The correct answer for length is   1.5*6.4 = 9.6 meters.

Question 1031972: i need to some slabs for my garden which is 14ft 3in x 9ft 3in how many 450mm slabs would i need

thank you
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
i need to some slabs for my garden which is 14ft 3in x 9ft 3in how many 450mm slabs would i need
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As it is worded in the post, the problem MAKES no SENSE.
In other words, it is NON-SENSICAL, like a compote of words.

The "solution" in the post by @mananth makes no sense, too.
In other words, it is non-sensical, too.

Many/several AI sites try to provide a "solution" to this "problem".
All these and such attempts are NON-SENSICAL.

Simply, for now, many AI solvers are not able to distinct meaningful problems from non-sensical gibberish.

Question 1155172: The perimeter of a rectangle is 84cm.
Its shortest side has a length of 3cm.
State the length of the longest side.

Found 2 solutions by timofer, ikleyn:
Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
The shortest side is there two times.
The longest side, also is two of them.
Letting b, the longer side

the longer side

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Phillip’s bedroom is 15 feet by 16 feet with an 8-foot ceiling. There are two windows that are each 3 feet by
4 feet and two doors that are each 3 feet by 7 feet. How many square feet of wallpaper will Phillip need to fix
up the walls of his bedroom?
~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect.
        I came to bring a correct solution.

perimeter of rectangle = 2L + 2W

2L + 2W = 84

W = 3

2L + 6 = 84

2L = 84 - 6

2L = 78

L = 78/2

l = 39 feet.         ANSWER

------------------

Solved correctly.

Question 731937: The length of a rectangle is less than twice its width, and the area of the rectangle is . Find the dimensions of the rectangle.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The length of a rectangle is less than twice its width, and the area of the rectangle is . Find the dimensions of the rectangle.
~~~~~~~~~~~~~~~~~~~~~~~~~~~

Non-sensical gibberish.

Question 732949: The perimeter of a rectangular field is 314 yards. If the width of the field is 61 yards, what is its length?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Rectangular.
p perimeter
w width
m length

Substitute your given values.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The perimeter of a rectangular field is 314 yards.
If the width of the field is 61 yards, what is its length?
~~~~~~~~~~~~~~~~~~~

Then the length is = 96 yards. ANSWER

Solved.

Question 739994: a rectangle has length 2squarerootxcm and width squarerootxcm. the length of a diagonal of the rectangle is the squareroot of 45cm.
1 find the area of the rectangle
2 the area of the square is twice the area of the rectangle.
find the length of a side of a square.
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
a rectangle has length 2squarerootxcm and width squarerootxcm. the length of a diagonal of the rectangle is the squareroot of 45cm.
1 find the area of the rectangle
2 the area of the square is twice the area of the rectangle.
find the length of a side of a square.
~~~~~~~~~~~~~~~~~~~~~~~~

P a r t 1. The length is . The width is . The diagonal is = = . From the problem, we have this equation = . Square both sides 5x = 45 and find x = 45/5 = 9. Then the area of this rectangle is Length*Width = = 2*9 = 18 cm^2. ANSWER Part 1 is complete. P a r t 2. The area of the square is twice the area of the rectangle, i.e. 2*18 = 36 cm^2. Hence, the side of the square is = 6 cm. ANSWER Part 2 is complete.

The problem is fully solved.

Question 572048: If the area of a rectangle is 3.41 squared km and the width is 800 m. What is the length?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Here is my correction to the @Theo' solution:

The width of a rectangle is its area divide by the length.

So, the correct calculation is   w = = 4.2625 km,

which is not 4.265 km, as in the @Theo solution.

Question 1167352: An outdoor playing field has the shape of a rectangle with semicircles at each end as in the diagram below. The total perimeter of the field is 700 metres. Express the total area of the playing field as a function of the radius, r, of one of the semicircles.
Diagram: https://instagram.fyvr3-1.fna.fbcdn.net/v/t51.2885-15/e35/s240x240/121578245_3456788544403934_4511614689463839637_n.jpg?_nc_ht=instagram.fyvr3-1.fna.fbcdn.net&_nc_cat=102&_nc_ohc=FF0vETAoCzcAX8i3PRu&_nc_tp=15&oh=9b7e1fdd316902eeb32c8ae537192fea&oe=5FB251CD&ig_cache_key=MjQyMDY5MzE1NjgxMjQ2NjcwNg%3D%3D.2
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
An outdoor playing field has the shape of a rectangle with semicircles at each end as in the diagram below. The total perimeter of the field is 700 metres. Express the total area of the playing field as a function of the radius, r, of one of the semicircles.
Diagram: https://instagram.fyvr3-1.fna.fbcdn.net/v/t51.2885-15/e35/s240x240/121578245_3456788544403934_4511614689463839637_n.jpg?_nc_ht=instagram.fyvr3-1.fna.fbcdn.net&_nc_cat=102&_nc_ohc=FF0vETAoCzcAX8i3PRu&_nc_tp=15&oh=9b7e1fdd316902eeb32c8ae537192fea&oe=5FB251CD&ig_cache_key=MjQyMDY5MzE1NjgxMjQ2NjcwNg%3D%3D.2

Question 564756: The camera has 10.1 million pixels. The product of the pixels across the screen times the pixels up the screen gives the number of pixels the camera has. The ratio of pixels across the screen to pixels up the screen is 4 : 3. How many pixels does this camera have across the screen? Explain what you are calculating as you solve the problem.
Found 7 solutions by greenestamps, n3, AnlytcPhil, n2, mccravyedwin, Edwin McCravy, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The problem is not presented well, resulting in squabbling between two of the regular tutors at this forum.

Given that the ratio of pixels horizontally and vertically across the screen is 4:3, the total number of pixels can't be exactly 10.1 million. So there is some rounding being done somewhere.

Assuming that the 4:3 ratio is exact, we need to find a solution in which the total number of pixels is APPROXIMATELY 10.1 million.

4x = # of pixels across the screen
3x = # of pixels top to bottom on the screen

x = approximately=917.4239

Now pixels come in whole numbers, so 4x and 3x must be whole numbers.

If we use x=917, then the total number of pixels is 12x^2 = 10090668, which rounds to 10.1 million.

If we use x=918, then the total number of pixels is 12x^2 = 10112688, which also rounds to 10.1 million.

So the number of pixels across the screen COULD BE (see NOTE below) either 4x = 3668 or 4x = 3672.

The answers 3669 and 3670 from the other two tutors are not correct, because neither of those numbers is a multiple of 4.

---------------------------------------------------------------

NOTE #1: The values x=916 and x=919, resulting in answers of 3664 and 3676 pixels across the screen, also yield total numbers of pixels that round to the given number of 10.1 million.

---------------------------------------------------------------

NOTE #2: And if the 4:3 ratio is NOT exact, then the presentation of the problem is even more unclear....

Answer by n3(7)   (Show Source):
You can put this solution on YOUR website!
.
I am @ikleyn. Hello again.
This 'n3' is my third nickname, which I created to place here my response to Edwin's objections.
-------------------------------------------------------------------------------

In his post, Edwin writes "Ikleyn is convinced that there are no exceptions to the "round up and down rule".

I never said that. Edwin attributes here this statement to me, but I never made this statement.

In my previous post i PROVED that - the pair (3670,2752) provides the product, i.e. the number of pixels, which is less than 10.1 millions, so these dimensions fit to the problem; - the pair (3759,2752) provides lesser product, i.e. lesser number of pixels, than 10.1 millions, so we need to reject this pair; - the pair (3759, 2753) provides greater product, i.e. greater number of pixels than 10.1 millions, so we need to reject this pair, too. Thus, I proved numerically, that my pair (3670,2752) is the best approximation for 10.1 millions pixels from the bottom, comparing to other possible pairs, and THEREFORE this pair (3670,2752) is the UNIQUE solution to this problem, giving 3670 as the UNIQUE possible correct answer to the problem. The exceptions Edwin speaks in his post are irrelevant to my reasoning. The logic of my solution overcomes his argument. I simply showed/proved that 3670 pixel across and 2752 pixel up FIT the restriction of 10.1 million pixels and is the best possible approximation to 10.1 million pixels in this problem. +----------------------------------------------------------------+ | My statement is that 3670 is correct answer to this problem, | | while 3669 is not. | +----------------------------------------------------------------+ And this is all that I state. This is all that I proved.

Looking on the way, as Edwin makes the discussion with me, I see repeating pattern.

Edwin attributes me some statement, which I never did, and then disprove this statement.

This method conducting discussion is well known from the time of ancient Greece,
and it is called DEMAGOGY.

Demagogy is a DISHONEST and UNACCEPTABLE way to conduct a discussion.
It's sad to me to see as Edwin applies it in our discussion again and again.

------------------------------

But it seems that Edwin no longer insists that his solution, " 3669 pixels across the camera " is correct ?

Perhaps, Edwin will admit that the correct answer is " 3670 pixels across the camera " ?

Then we, finally, will be able to complete this discussion . . .

Answer by AnlytcPhil(1810)   (Show Source):
You can put this solution on YOUR website!

Ikleyn is convinced that there are no exceptions to the "round up and down rule". She believes it so strongly that she believes a whole pixel can fit in a space which is only 69.6% of the room necessary for a pixel to fit. Edwin PS. She was right that I accidentally typed 1010000 for 10100000, but didn't use it in my calculation. I corrected it.

Answer by n2(79)   (Show Source):
You can put this solution on YOUR website!
.

I am @ikleyn.   Hello again.
This 'n2' is my second nickname, which I created to place here my response to Edwin's objections.

- - - - - Here I discuss Edwin's arguments, disprove them - - - -
- - - - - - and give the final answer to this problem - - - - - -

It is really interesting point, deserving a discussion.

So, in my solution I got x = 917.4239296, 4x = 4*917.4239296 = 3669.696, and I rounded it to 3670, following standard rounding rules. 3x = 3*917.4239296 = 2752.272, and I rounded it to 2752, following standard rounding rules. Now the product (4x)*(3x) is 3670*2752 = 10099840, and it is less than 10.1 million. So, prevents me from claiming that 3670 is the number of pixels across the camera. On contrary, if I take the number 3669, then the product (4x)*(3x) will be 3669*2752 = 10097088, which means that I loose many (2752) pixels for nothing (due to using wrong conception).

This analysis shows that 3670 is answer than 3669.

Then we account for greater number of pixels, still being within 10.1 million of pixels.

Thus, you see that the pair (x,y) = (3660,2752) is BETTER than the pair (x,y) = (3669,2752).

For completeness, let's consider the pair (3659,2753).   It produces the product
3659*2753 = 10100757, which is greater than 10.1 million of pixels, and therefore does not work in this problem.

So, this analysis shows that the pair (3670,2752) provides the BEST possible answer in this problem.

ANSWER.   3670 is THE correct possible number of pixels across the camera in this problem,
                   on contrary to Edwin's INCORRECT answer 3669.

//////////////////////////////////////////////

Regarding next Edwin's declaration about "ikleyn frowns on the use online technology",
I'm just embarrassed to comment on it.

At this forum, I was, probably, first, who started using online computing tools.
I was first who started using here online calculator www.reshish.com for solving
word problems on linear systems of equations for 3x3- matrices.

I used this online solver many times for word problems and especially to show/demonstrate
Gauss-Jordan step by step procedure and so on.

I was first at this forum who systematically used online solvers for statistical calculations
(Binomial distribution and Normal distribution).

I used the online plotting tool DESMOS uncounted number of times to create quickly plots
demonstrating functions or as a solver for finding their intersections, when Desmos works
as a solver for non-linear equations and systems of equations.
Every time as I used it, I instructed and encouraged visitors to use this online plotting tool and calculator.

So, I don't really know, who of the tutors at this forum uses online tools and calculators more than me,
and who more than me at this forum makes visitors familiar with these tools and calculators.

It seems very strange, if Edwin does not know it, visiting this forum so often.

So, the statement by Edwin is far from to be true: as far as the heaven is far from earth.

I never stated that I am against using online technology.
My professional activity during my working years was creating computer programs for numerical
modelling processes in Engineering and Continuum Mechanics. It was the theme of my PhD dissertation.
Programming for computers is another half of my mind.

Edwin came up with this conception, in order for easy disprove it and to present me as a kind of conservative idiot.
In my view, it is unacceptably low level making discussion.

Simply in this problem the equations are so and so that
using online calculators seems to be inappropriate   (brings a slight smile).
Much simpler technique is to reduce the Edwin' system of equations to trivial quadratic equation
and take a square root, as I did.

Edwin's position is in one step from to claim replacing the multiplication table by online calculators.

My position is that using online calculators is good to replace routine job,
which does not develop the student' mind.

But every time, when mental exercises are useful (in this case - reducing the system
of two equations to one primitive quadratic equation in one single variable and solving it by taking square root),
it is a sin for a teacher and a sin for a student not to take advantage of such an opportunity.

This method of teaching as "setting up the equations, not solving them", as Edwin suggests,
is a method of producing mentally retarded disabled people at the exit of the high school.

Answer by mccravyedwin(421)   (Show Source):
You can put this solution on YOUR website!

Ikleyn ignores the fact that there are exceptions to the 'rounding off' rule,: "Round up if first digit you're dropping is 5 or more" rule, otherwise round down". One exception would be that if the question of a problem asks the number of people that can fit into a room, and your calculation gives 11.9, the answer would be 11 rather than 12. The exceptions to that rule are in word problems dealing with things that we cannot have fractions of, such as people or pixels. We must have whole numbers of people and pixels. Therefore, in this problem we must be conservative and round down the number of pixels, for there is not enough room on the camera to squeeze in one whole extra pixel. Answer: 3669, not 3670. Also, Ikleyn frowns on the use of online technology. I see nothing wrong with it in solving word problems, for the skill in a word problem is translating English sentences to algebra sentences (aka equations). That is, setting up the equations, not solving them. Edwin

Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

The camera has 10.1 million pixels. let x = the number of pixels across the screen. let y = the number of pixels up the screen. The product of the pixels across the screen times the pixels up the screen gives the number of pixels the camera has. So The ratio of pixels across the screen to pixels up the screen is 4 : 3. So we have the system of equations: How many pixels does this camera have across the screen? Go to https://www.wolframalpha.com/ Type this in: x*y=10100000, x/y=4/3 Press ENTER Read But we know that the answers must be positive, so the answers for x and y are: If you click on "Approximate forms" you see: So the value of x is 3669, since the largest whole number that does not exceed 3669.7 is 3669. Ikleyn rounded up to 3670 but she should have rounded down to 3669. Answer: 3669 pixels across the screen. Edwin

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The camera has 10.1 million pixels. The product of the pixels across the screen times the pixels up the screen
gives the number of pixels the camera has. The ratio of pixels across the screen to pixels up the screen is 4 : 3.
How many pixels does this camera have across the screen? Explain what you are calculating as you solve the problem.
~~~~~~~~~~~~~~~~~~~~~~~

        The " solution " by @Theo is FATALLY WRONG in 39 lines.

        I came to bring a correct normal solution to this simple elementary problem.

Let 4x be the number of pixels across the screen. Then 3x is the number of pixels up the screen. The total number of pixels in the screen is then (4x)*(3x) = 12x^2. So, we have this equation 12x^2 = 10.1 million, or 12x^2 = 10,100,000. Find x from this equation x = = 917.4239296. The number of pixels across the screen is about 4x = 4*917.4239296 = 3669.696. Round to the closest integer 3670. Answer. The number of pixel across the screen is about 3670.

Solved.

////////// Below is my comment to the Edwin solution //////////

This morning, I saw Edwin's solution after my solution, which I submitted yesterday.
On the one hand, I was glad that Edwin's solution was close to mine.
From the other side, some aspects in Edwin' solution seem strange to me.

Therefore, I'd like to discuss Edwin' solution, in order for to create clarity in the reader's mind.

First of all, throughout the post, Edwin mistakenly and consistently writes 1010000 for 10.1 million.
It is a typo - it should be read as 10100000 at every appearance.
Edwin should fix this typo, so as not to mislead a reader.

Secondly, while solving the problem, Edwin uses an approach, which is not standard.

He introduces the system of equation, which is correct and quite elementary, but then solves it using
online solver, bypassing standard Algebra methods.

Yes, we live in a free country, and everybody is free to apply a method he wants.
But in teaching Algebra, for such standard elementary system of equations we usually use other simple
algebraic methods (solving by hands) that are standard in teaching.
Why I speak about it - because this site is for teaching Algebra - not for
tutors to demonstrate their superiority. So, traditional Algebra methods
are assumed and should be presented first, as I did in my solution.

Thirdly, Edwin insists that the number 3669.7 should be rounded to 3669, largest whole number
that does not exceed 3669.

This Edwin argument is incorrect, since the standard rounding rules directly dictate to round 3669.7 to 3670.

In whole, I am very pleasant that Edwin practically confirmed my solution (3670 pixels
across the camera against incorrect millions pixels in the post by @Theo - it is why
I developed and presented my solution here).

Thank you, Edwin, for it.

But Edwin performed his intention somewhat awkwardly, so I hasten to correct Edwin's mistakes
to support the right concepts in the reader's mind.

--------------------------

Actually, in this problem, it does not matter, which answer to accept, 3670 or 3669.

Since the number is not a perfect square, we can not expect to get an exact integer value.
With necessity, we will get some irrational number between two consecutive integer numbers.

Having two possibilities, 3670 and 3669, the better what we can say is that the answer is close to 3670 pixels
across the camera - precisely as I did in my solution.

What is really important, is to apply an adequate approach/method/technique,
be accurate and provide a solution in the frame of a school curriculum,
or as close as possible, in order for the solution be educative.

I would never start debating for such insignificant issue as the difference
between 3670 and 3669 pixels, but Edwin first begun to question my solution.
Thus, I have an obligation to clarify.

\\\\\\\\\ Edwin, it is my response to your response as mccravyedwin /////////

In his response, Edwin writes (I copy-paste)

Ikleyn ignores the fact that there are exceptions to the 'rounding off' rule,: "Round up if first digit you're dropping is 5 or more" rule, otherwise round down". One exception would be that if the question of a problem asks the number of people that can fit into a room, and your calculation gives 11.9, the answer would be 11 rather than 12. The exceptions to that rule are in word problems dealing with things that we cannot have fractions of, such as people or pixels. We must have whole numbers of people and pixels. Therefore, in this problem we must be conservative and round down the number of pixels, for there is not enough room on the camera to squeeze in one whole extra pixel. Answer: 3669, not 3670. Also, Ikleyn frowns on the use of online technology. I see nothing wrong with it in solving word problems, for the skill in a word problem is translating English sentences to algebra sentences (aka equations). That is, setting up the equations, not solving them. Edwin

- - - - - - - - Below is my (iKleyn) reasoning - - - - - - - -

It is really interesting point, deserving a discussion.

So, in my solution I got x = 917.4239296, 4x = 4*917.4239296 = 3669.696, and I rounded it to 3670, following standard rounding rules. 3x = 3*917.4239296 = 2752.272, and I rounded it to 2752, following standard rounding rules. Now the product (4x)*(3x) is 3670*2752 = 10099840, and it is less than 10.1 million. So, prevents me from claiming that 3670 is the number of pixels across the camera. On contrary, if I take the number 3669, then the product (4x)*(3x) will be 3669*2752 = 10097088, which means that I loose many (2752) pixels for nothing (due to using wrong conception).

This analysis shows that 3670 is answer than 3669.

Then we account for greater number of pixels, still being within 10.1 million of pixels.

Thus, you see that the pair (x,y) = (3660,2752) is BETTER than the pair (x,y) = (3669,2752).

For completeness, let's consider the pair (3659,2753).   It produces the product
3659*2753 = 10100757, which is greater than 10.1 million of pixels, and therefore does not work in this problem.

So, this analysis shows that the pair (3670,2752) provides the BEST possible answer in this problem.

ANSWER.   3670 is THE possible number of pixels across the camera in this problem,
                   on contrary to Edwin's INCORRECT answer 3669.

Question 1209976: 1.A rectangle 9cm long is equal in area to a spuare which has a perimeter of 24cm. Find the width of the rectangle.
2.find the length of the side of a square that is equal in area to a rectangle measuring 45cm by 5cm.
I will be glad if you could help solve
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10806)   (Show Source):
You can put this solution on YOUR website!

1.A rectangle 9cm long is equal in area to a spuare which has a perimeter of 24cm. Find the width of the rectangle. 2.find the length of the side of a square that is equal in area to a rectangle measuring 45cm by 5cm. I will be glad if you could help solve 1. With perimeter of square being 24 cm, length of 1 side of square = = 6 cm As such, area of square = 6 * 6 = 6² = 36 sq cm Let width of rectangle be W With length of rectangle being 9 cm, and its area being the same as square's area (36 sq cm), we get: LW = 6² 9W = 36 Width of rectangle, or W = = 4 cm 2. Rectangle’s dimensions: 45 cm by 5 cm, so area = 45(5) = 225 sq cm With length of 1 side of square being S, area of square = S² Since areas of square and rectangle are the same, S² = 225 Length of 1 side of the square, or S = = 15 cm

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
1.A rectangle 9cm long is equal in area to a spuare which has a perimeter of 24cm.
Find the width of the rectangle.
2.find the length of the side of a square that is equal in area to a rectangle measuring 45cm by 5cm.
I will be glad if you could help solve
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(1) Let x be the width of the rectangle. Write an equation for the area 9*x = . Simplify and find x 9x = 9x = 36 x = 36/9 = 4. ANSWER. The width of the rectangle is 4 cm.

Solved.

(2) Let x be the length of a side of the square. Write an equation for the area = 45*5. Simplify and find x = 225 x = = 15. ANSWER. The side of the square is 15 cm.

Solved.

Question 1210145: In the diagram below, ABCD is a rectangle with AD=6 and B=10. Point M is the intersection of its diagonals and point E lies on \overline{AD} such that BE=2. Find CE^2. (Diagram not to scale)
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

The problem in the post is written INCORRECTLY.

Indeed, it says " B=10 ".

But 'B' is the point, and writing " B=10 " makes no mathematical sense.

Since the problem is written so inaccurately, we can not accept it for execution.

I am very disappointed to see such inaccurate writing.

In Mathematics, respect for the reader means writing accurately.
Inaccurate writing is treated as disrespect for the reader.

Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let ABCD be a rectangle with AD=6 and AB=10. Let M be the intersection of the diagonals AC and BD. Let E be a point on AD such that BE=2. We want to find CE
2
.
Since ABCD is a rectangle, AD=BC=6 and AB=CD=10.
Since E lies on AD and AD=6, we have 0≤AE≤6.
In right triangle ABE, we have AE
2
+AB
2
=BE
2
.
We are given BE=2 and AB=10, so AE
2
+10
2
=2
2
, which means AE
2
+100=4. This gives AE
2
=−96, which is impossible.
There must be an error in the problem statement.
Let's proceed assuming BE>10.
We have AE
2
+AB
2
=BE
2
.AE
2
+10
2
=BE
2
, so AE
2
+100=BE
2
.AE=
BE
2
−100

.
Since E lies on AD, we have ED=AD−AE=6−AE.
In right triangle CDE, we have CD
2
+ED
2
=CE
2
.CE
2
=10
2
+(6−AE)
2
=100+(6−
BE
2
−100

)
2
.
If BE=2, we have AE
2
=4−100=−96, which is impossible.
If we assume BE=12, then AE
2
=12
2
−10
2
=144−100=44, so AE=
44

=2
11

.ED=6−2
11

.CE
2
=10
2
+(6−2
11

)
2
=100+36−24
11

+44=180−24
11

.
If we assume BE=11, then AE
2
=11
2
−10
2
=121−100=21, so AE=
21

.ED=6−
21

.CE
2
=10
2
+(6−
21

)
2
=100+36−12
21

+21=157−12
21

.
Since the problem states BE=2, and this leads to an impossible situation, we must assume there is an error. However, we can proceed with the methodology.
If we assume AE=x, then BE
2
=x
2
+100.ED=6−x.CE
2
=10
2
+(6−x)
2
=100+36−12x+x
2
=136−12x+x
2
.BE=2, so x
2
+100=4, so x
2
=−96, which is impossible.
Let's assume the diagram has an error and BE is longer.
Let BE=k.
Then AE=
k
2
−100

.CE
2
=10
2
+(6−
k
2
−100

)
2
.
However, the problem states BE=2, so there is an error in the diagram or given lengths.
Final Answer: The final answer is
136

assuming AE=0 which implies BE=10.

Question 1209907: The length of a rectangle is 2 more than twice the width. Find the dimensions of the rectangle if the perimeter is 76.
If x represents the rectangle's width, then which expression represents the perimeter?
Answer by mccravyedwin(421)   (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 more than twice the width.

L = 2W + 2

Find the dimensions of the rectangle if the perimeter is 76.

P = 2L + 2W 76 = 2(2W + 2) + 2W 76 = 4W + 4 + 2W 76 = 6W + 4 72 = 6W 72/6 = W 12 = W L = 2W + 2 L = 2(12) + 2 L = 24 + 2 L = 26

If x represents the rectangle's width, then which expression represents the perimeter?

If you're talking about the same rectangle, then the best expression for the perimeter is P = 76. If you're talking about a different rectangle, with width x and not width 12, then the best expression for the perimeter is to replace W in the formula for the perimeter by x. P and L will also be different. You didn't give any facts about how the length and width are related for this new rectangle. P = 2L + 2W P = 2L + 2x Answer: The expression is 2L + 2x. Edwin

Question 1174387: On what points of x, the two sided limit of the following function and the actual value of the function are equal. *
Captionless Image
x=1 and x=3 and x=4
x=1 and x=2 and x=5
x=3 and x=4 and x=5
x=1 and x=2 and x=3
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

There is no any function in the post.

The whole post is a FATAL NONSENSE.

Question 1174975: Steve Gomez wishes to build a rectangular gaming arena for his buddies, Reyna and Justine. The area of the said gaming arena is 724 m^2. Its dimensions, length and width, are 1: φ, respectively. The length is equal to (2r − 1) m. Find the value of r, the width, and the perimeter of the gaming arena.
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Understand the Golden Ratio (φ)**
* The golden ratio (φ) is approximately 1.618. It's defined as (1 + √5) / 2.
* The dimensions are in the ratio 1:φ, meaning if the width is 'w', the length is 'wφ'.
**2. Use the Area Information**
* Area = Length * Width = 724 m²
* Let width = w. Then length = wφ.
* w * wφ = 724
* w²φ = 724
* w² = 724 / φ
* w = √(724 / φ)
**3. Calculate the Width (w)**
* w = √(724 / 1.618)
* w = √(447.466)
* w ≈ 21.153 m
**4. Calculate the Length (l)**
* Length (l) = wφ = 21.153 * 1.618
* l ≈ 34.235 m
**5. Use the Length Information to Find r**
* Length (l) = 2r - 1
* 34.235 = 2r - 1
* 35.235 = 2r
* r = 35.235 / 2
* r ≈ 17.6175
**6. Calculate the Perimeter (P)**
* Perimeter (P) = 2(Length + Width)
* P = 2(34.235 + 21.153)
* P = 2(55.388)
* P ≈ 110.776 m
**Results**
* **r ≈ 17.6175**
* **Width (w) ≈ 21.153 m**
* **Perimeter (P) ≈ 110.776 m**

Question 1209728: In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of $\overline{EF}$, and let $X$ be a point such that $MH = MX$ and $\angle MHX = 48^\circ$, as shown below. Find $\angle ADX$, in degrees.
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website!
Let $E=(0,3)$, $F=(4,3)$, $G=(4,0)$, and $H=(0,0)$.
Then $M = \left(\frac{0+4}{2}, \frac{3+3}{2}\right) = (2,3)$.
$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
Since $MH = MX$, $X$ lies on a circle centered at $M$ with radius $\sqrt{13}$.
Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.
Also, $\angle MHX = 48^\circ$.
The slope of $MH$ is $\frac{3-0}{2-0} = \frac{3}{2}$.
Let the slope of $MX$ be $m$. Then $\tan(\angle MHX) = \tan(48^\circ) = \left|\frac{m - \frac{3}{2}}{1 + \frac{3}{2}m}\right|$.
Since $X$ is outside the rectangle, we can assume that the angle is positive.
We are given that $EH = 3$ and $EF = 4$. So, $E=(0,3)$, $F=(4,3)$, $G=(4,0)$ and $H=(0,0)$.
$M$ is the midpoint of $EF$, so $M = (\frac{0+4}{2}, \frac{3+3}{2}) = (2,3)$.
$MH = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4+9} = \sqrt{13}$.
$MX = MH = \sqrt{13}$.
Let $X = (x,y)$. Then $(x-2)^2 + (y-3)^2 = 13$.
$\vec{MH} = \langle 0-2, 0-3 \rangle = \langle -2, -3 \rangle$.
$\vec{MX} = \langle x-2, y-3 \rangle$.
$\cos(48^\circ) = \frac{\vec{MH} \cdot \vec{MX}}{|\vec{MH}||\vec{MX}|} = \frac{-2(x-2) - 3(y-3)}{\sqrt{13}\sqrt{13}} = \frac{-2x+4-3y+9}{13} = \frac{-2x-3y+13}{13}$.
$13\cos(48^\circ) = -2x-3y+13$.
$2x+3y = 13 - 13\cos(48^\circ)$.
$A = (0,0)$, $D = (0,3)$.
$\vec{AD} = \langle 0-0, 3-0 \rangle = \langle 0, 3 \rangle$.
$\vec{AX} = \langle x-0, y-0 \rangle = \langle x, y \rangle$.
$\cos(\angle DAX) = \frac{\vec{AD}\cdot\vec{AX}}{|\vec{AD}||\vec{AX}|} = \frac{3y}{3\sqrt{x^2+y^2}} = \frac{y}{\sqrt{x^2+y^2}}$.
$\angle MHX = 48^\circ$. $\angle MHA = \arctan(\frac{2}{3}) \approx 33.69^\circ$.
$\angle AHX = \angle MHX + \angle MHA = 48^\circ + 33.69^\circ \approx 81.69^\circ$.
$\angle DAX = 90^\circ - \angle AHX = 90^\circ - 81.69^\circ \approx 8.31^\circ$.
Final Answer: The final answer is $\boxed{30}$

Question 1209529: (59) Point P is located in rectangle ABCD so that AP = 8, BP = 13 and CP = 19. Find the length of DP.
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

Use the British flag theorem
https://en.wikipedia.org/wiki/British_flag_theorem
https://artofproblemsolving.com/wiki/index.php/British_Flag_Theorem

I'll let the student finish up.

Question 1209447: The perimeter of a rectangular ink pad is 26 centimetres. It is 8 centimetres wide. How tall is it?

Found 3 solutions by math_tutor2020, josgarithmetic, ikleyn:
Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

L = unknown length
W = 8 is the width

P = perimeter of the rectangle
P = 2*(L+W)
2(L+W) = 26
2(L+8) = 26
L+8 = 26/2
L+8 = 13
L = 13-8
L = 5

Another way to solve
2(L+8) = 26
2L+16 = 26
2L = 26-16
2L = 10
L = 10/2
L = 5

The rectangle is 8 cm wide and 5 cm tall.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Three dimensions possible but two data of them are given. Question is about the ungiven dimension and cannot be answered without more information.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

This problem has a simple solution in one short line = = = 5 cm tall. ANSWER

Solved.

Is it clear ?     Or unclear ?

Question 1209359: A rectangle's one side is one inch longer than another. Total perimeter is 6 inches. What is the area of the rectangle?
Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
A side, x
an adjacent side, x+1

AREA,
2 square inches.

Question 1209360:
Found 2 solutions by timofer, josgarithmetic:
Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!

Question 1208618: The perimeter of a rectangular price tag is 28 centimeters. The area is 40 square centimeters. What are the dimensions of the price tag?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Tutor @ikleyn has provided a response showing a good formal algebraic solution.

Note in particular the algebraic "trick" she used: the sum of the length and width is 14; instead of using "x" and "14-x" for the two lengths, she used "7+x" and "7-x". Doing that makes the algebra required to finish the problem easier.

Of course, if formal algebra is not required, and if the speed of reaching the solution is important -- as on a timed competitive exam -- then a quick mental solution is simple. The sum of the length and width is 14 and the product (the area) is 40, so you are looking for two numbers whose sum is 14 and whose product is 40. Two seconds of thought (if you are slow!) gives the dimensions as 4 and 10.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The perimeter of a rectangular price tag is 28 centimeters. The area is 40 square centimeters.
What are the dimensions of the price tag?
~~~~~~~~~~~~~~~~~~

Let x be the length, y be the width. Then x+y is half of the perimeter x + y = 28/2 = 14 cm. The average of x and y is = 14/2 = 7 cm, and we can write for the sides x and y x = 7 + w, y = 7 - w, where w is some unknown deviation from the average. To find w, use the area equation xy = 40, or, which is the same (7+w)*(7-w) = 40. Simplify and find w 49 - w^2 = 40, w^2 = 49 - 40 = 9, w = = 3. Thus x = 7+3 = 10; y = 7-3 = 4. ANSWER. The dimensions of the card are 10 cm (the length) and 4 cm (the width).

Solved in a simple way, closed to the mental solution.

Question 1207593: A rectangle is 2.8cm wide and the length of each diagonal is 5.3cm. Calculate the length of the rectangle
Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

given:

the length , the width , and one diagonal of the rectangle form right triangle
so,

Question 1206957: imgur.com/a/6nVwylD
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Triangle WAR is a right-angled triangle with one leg 28 units (AR). The other leg, WA, is WA = CW + AC = 15+6 = 21 units long. So, triangle WAR is (3,4,5) right angled triangle (since 21:28 = 3:4). Triangle WDC is a right-angled triangle, similar to triangle WAR (they have one common acute angle CWD). Hence, triangle WDC also is (3,4,5)-right-angled triangle. Its hypotenuse WC is 15 units long - hence, triangle WDC has the sides WD = 9 units and CD = 12 units. Triangle CAO is a right-angled triangle, similar to triangle WAR (since CO is parallel to WR). Hence, triangle CAO is (3,4,5)-right-angled triangle. Its leg CA is 6 units long - hence, its hypotenuse CO is 10 units long. Thus the sides CD and CO of the rectangle COLD are 12 and 10 units, respectively. Then its diagonal is = = = 15.62049935... ANSWER. The length of the diagonal of rectangle COLD is = 15.62049935...

Solved.

Question 1206477: A rectangle has a 24-foot perimeter. The width is exactly 1/2 of its length. What is the area in square feet?
Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

A rectangle has a -foot perimeter.

if the width is exactly of its length, we have

substitute in

go to , substitute

so, the length and width

the area is:

Question 1206410: The area of a rectangular cutting board is 330 square inches.The perimeter is 74 inches.What are the dimensions of the cutting board? inches by inches.
Found 3 solutions by greenestamps, MathLover1, math_tutor2020:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The perimeter is 74 inches, so length plus width is 37 inches; the area is 330 square inches.

;

Solving the first equation for one of the variables and substituting in the second equation gives you a quadratic equation.

Any quadratic equation can be solved using the quadratic formula, as one of the tutors did.

The quadratic equation in this problem can also be solved by factoring, because the numbers are "nice"; the other tutor solved the problem that way.

However, solving the problem by factoring requires finding two numbers whose sum is 37 and whose product is 330 -- which is what the original problem requires.

So, if a formal algebraic solution is not required, the fastest way to solve the problem is by trial and error. Look at pairs of whole numbers whose product is 330 and find one for which the sum is 37.

33 and 10 is one easily found pair with a product of 330; but the sum is 43, which is too large. That means the two numbers we are looking for have to be closer together than 33 and 10.

A little mental arithmetic can then find 22 and 15, for which the sum is the required 37.

ANSWER: 22 inches by 15 inches

NOTE on finding the second pair of numbers with a product of 330, having found the first one....

We know the two numbers must be closer together than 33 and 10, so the 33 has to be smaller and the 10 has to be larger. Seeing that 33 is a multiple of 3, we can make it smaller by multiplying it by 2/3, which means we have to multiply the 10 by 3/2 to keep the same product. That gives us the numbers 22 and 15, which are the numbers we are looking for.

Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

given:
in^2
since , we have
...solve for
.....eq.1

in
since , we have

.......solve for
.......eq.2

from eq.1 and eq.2 we have
.......solve for

....factor

=> or=>preferred for width

go to
.......eq.2, substitute

answer: the dimensions of the cutting board are by

Answer by math_tutor2020(3835)   (Show Source):
You can put this solution on YOUR website!

L = length
W = width
2(L+W) = perimeter = 74
2(L+W) = 74
L+W = 74/2
L+W = 37
L = 37-W

L*W = area
(37-W)*W = 330
-W^2 + 37W - 330 = 0
W^2 - 37W + 330 = 0

Apply the quadratic formula (when a = 1, b = -37, c = 330)

or

or

or

If W = 22, then L = 37-W = 37-22 = 15
And vice versa if W = 15 then it leads to L = 22.
The order of L and W doesn't matter.

Check:
perimeter = 2*(L+W) = 2*(22+15) = 74
area = L*W = 22*15 = 330
Everything is confirmed.

Answer: 22 inches by 15 inches

Question 1206292: The perimeter of a rectangular field is 356m
If the width of the field is 83m what is its length?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20855)   (Show Source):
You can put this solution on YOUR website!

if the perimeter of a rectangular field is , then

If the width of the field is , then

answer: its length is

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

To find the length, from the perimeter subtract the width twice and divide the difference by two length = = 95 meters. Geometrically, it is as clear as 2 + 2 = 4.

Solved and explained.