Guide :

# what is the diagonal of 10' x 16'

~~what is the diagonal of 10 feel by 16 feet

## Research, Knowledge and Information :

### How do your find the diagonal measurement for a 10 ft square

... conversions and physics are my interests on Answers.com. ... 10' x 16' is not a square but a rectangle and the diagonal is square root of (10^2 + 16^2) ...

### What is the diagonal measurement for a square 10' x 16'

What is the diagonal measurement for a square 10' x 16'? ... What is the diagonal measurement of a 12 x 16 rectangle? The diagonal is 20.

### Diagonal - Wikipedia

In geometry, a diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge. Informally, any sloping line ...
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### Calculate area of a rectangle, its perimeter and diagonal

Rectangle Calculator Calculate area of a rectangle perimeter, diagonal, and two angles between the diagonal and the rectangle sides. ... = 16 Diagonal d = ...
Read More At : www.aqua-calc.com...

### Rectangle Area, Perimeter & Diagonal Length Calculator

Rectangle Area, Perimeter & Diagonal Length Calculator, ... Rectangle Area, Perimeter & Diagonal Length Calculator: Length of Rectangle: Width of Rectangle: Unit
Read More At : ncalculators.com...

### How to Calculate a Diagonal of a Square - wikiHow

How to Calculate a Diagonal of a ... The diagonal of a square is the line stretching ... 15^2 + 7^2 = c^2 225 + 49 = c^2 274 = c^2 sqrt*274 = sqrt*c^2 16.55 = c ...
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### 16:10 - Wikipedia

The shift from 16:10 to 16:9 ... there was criticism towards the lack of vertical screen real estate when compared to 16:10 displays of the same screen diagonal ...
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### What is the diagonal of a rectangle with a 16:9 ratio (width ...

What is the diagonal of a rectangle with a 16:9 ratio ... Each side of a cube is 5 inches long, how do you find the lengths of a diagonal of the cube?
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### Diagonals of Polygons - Maths Resources - Math Is Fun

Diagonals of Polygons A square has 2 diagonals: An octagon has 20 diagonals: A polygon's diagonals are line segments from one corner to another (but not the edges).
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## Suggested Questions And Answer :

### polygon ABCD is equilateral. prove that AC and BD bisect eachother and are perpendicular

4-sided equalateral be a ROMBUS...squashaed square all 4 sides the same draw 1 diagonal...2 sides the same, so it be isosaleez triangel , so base angels the same the diagonal make 2 triangels with 1 kommon side & other 2 sides the same so em 2 triangels the same that meen the diagonal bisekted the angels at the 2 ends draw other diagonal & it bisekt the other 2 angels & form 2 idetikal triangels with both diagonals, yu hav 4 triangels...all  hav 1 side=side av rombus=the same all hav 1 kommon side all hav 1 angel=weer 2 lines kross, so angels on 2 sides av the kros =the same Thus, all 4 triangel the same Thus, 1 side=half av 1 diagonal & other side=half av other diagonal Thus diagonals bisekt eech other

### How long is the diagonal of a 9in by 40 in rectangular

How long is the diagonal of a 9in by 40 in rectangle? Using Pythagoras' theorem, diagonal-squared = base-squared + length-squared diagonal-squared = 9^2 + 40^2 = 81 + 1600 = 1681 diagonal = sqrt(1681) = 41 diagonal = 41 in

### How to find c and y so that each quadrilateral is a parallelogram

The opposite internal angles of a parallelogram are equal, and adjacent angles are supplementary, but which of the given angles are opposite and which are adjacent? We know that all angles must be positive, so 7x-11>0 and 5y-9>0 so x>11/7 or 1.57 and y>1.8. We also know that all the angles of a parallelogram can be determined if just one is known, because of the relationships. This means that if we take any pair of angles we know that they are (a) equal or (b) supplementary. Take the first pair: 5x+29 and 5y-9: if (a), 5x+29=5y-9, so 5(y-x)=38 and y=(38+5x)/5; or if (b), 5(y+x)=160, or y+x=32 and y=32-x. Also, the remaining pair 3y+15 and 7x-11: if (a), 7x-3y=26, y=(7x-26)/3; or if (b), 7x+3y=176 and y=(176-7x)/3. We can see that if (a) is applied to the first pair at least one of x or y will contain a fraction. If (a) is applied to the second pair, which can be written 2x-8+(x-2)/3, x must be x=5, 8, 11, ..., 3n+2 for y to be an integer (where n is a positive integer) and y is 3, 10, 17, ..., and if (b), which can be written 58-2x-(x-2)/3, x must be 3n+2 for y to be an integer (where n is an integer 0 Read More: ...

### area of square problem

You are using squared graph paper, right? When you have drawn a 2x2 square you just count how many squares of graph paper are enclosed. If the side of the square is 2 then the area enclosed is 4 squares. But how do you draw a square with area 2? What you do is draw the diagonals of each of the 4 enclosed squares. For the top left and bottom right squares you draw the diagonal from the bottom left corner to the top right corner; and for the other two squares you draw the other diagonal. This gives you a square inside the bigger square tilted by 45 degrees. You have also divided the area of 4 squares into 8 triangles. So 8 triangles have a total area of 4 little squares. Each triangle has an area of ½. The tilted square contains 4 of these triangles. So if 8 triangles is equivalent to 4, then 4 triangles is equivalent to 2. That's how you know the area of your tilted square is 2. So its sides have length √2, the length of each diagonal. No Pythagoras! Looking at squares geometrically probably also helps you to answer your other question, when you start with the tilted square. Because the area of the tilted square is 2, its side length is √2. But the side is the diagonal of a unit square. This means to get the length of the side of the square knowing the length of its diagonal, you just divide the length of the diagonal by √2.

### what is the diagonal of a 56 ft by 26 ft 3 inch rectangle?

Because the diagonal bisects the the rectangle into two identical right-angled triangles you can apply the Theorem of Pythagorus: The square of the hypotenuse (diagonal) = the sum of the squares of the length & breadth of the rectangle. This means: diagonal (squared) = (26.25 x 26.25) + (56 x 56) = 689.0625 + 3136 = 3825.0625 to calculate the diagonal, you need to find the reverse of 'squared', which is 'square root' diagonal = square root of 3825.0625 = 61.84709 Note: this is not an exact answer, since the above number is an irrational number with an unending number of decimals, without any pattern. But it is accurate enough considering the number of decimals in the dimensions you supplied.

### magic square 16 boxes each box has to give a number up to16 but when added any 4 boxes equal 34

The magic square consists of 4X4 boxes. We don't yet know whether all the numbers from 1 to 16 are there. Each row adds up to 34, so since there are 4 rows the sum of the numbers must be 4*34=136. When we add the numbers from 1 to 16 we get 136, so we now know that the square consists of all the numbers between 1 and 16. We can arrange the numbers 1 to 16 into four groups of four such that within the group there are 2 pairs of numbers {x y 17-x 17-y}. These add up to 34. We need to find 8 numbers represented by A, B, ..., H, so that all the rows, columns and two diagonals add up to 34. A+B+17-A+17-B=34, C+D+17-C+17-D=34, ... (rows) A+C+17-A+17-C=34, ... (columns) Now there's a problem, because the complement of A, for example, appears in the first row and the first column, which would imply duplication and mean that we would not be able to use up all the numbers between 1 and 16. To avoid this problem we need to consider other ways of making up the sum 34 in the columns. Let's use an example. The complement of 1 is 16 anewd its accompanying pair in the row is 2+15; but if we have the sum 1+14 we need another pair that adds up to 19 so that the sum 34 is preserved. We would perhaps need 3+16. We can't use 16 and 15 because they're being used in a row, and we can't use 14, because it's being used in a column, so we would have to use 6+13 as the next available pair. So the row pairs would be 1+16 and 2+15; the column pairs 1+14 and 6+13; and the diagonal pairs 1+12 and 10+11. Note that we've used up 10 of the 16 numbers so far. This type of logic applies to every box, apart from the middle two boxes of each side of the square (these are part of a row and column only), because they appear in a row, a column and diagonal. There are only 10 equations but 16 numbers. In the following sets, in which each of the 16 boxes is represented by a letter of the alphabet between A and P, the sum of the members of each set is 34: {A B C D} {A E I M} {A F K P} {B F J N} {C G K O} {D H L P} {D G J M} {E F G H} {I J K L} {M N O P} The sums of the numbers in the following sets in rows satisfy the magic square requirement of equalling 34. This is a list of all possibilities. However, there are also 24 ways of arranging the numbers in order. We've also seen that we need 10 out of the 16 numbers to satisfy the 34 requirement for numbers that are part of a row, column and diagonal; but the remaining numbers (the central pair of numbers on each side of the square) only use 7 out of 16. The next problem is to find out how to combine the arrangements. The vertical line divides pairs of numbers that could replace the second pair of the set. So 1 16 paired with 2 15 can also be paired with 3 14, 4 13, etc. The number of alternative pairs decreases as we move down the list. 17 X 17: 1 16 2 15 | 3 14 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 2 15 3 14 | 1 16 | 4 13 | 5 12 | 6 11 | 7 10 | 8 9 3 14 4 13 | 1 16 | 2 15 | 5 12 | 6 11 | 7 10 | 8 9 4 13 5 12 | 1 16 | 2 15 | 3 14 | 6 11 | 7 10 | 8 9 5 12 6 11 | 1 16 | 2 15 | 3 14 | 4 13 | 7 10 | 8 9 6 11 7 10 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 8 9 7 10 8 9 | 1 16 | 2 15 | 3 14 | 4 13 | 5 12 | 6 11 16 X 18: 1 15 2 16 | 4 14 | 5 13 | 6 12 | 7 11 | 8 10 2 14 3 15 | 5 13 | 6 12 | 7 11 | 8 10 3 13 4 14 | 2 16 | 6 12 | 7 11 | 8 10 4 12 5 13 | 2 16 | 3 15 | 7 11 | 8 10 5 11 6 12 | 2 16 | 3 15 | 4 14 | 8 10 6 10 7 11 | 2 16 | 3 15 | 4 14 | 5 13 7 9 8 10 | 2 16 | 3 15 | 4 14 | 5 13 | 6 12 15 X 19: 1 14 3 16 | 4 15 | 6 13 | 7 12 | 8 11 2 13 4 15 | 3 16 | 5 14 | 7 12 | 8 11 3 12 5 14 | 4 15 | 6 13 | 8 11 4 11 6 13 | 3 16 | 5 14 | 7 12 5 10 7 12 | 3 16 | 4 15 | 6 13 | 8 11 6 9 8 11 | 3 16 | 4 15 | 5 14 | 7 12  7 8 9 10 | 3 16 | 4 15 | 5 14 | 6 13 14 X 20: 1 13 4 16 | 5 15 | 6 14 | 8 12 | 9 11 2 12 5 15 | 4 16 | 6 14 | 7 13 | 9 11 3 11 6 14 | 4 16 | 5 15 | 7 13 | 8 12 4 10 7 13 | 5 15 | 6 14 | 8 12 | 9 11 5 9 8 12 | 4 16 | 6 14 | 7 13 | 9 11 6 8 9 11 | 4 16 | 5 15 | 7 13 13 X 21: 1 12 5 16 | 6 15 | 7 14 | 8 13 | 10 11 2 11 6 15 | 5 16 | 7 14 | 8 13 | 9 12 3 10 7 14 | 5 16 | 6 15 | 8 13 | 9 12 4 9 8 13 | 5 16 | 6 15 | 7 14 | 10 11 5 8 9 12 | 6 15 | 7 14 | 10 11 6 7 10 11 | 5 16 | 8 13 | 9 12 12 X 22: 1 11 6 16 | 7 15 | 8 14 | 9 13 | 10 12 2 10 7 15 | 6 16 | 8 14 | 9 13 | 10 12 3 9 8 14 | 6 16  4 8 9 13 | 6 16 | 7 15  5 7 10 12 | 6 16 11 X 23: 1 10 7 16 | 8 15 | 9 14 2 9 8 15 | 7 16  3 8 9 14 | 7 16  10 X 24: 1 9 8 16 To give an example of how the list could be used, let's take row 3 in 17 X 17 {3 14 4 13}. If 3 is the number in the row column and diagonal, then we need to inspect the list to find another row in a different group containing 3 that doesn't duplicate any other numbers. So we move on to 15 X 19 and we find {3 12 8 11} and another row in 13 X 21 with {3 10 5 16}. So we've used the numbers 3, 4, 5, 8, 10, 11, 12, 13, 14, 16. That leaves 1, 2, 6, 7, 9, 15. In fact, one answer is: 07 12 01 14 (see 15x19) 02 13 08 11 16 03 10 05 09 06 15 04

### What is the probability that the difference of the two (2) numbers drawn is 10 or less?

After one number, a, is drawn, 49 remain. Let's suppose a=1, then the numbers 2-11 (10 tickets) would satisfy the criterion that b, the second number, satisfies |b-a|≤10. So that's 10/49. If a=2, then if b=1 or 3≤b≤12 |b-a|≤10. That's 11 tickets, 11/49. If a=25, 15≤b≤24 or 26≤b≤35, that's 20 tickets, 20/49. If a=30, then 20≤b≤29 or 31≤b≤40, 20 tickets again, 20/49. The rule is then that for 11≤a≤40 the probability for b is 20/49. The probability of picking a in range for this probability for b is 30/50. The combined probability is 30/50*20/49=12/49.  At the other end of the scale for 50≥a≥41 we have the same probabilities as for 1≤a≤10; when a=50, b can be 40-49 (10 tickets); when a=49, b can be 50 or 39-48 (11 tickets), and so on. We can put together a 50X50 table, but we can also summarise it: ("Y" means "yes, the values of 'a' (row number) and 'b' (column number) represented by this little square satisfy |a-b|≤10".) The small square with the red diagonal represents the whole of the large table on the left, which is only shown up to its 40th row. The red diagonal, containing X's for no-go, or disallowed combinations, excludes a=b. One way of calculating probability is to work out the fraction (number of "successful " events) ÷ (total possible number of events). This is the same as dividing the "area" occupied by Y's in the table by the "area" of the table itself, excluding the red diagonal. This is better represented by the small square with the red diagonal. This approach of using areas is the fastest way to arrive at the probability and it's easy to calculate: Area of table=50^2-50=2450=50*49 (there are 50 red squares in the red diagonal to be removed). Area of a trapezoidal corner piece=19+18+...+10=145. Total area of both trapezoids=290. (The right-angled trapezoids are formed by the Y's in rows a=1 to 10 and b=1 to 20, and a=41 to 50 and b=31 to 50. The trapezoid consists of a 10X10 square=100, less the diagonal of 10, which makes 90; and a right-angled triangle containing 55 Y's. Total 145. Not the usual way of calculating the area of a triangle!) Area of central "parallelogram"=30*20=600. Total area of Y's (central band)=890. So the probability of picking two numbered tickets, the difference of which is 10 or less, is 890/2450=89/245. In case you're not convinced that this simple solution is correct, let's go back to the calculations at the beginning. For any particular value of a there's always a 1/50 probability of picking it, but, depending on the value we have varying probabilities for b. Take a=1: the combined probability of |a-b|≤10 is 1/50*10/49; for a=2 it's 1/50*11/49, giving us the progression: 1/50*10/49+1/50*11/49+1/50*12/49+...+1/50*19/49. This also applies to a=50, a=49, ..., a=41. The total probability for the "ends" is (2/2450)(10+11+12+...+19)=2*145/2450. In the "middle section" we have constant 20/49 for b in the range 11≤a≤40=30/50*20/49=600=600/2450. The total is (290+600)/2450=890/2450=89/245. So the figures match the "areas".

### In a rhombus ABCD, AB=18 and AC=28. Find the area of the rhombus to the nearest tenth.

The sides of a rhombus are all equal and the diagonals cross at right angles and bisect each other. If the diagonals are drawn the rhombus can be seen to consist of 4 right-angled triangles, with the sides of the rhombus being the hypotenuses. The side of the rhombus is 18 and the length of one diagonal is 28 so the four triangles have a height of 28/2=14 and a hypotenuse of length 18. The third side=sqrt(18^2-14^2)=sqrt(128)=8sqrt(2)=11.3137. The area of each triangle is 0.5*8sqrt(2)*14=56sqrt(2), so the area of the rhombus is 4*56sqrt(2)=224sqrt(2)=316.8. The area of the trapezoid is made up of a central rectangle and two triangles, one on each side of the rectangle. If the length of the shorter side is x then the area of the rectangle is 11x. Let the base of one of the triangles be b then the base of the other will be 22-x-b. The combined area of the triangles is (11b+11(22-x-b))/2=11(22-x)/2. The combined area of the triangles and rectangle is 11x+11(22-x)/2=190. So 22x+231-11x=380, 11x=380-231=149 and x=149/11=13.5 approx. A. The diagonals of a parallelogram are not always perpendicular. Call the parallelogram ABCD where A is (2,3), C is (3,1) and D is (0,0). To get from D to C we go along 3 and up 1. To go from A to B we do the same, so B is (5,4).

### show that acute angle between two diagonals of cube has cosine 1/3

Call the corners or vertices of the unit cube A, B, C, D, E, F, G, H. If A to D clockwise is the top face and E to H clockwise is the bottom face where AE form an edge of the cube, then two diagonals are AG and BH. They intersect at X, bisecting one another. The base diagonal EG has length sqrt(1+1)=sqrt(2) by Pythagoras, and this is the length of all face diagonals. The length of AG is sqrt(AE^2+EG^2)=sqrt(1+2)=sqrt(3). AX=BX=(1/2)AG=sqrt(3)/2. In triangle ABX, we apply the cosine rule: AB^2=AX^2+BX^2-2AX*BXcosX. So 1=3/4+3/4-2*3/4cosX. -(1/2)=-(3/2)cosX and cosX=(2/3)(1/2)=1/3.

### What's the length of a diagonal of a cube that haves edges of length 6 in.

That depend on wot sorta "diagonal" yu look at Diagonal av a faes=2D so leng=sqrt(36+36) If yu want the 3D diagonal=tween opposit points, then leng=sqrt(36+36+36)