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There is no such triangle, due to

Without loss of generality, one of the vertices is at

Let the vertex anticlockwise from

The midpoint of the corresponding side is

The line through

The third vertex of the triangle lies on the line through the midpoint

In fact it will lie at the point:

#(m/2, n/2) + sqrt(3)/2 (-n, m) = ((m-sqrt(3)n)/2, (n+sqrt(3)m)/2)#

since the height of an equilateral triangle is

So we require

Since

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The trapezoid described is somewhat as the figure below:

The formula of the area of the triangle is

Since AF is perpendicular to the triangle's side DE, and AF=BC=12, we have:

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Perimeter

This is Geometry, so lets look at at a picture of what we are dealing with:

We are told

and to use

If

and

Perimeter

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The area of any triangle is given by

Therefore the area here is

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The hypotenuse

The above triangle is a right angled isosceles triangle , with

The length of the leg given

So,

The value of the hypotenuse

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In a regular polygon each interior angle can be obtained in this way:

From the conditions of the problem:

That's the conjugation of this two inequations:

Resolving the first inequation

Resolving the second inequation

Conjugating the two inequations

Since

By the way

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Calculate the area of the circular sector, from which subtract the area of the triangles whose base is the circles' chord defined by the intersection points; finally sum the results.

Consider Figs. 1 and 2

**Figure 1** shows two circles (with centers

The area of interest is encompassed by arcs ADB and AEB.

In such a problem, most probably

We only need one more information to determine the area of the region ADBE: the proportion of any of the areas of circular sector to the total area of its circle or the angle of the circular sector (

If we have the coordinates of the center points C1 and C2 as well as the radii

If we get the chord AB (called "x" in **Figure 2** ) we can obtain

Knowing

Next we can easily find

Then with the chord AB (

Next we need just to subtract from each circular sector the area of its triangle.

Finally, we sum the results of the aforementioned subtractions obtaining the area of interest. Job done!

OR we could integrate the area between the equations of the two circles from the point A (

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C)

To ease the comprehension of the problem refer to the figure below:

If the sides BC and AD are extended the

First, in addition to the angles informed it was possible to determine other interior angles of the triangles ABC and ABD, as shown in the figure.

Then using the Law of Sines on these triangles:

#triangle_(ABC)#

#"AB"/sin (120^@-alpha)="BC"/sin alpha="AC"/sin 60^@#

#triangle_(ABD)#

#"AB"/sin (60^@+alpha)="AD"/sin (60^@-alpha)="BD"/sin 60^@#

To focus on viable hypotheses I suggest to begin trying to prove a specific case.

For instance

From

From

Note that

Finally,

#AD+BC=5*(sin 15^@+sin 45^@)/(sin 75^@)=5*1=AB#

So in this case hypothesis (C) is true (I tried the others and verified that they are false.)

Now for the general proof of the hypothesis (C), following the same steps that worked for the specific proof:

From

Observation:

#sin (120^@-alpha)=sin (180^@-(120^@-alpha))=sin(60^@+alpha)#

Therefore,

# "BC" ="AB" * (sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

From

This will get you

#AD="AB"*sin(60^@-alpha)/sin (60^@+alpha)#

#=AB((sqrt(3)/2)*cos alpha-(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

Finally,

#BC+AD="AB"(sin alpha+(sqrt(3)/2)*cos alpha-(1/2)*sin alpha)/((sqrt(3)/2)*cos alpha+(1/2)*sin alpha)#

#BC + AD ="AB" * 1#

Therefore,

#AB=BC+AD# and the hypothesis (C) is true.**

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For a rectangular prism with sides

#"SA"=2(wl+lh+hw)#

This occurs since there are two pairs of three different faces on every rectangular prism.

Each pair of faces is a different rectangle using two of the the three dimensions of the prism as its own side.

One side is just

This could also be imagined as a series of flattened-out rectangles:

The blue rectangles are

The yellow rectangles are

The red rectangles are

Again, the surface area would be

#"SA"=2wl+2lh+2hw#

#=2(wl+lh+hw)#