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Featured 3 months ago

Sides:

Side

has a length of

Not that

#color(red)(a)# can not be one of the equal length sides of the equilateral triangle since the maximum area such a triangle could have would be#(color(red)(2sqrt(2)))^2/2# which is less than#15#

Using

Using the Pythagorean Theorem:

and since the triangle is isosceles

Featured 3 months ago

Radius of

Perimeter of each outer disc =

If we sketch a picture of the situation, we can see that the centers of the six discs lie on the vertices of a regular hexagon inscribed in the circle.

As the triangles formed by connecting the vertices of the hexagon to its center are equilateral, we know that the sides of the hexagon are also of length

As the sides of the hexagons serve as distances between the centers of the discs, we know each disc must intersect at a midpoint of a side, meaning each disc has a radius of

Given that the discs each have radius

Finally, the perimeter of each of the outside discs is given by the product of twice its radius and

Featured 2 months ago

As per given figure

Since

( sides of equilateral

So

**Using trigonometry**

In triangle BCR

and

If the length of each side of the equilareral triangle BPR be **a**,then

For

So

Now

**Without using trigonometry**

We have shown above

So CR=AP

Now

Applying Pythagoras theorem for

Niw applying Pythagoras theorem for

So

Featured 2 months ago

The farthest distance that between to points on the circle is the sum of the distance between the centers and the two radii:

Here is a graph of the two circles:

The circles overlap but the larger circle does not contain the other. This could have been deduced by the following process.

We assign

For circle 1:

For circle 2:

Compute the distance,

If the larger circle contained the smaller, then the distance between the centers would be less than difference between the two radii, 1. This is not the case.

The farthest distance that between two points on the circle is the sum of the distance between the centers and the two radii:

Featured 2 months ago

The coordinates of C are either (

This question took a lot longer to answer than I first anticipated!

There are two possible solutions for point C, both of which must lie on the line bisecting AB, as ABC is isosceles and all the points on this perpendicular line are equidistant from points A and B.

The mid-point of AB is straightforward enough:

AB has a slope of 6, because an increase of 1 on the

Now triangle ABC has an area of 12

The area

AB =

So with this value for

So point C must lie on one side or the other of AB, a distance of

In order to calculate the coordinates, we can imagine a right-angled triangle with a hypotenuse of length

From Pythagorus, we know that

So

Point

So

Point

So

Featured 1 month ago

Given two conics

The tangent space to those conics is

the tangency condition reads

Solving now

for

Attached a plot focusing the main elements

The area is

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