Dragan

The medians of a right triangle that are drawn from vertices of the acute angles have lengths of 2√13 and √73. Find the length of the hypotenuse.

AC => x       AB => y

From the point inside a square,  the distance to three corners is 4, 5, and 6 cm, respectively.
Find the length of the side of a square.

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https://web2.0calc.com/questions/help-please-thx_4#r4

Important:  triangle WXY can be any type of triangle and the result will be the same!!!

I used the right triangle, and I got this answer:  YZ = 9.428571429

I'll let you convert that decimal into a fraction. 

Here's a very clever method to solve this problem:

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Let a, b, and c be the radii of circles A, B, and C respectively. From the, Segment Addition Postulate we know that:

AB = a + b = 14

BC = b + c = 18

AC = a + c = 16

We have three equations and three unknowns:

a + b = 14

b + c = 18

a + c = 16

Let's eliminate the a by multiplying the first equation by -1 then adding them together: I'll line up the a's b's and c's, putting in a 0 if one of them is missing:

-a - b + 0 = -14

0 + b + c = 18

a + 0 + c = 16

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0 + 0 + 2c = 20

2c = 20

c = 10 The radius of circle C is 10.

To find b:

b + c = 18

b + 10 = 18

b = 8 The radius of Circle B is 8

To find a, use a + b = 14, knowing that b = 8.

(Solved by Philip P)

I've posted this question nearly 6 years ago, and my son answered it. He told me about that 2 days ago. (Beautiful communication)

His answer is very close to mine. (Dragan ≅ jugoslav ) Here's his answer:

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This one was a bit tricky, but I think it's because I made it more complicated than it needed to be.

The most important first step was to determine the center of our fourth circle. How far away is it from the centers of the other three circles?

I set the center of the bottom left circle as my "origin." Then, creating a triangle between the three centers, I solved for the centroid of the triangle (which had an effective coordinate of (30,17.32)). This means that the distance between the fourth circle and the other three circles was 34.64 cm.

Now, knowing this, I generated an independent problem: what is an overlapping area of two circles 34.64 cm apart, with radii of 30 cm? Once I had this number, it was simply a matter of multiplying it by three and subtracting it from the overall area of one circle.

You probably don't want to see my math...I used an integral in the cartesian coordinate system. I basically found the midpoint of their intersecting area and set it as my lower limit (17.32), set 30 as my upper limit, and integrated sqrt(30^2-x^2)dx. Then I multiplied that area by 4, since integrating circles in cartesian coordinates is weird. It was messy, but the answer: the overlapping area was 871.08 cm^2.

Since the area of the fourth circle is pi*r^2 (or 2827.43 cm^2), you simply subtract 871.08*3 (or 2613.24 cm^2, which is the area of intersection between the fourth circle and each of the other three circles).

This yields the final answer of 214.22 cm^2...give or take for rounding error. ;)

-KMM (May 19, 2015)

There's more than one way to... 

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We can use this formula to calculate the circumradius. (R-radius, s-semiperimeter, a,b,c,d-sides)

When we have the length of a circumradius, we can calculate the angles B and C or A and D, and then using the law of cosines we can find diagonals AC and BD.

This is perhaps the longer way, but it would work. 

Excellent work, Melody!!!  

Let A, B, C be points on circle O such that AB is a diameter, and CO is perpendicular to AB. Let P be a point on OA, and let line CP intersect the circle again at Q. If OP = 20 and PQ = 7, find r^2, where r is the radius of the circle.

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OP = 20        PQ = 7         CP = sqrt(r² + 20²)        AP = r - 20         PB = r + 20

PQ * CP = AP * PB

7 * sqrt(r² + 20²) = (r - 20)(r + 20)

r = 4√39           r² = 624

3x - 5 + x + 1 = 180         x = 46

To answer the question... ∠RQT ≅ ∠SQT = 46/2 = 23 degrees 

[ABC] = √3 / 4 * AB²

Calculating the area of a hexagon:

AC = 3        MC = 1.5        A'B' = 4         C'N = √12         QN = 1/3(C'N)

CQ = cos30º * MC

CN = CQ - QN

DN = tan30º * CN

[CDE] = DN * CN

Hexagon area = [ABC] - 3[CDE]

PQ / 2  ==>  x

x*( x + x / tan30º ) = 6 + 6√3        x = 2.449491

Radius  AP = x / sin45º = 2√3

Radius  BP = x / sin30º = 4.898979486

Area = [AP²*pi (270/360)] + [BP²*pi (300/360)]

Area = 29pi

Here's a similar question: 

https://web2.0calc.com/questions/help_54234#r3

That's gonna look like this:

You are right! Angle B = 23 degrees  

jugoslav