geno3141

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 #1
avatar+23254 
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I'll try the first one.

Since the triangle has side lengths of 10, 24, and 26, this triangle is a right triangle (102 + 242 = 262).

 

The center of the circumcenter of a right triangle is the midpoint of the hypotenuse.

Since the hypotenuse has a length of 26, the distance from this point to any of the vertices is 13.

Therefore, the area of the circumcircle is:  pi·132  =  169·pi.

 

The center of the incircle is the point of intersection of the angle bisectors.

Call the vertex between the 10 and 24 sides = C (it is a right angle).

Call the vertex between the 24 and 26 sides = A.

Call the center of the incircle X.

Draw the perpendicular from X to side AC; call this point Y; this is a radius of the incircle;

call its value 'h'.

Angle C = 90o; therefore, angle(XCY) = 45o.

I will need to find the size of angle(XAY).

I will use the sin(A) in triangle ABC) and find one-half of that value.

sin(A) = 10/26   --->   A  =  sin-1(10/26)  =  22.61986o   --->   angle(XAY)  =  11.31o.

 

Now, to find the value of h:

Call the distance from C to Y 'x'; therefore, the distance from A to Y is '24 - x'.

 

In triangle(XCY), tan(45o)  =  h/x   --->   h  =  x·tan(45o).

In triangle (XAY), tan(11.31o)  =  h/(24 - x)   --->  h  =  (24 - x)·tan(11.31o). 

 

Setting these two equation equal to each other:  x·tan(45o)  =  (24 - x)·tan(11.31o)

--->   x·tan(45o)  = 24·tan(11.31o) - x·tan(11.31o)

--->   x·tan(45o) + x·tan(11.31o)  =  24·tan(11.31o)

--->   x·( tan(45o) + tan(11.31o) )  =  24·tan(11.31o)

--->   x  =  24·tan(11.31o) / [ tan(45o) + tan(11.31o) ]

--->   x  =  4

 

Therefore, the area of the incircle is:  pi·42  =  16·pi.

 

169·pi  -  16·pi  =  153·pi 

22 февр. 2020 г.