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A triangle has side lengths of 10, 24, and 26. Let a be the area of the circumcircle. Let b be the area of the incircle. Compute a - b.

 

Thank you so much!!!

 Apr 24, 2020
 #1
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Since these sides form a right triangle, the center of the circumcircle will be the midpoint of the hypotenuse. 

This makes the radius of the circumcircle one-half the length of the hypotenuse, or 13.

 

I can't see any easy way to find the radius of the incircle, so I did this.

I placed this on a coordinate axis with vertex C to be at the origin:  C  =  (0, 0).

I placed vertex A at (24, 0) and vertex B at (0,10).

I call the center of the incircle D.

I found the size of anlge A by using the inverse tangent of BC over CA. and then divided the angle by 2 to get

the size of angle(CAD) to be 11.30993247o.

Using this, the equation of AD is:  y  =  -0.2x + 4.8.

Also the equation of CD is:  y  =  x.

Solving these two simultaneously, the point D becomes (4, 4); making the radius of the incircle 4.

From this, I can find the area of the circumcircle and the area of the incircle.

 Apr 24, 2020

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