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In the right triangle, an altitude is drawn from the right angle to the hypotenuse. Circles are inscribed within each of the smaller triangles.  What is the distance between the centers of these circles?

 

 Dec 25, 2020

Best Answer 

 #2
avatar+1641 
+3

Nope, it's not 7.

The distance between the centers is a hypotenuse of a triangle, and the legs are 1 and 7 units.

Distance is 5√2 units.

 Dec 25, 2020
 #1
avatar+129907 
+1

The hypotenuse  = 25

 

The   altitude length is    product of the legs  /hypotenuse = 15 *20 / 25 =  12

 The other leg  of  the right triangle in the  upper left of the figure  is 9

So...the area of this triangle  = (1/2) (product of the legs)  (1/2(9 *12)  = 54

And we can find the radius, R, of the inscribed circle as follows

 

54  =  (1/2) ( 9 + 12 + 15)  R

54 =  18  R

R  =  3

 

 

Similarly the  larger right triangle is a 12 - 16 - 20 right triangle

Its area  =  (1/2) (12 * 16)  =  96

 

And we can find the radius, R,  of the inscribed circle in this triangle as

 

96=  (1/2)  (12 + 16 + 20 ) R

96  = 24 R 

R = 4

 

The distance  between the  centers  = 3 + 4  =  7

 

INCORRECT>>>>>SEE JUGOSLAV'S ANSWER BELOW....

 

 cool cool cool

 Dec 25, 2020
edited by CPhill  Dec 25, 2020
 #2
avatar+1641 
+3
Best Answer

Nope, it's not 7.

The distance between the centers is a hypotenuse of a triangle, and the legs are 1 and 7 units.

Distance is 5√2 units.

jugoslav  Dec 25, 2020
 #3
avatar+129907 
+1

Thanks for the correction , jugoslav !!!!

 

 

cool cool cool

CPhill  Dec 25, 2020

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