Question

50 points,
In triangle DEF, DE = 5, DF = 6, and EF = 7. Circles are drawn centered at D, E, and F, so that the circles are pairwise externally tangent. Find the sum of the areas of the three circles.

Answer 1

Answer:

Sum of the areas of the three circles is 91.06 units².

Step-by-step explanation:

Given:

Let the  radius of the circle with center D  = x

Let the radius  of the circle with center E  = y  

Let the radius  of the circle with center F  = z

To Find:

Sum of the areas of the three circles = ?

Solution:

So we have these equations

$DE =x + y = 5............( 1 )\\EF = x + z = 6..........( 2 )\\DF =y + z = 7...........( 3 )$

Subtract the second equation from the  first  and we have that

$y- z=-1$

Add this to equation to the third equation  and we have that

$\therefore 2y= 6\\\\\therefore y= 3\\\\\therefore x = 2\\\\\therefore z = 4$

Now we have Area of Circle

$\textrm{Area of Circle}=\pi (Radius)^{2}$

Substituting Radius we get

$\textrm{Area of Circle with center D}=3.14\times 2^{2}=12.56\ units^{2}$

$\textrm{Area of Circle with center E}=3.14\times 3^{2}=28.26\ units^{2}$

$\textrm{Area of Circle with center F}=3.14\times 4^{2}=50.24\ units^{2}$

∴ $\textrm{Sum of the areas of the three circles}=12.56+28.26+50.24=91.06\ units^{2}$