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

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

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* 2^(2)=12.56\ units^(2)`

`\textrm{Area of Circle with center E}=3.14* 3^(2)=28.26\ units^(2)`

`\textrm{Area of Circle with center F}=3.14* 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)`