Question

Determine the length of DC.

Need the answer urgently.

Answer 1

Use Pythagorean theorem

• DB²=DE²+EB²
• DB²=10²+24²
• DB²=100+576
• DB²=676
• DB=26

Now use proportion

• DE/DB=AC/BC
• 10/26=5/BC
• 5/13=5/BC
• BC=13

Now

DC=DB+BC=26+13=39

Answer 2

DC = 39

Explanation:

From inspection of the diagram:

• EA is tangent to both circles
• DE is the radius of circle D
• CA is the radius of circle C

The tangent of a circle is always perpendicular to the radius, therefore:

DE ⊥ EA and CA ⊥ EA

As ∠DEB and ∠BAC are both 90°, then DE is parallel to CA.

Therefore, ∠DBE and ∠ABC are vertically opposite angles, and are therefore equal.

As triangles ΔBED and ΔBAC have two pairs of corresponding congruent angles, the triangles are similar.

Therefore:

`\implies \sf (DE)/(CA)=(EB)/(BA)`

`\implies \sf (10)/(5)=(24)/(BA)`

`\implies \sf BA=12`

Using Pythagoras' Theorem for ΔBED to find DB:

`\implies \sf DE^2+EB^2=DB^2`

`\implies \sf 10^2+24^2=DB^2`

`\implies \sf DB^2=676`

`\implies \sf DB=√(676)`

`\implies \sf DB=26`

Using Pythagoras' Theorem for ΔBAC to find BC:

`\implies \sf CA^2+BA^2=BC^2`

`\implies \sf 5^2+12^2=BC^2`

`\implies \sf BC^2=169`

`\implies \sf BC=√(169)`

`\implies \sf BC=13`

Therefore, the distance between the center of the circles DC is:

`\begin{aligned} \implies \sf DC & = \sf DB + BC\\& = \sf 26 + 13\\& = \sf 39\end{aligned}`