If △CDE
has sides that are tangent to the circle, and CF=15 cm
, BD=12 cm
, and EG=4 cm
, then what is the perimeter of △CDE
?
To find the perimeter of triangle △CDE, we first need to find the length of the sides CE, DE, and CD.
Let x be the length of CE, y be the length of DE, and z be the length of CD.
Since CE and DE are tangent to the circle, we can use the tangent-tangent theorem which states that the length of the tangents drawn from an external point to a circle are equal.
So, CF = BD
15 cm = 12 cm
This means CE = DE = 15 cm.
Now, we draw perpendiculars from C, D, and E to the center of the circle O. Let the perpendiculars be h1, h2, and h3 respectively.
Now using Pythagoras theorem in triangles CFE, DCE, and DBE:
CE^2 = CO^2 + OE^2
15^2 = (x+4)^2 + 4^2
225 = x^2 + 8x + 16
CD^2 = CO^2 + DO^2
z^2 = (y+4)^2 + 4^2
z^2 = y^2 + 8y + 16
ED^2 = EO^2 + DO^2
15^2 = y^2 + 16
y = 15
Using the values of x and y in the equation of CD:
z^2 = 225 + 8*15 + 16
z^2 = 225 + 120 + 16
z = √361
z = 19
Therefore, the sides of the triangle are CE = DE = 15 cm, CD = 19 cm.
The perimeter of the triangle △CDE is:
Perimeter = CE + CD + DE
Perimeter = 15 + 19 + 15
Perimeter = 49 cm
Therefore, the perimeter of the triangle △CDE is 49 cm.