Question

in the figure radius of circle with centre o is 3cm and oa 5cm then find ab+bc+ca​

Answer 1

given info : In the figure, radius of the circle with centre O is 3 cm and OA is 5 cm.

To find : the value of (AB + BC + CA)

Solution : see attached diagram,

Here circle touches the triangle at D, E and F as shown in figure.

We know, tangents from a point to a circle must be same.

So, AF = AE = x (let)

BF = BD = y (let)

CE = CD = z (let)

See ∆AOE (a right angled triangle)

So, AE = √(5² - 3²) = 4cm

so, AE = AF = x = 4 cm

Now from ∆ABD (a right angled triangle)

AB² = BD² + AD² [ Pythagoras theorem]

⇒(y + x)² = y² + (5cm + 3cm)²

⇒(y + 4)² = y² + 8²

⇒y² + 8y + 16 = y² + 8²

⇒8y = 64 - 16 = 48

⇒y = 6

Similarly, we get z = 6 ( you can try to solve ∆ADC using Pythagoras theorem)

now AB = y + x = 6 + 4 = 10cm

BC = y + z = 6 + 6 = 12 cm

CA = z + x = 6 + 4 = 10 cm

Now, AB + BC + CA = 32 cm