Question

A circle is inscribed in a ∆DEF, such that it touches the sides DE, EE and

DF at points A, B and C respectively. If the lengths of sides DE, EF and DF are 9 cm, 13 cm

and 11 cm respectively, find the length of BE, CF and AD.​

Answer 1

Answer:

We know that tangent drawn from external point to a circle are equal

So AD=AF=x

Or BD=BE=y

And CF=CE=z

Now, AB=AD+DB=x+y=12 cm ...(1)

Or BC=BE+EC=y+z=8 cm ...(2)

And AC=AF+CF=x+z=10 cm ...(3)

Adding the above three equation, we get

2(x+y+z)=12+8+10=30 cm

Or x+y+z=15 cm

As per equation (1),

x+y=12 cm

Then, z=15−10=5 cm =CF

As per equation (3),

x+z=12 cm

Then, y=15−12=3 cm =BE

As per equation (2),

y+z=8 cm

Then, x=15−8=7 cm =AD

Answer 2

Answer:

Here is the answer

Step-by-step explanation:

Tangent are equal when they are drawn from same external point

so AD=DC=a

AE=EB=b

BF=FC=c

DE can be written as sum of AD+AE=a+b=9 …   (i)

Similarly,

EF=EB+BF=b+c=13 … (ii)

DF=CF+DC=c+a=11 …(iii)

Add (i),(ii),(iii) we get

a+b+b+c+c+a=9+13+11

2(a+b+c)=33

a+b+c=33/2 ( let it be in fraction so calculation can be done easily)

Substitute a+b+c value in (i),(ii),(iii)

(i) --- a+b=9 ( since c value is to be found)

     c=33/2-9=7.5

(ii) --- b+c=13 ( since a value is to be found)

      a=33/2-13=3.5

(iii) --- a+c=11 ( since b value is to be found)

         b=33/2-11=5.5

Therefore BE=7.5cm, CF=3.5, AD=5.5cm

Hope you can understand