Q) Varun has been selected by his School to design logo for Sports Day T-shirts for students and staff . The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.

1. Find the length of AD
a) 7              b) 8                 c) 5                          d) 9

2. Find the Length of BE
a) 8              b) 5                 c) 2                          d) 9

3. Find the length of CF
a) 9             b) 5                 c) 2                          d) 3

4. If radius of the circle is 4cm, Find the area of ∆ OAB
a) 20           b) 36               c) 24                        d) 48

5. Find area of ∆ ABC
a) 50           b) 60               c) 100                     d) 90

Ans: 

1. Length of AD: 

Step 1: Since tangents drawn from a point on a circle are equal

∴ AD = AF

Similarly, BD = BE

and CE = CF

Step 2: let’s consider AD = a,

∴ AD = AF = a

Similarly, if BE = b, then BD = BE = b

and if CF = c, then CE = CF = c

Step 3: Let’s mark these sides on the diagram:

Now from the diagram,

side AB = AD + BD = a + b

Similarly, BC = BE + CE = b + c

and CA = CF + AF = c + a

Step 4: By adding all the three sides, we get:

AB + BC + CA = (a + b) + (b + c) + (c + a)

∴ 12 + 8 + 10 = a + b + b + c + c + a

∴ 30 = 2 ( a + b + c)

∴ (a + b + c) =

∴ (a + b + c) = 15 cm

Step 5: Now we need to find the length of AD

Since AD = a, let’s find the value of a

Since a =  a + (b + c) – (b + c)

∴ a = (a + b + c) – (b + c)

∵ b + c = BC = 8 cm

∴ a = 15 – 8 = 7 cm

∴ AD = 7 cm

Therefore, option (a) is correct.

2. Length of BE:

Step 6: Since BE = b, let’s find the value of b

Since b =  b + (a + c) – (a + c)

∴ b = (a + b + c) – (a + c)

∵ a + c = AC = 10 cm

∴ b = 15 – 10  = 5 cm

∴ BE = 5 cm

Therefore, option (b) is correct.

3. Length of CF:

Step 7: Since CF = c, let’s find the value of c

Since c =  c + (a + b) – (a + b)

∴ c = (a + b + c) – (a + b)

∵ a + b = AB = 12 cm

∴ c = 15 – 12 = 3 cm

∴ CF = 3 cm

Therefore, option (d) is correct.

4. Area of Δ AOB:

Step 8: Let’s connect Point O to A and B by drawing OA and OB

Also, connect OD

Given the radius of circle = 4 cm, ∴ OD = 4 cm

Step 9: Since the side AB touches the circle at point D, hence the ∠ ODB = 90⁰

Or, we can say that OD is the altitude of Δ AOB

∴ Height of Δ AOB = OD = 4 cm

Step 10: Now Area of Δ AOB = x Base x height

∴ Δ AOB = x 12 x 4

∴ Δ AOB = 24 cm²

Therefore, option (c) is correct.

5. Area of Δ ABC:

Step 11: We have calculated the Area of Δ AOB in step 10

Similarly, let’s calculate the area of Δ AOC  and Δ BOC

Let’s connect OC, OF, and OE … here’s the revised diagram as shown below:

Step 12: Since the side AC touches the circle at point F, hence the ∠ OFA = 90⁰

Or, we can say that OF is the altitude of Δ AOC

∴ Height of Δ AOC = OF = 4 cm

Now Area of Δ AOC = x Base x height

∴ Δ AOC = x 10 x 4

∴ Δ AOC = 20 cm²

Step 13: Since the side BC touches the circle at point E hence the ∠ OEB = 90⁰

Or, we can say that OE is the altitude of Δ BOC

∴ Height of Δ BOC = OE = 4 cm

Now Area of Δ BOC = x Base x height

∴ Δ BOC = x 8 x 4

∴ Δ BOC = 16 cm²

Step 14: 

∵ Δ ABC = Δ AOB + Δ AOC + Δ BOC

Here, we have Δ AOB = 24 cm²       (from Step 10)

∴ Δ AOC = 20 cm²                  (from Step 12)

∴ Δ BOC = 16 cm²                  (from Step 13)

∴ Δ ABC = Δ AOB + Δ AOC + Δ BOC

∴ Δ ABC = 24 + 20 + 16 = 60 cm²

Therefore, option (b) is correct.

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