Question

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.

Answer 1

Complete Question :- 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 ΔOAB

a) 20

b) 36

c) 24

d) 48

5. Find area of ΔABC

a) 50

b) 60

c) 100

d) 90

Solution :- (from image) .

Since, tangent from same external points to the circle are equal in length .

Let,

• AD = AF = x cm .
• BD = BE = y cm .
• CF = CE = z cm .

so,

• AB = x + y = 12 cm .
• BC = y + z = 8 cm .
• CA = z + x = 10 cm .

adding all,

→ AB + BC + CA = 12 + 8 + 10

→ (x + y) + (y + z) + (z + x) = 30

→ 2(x + y + z) = 30

→ x + y + z = 15 cm .

then,

→ (x + y + z) - (y + z) = x => 15 - 8 = 7 cm = AD (Ans.1)

→ (x + y + z) - (x + z) = y => 15 - 10 = 5 cm = BE (Ans.2)

→ (x + y + z) - (x + y) = z => 15 - 12 = 3 cm = CF (Ans.3)

now, given that,

• Radius of circle = OD = 4 cm.

therefore,

→ Area ∆OAB = (1/2) * perpendicular height * Base = (1/2) * OD * AB = (1/2) * 4 * 12 = 24 cm² (Ans.4)

hence, using same formula we get,

→ Area ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA

→ Area ∆ABC = [(1/2) * radius * AB] + [(1/2) * radius * BC] + [(1/2) * radius * CA]

→ Area ∆ABC = (1/2) * radius * (AB + BC + CA)

→ Area ∆ABC = (1/2) * 4 * 30

→ Area ∆ABC = 2 * 30

→ Area ∆ABC = 60 cm² (Ans.5)

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Answer 2

Answer:

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