Question

Six circles each of radius 3 cm are inscribed in an equilateral triangle ABC such that they touch each other and also touch the sides of the triangle as shown in the adjacent figure. Then height of triangle ABC is​

Answer 1

Step-by-step explanation:

⇒ Here, AB=BC=AC=12cm

⇒ Let OP=OR=OQ=r

⇒ We have O as the incenter and OP,OQ and OR are equal.

⇒ ar(△ABC)=ar(△OAB)+ar(△OBC)+ar(△OCA)

43×(side)2=(21×OP×AB)+(21×OQ×BC)+(21×OR×AC)

⇒ 43×(12)2=(21×r×12)+(21×r×12)+(21×r×12)

⇒ 43×(12)2=3(21×12×r)

∴ r=18363

∴ r=23cm

⇒ Area of the shaded region = Area of △ABC - Area of circle.

⇒ Area of the shaded region =43×(12)2−722×(23)2

⇒ Area of the shaded region =(62.35−37.71)cm2=24.64cm2

Answer 2

Answer:

24.64cm^2

Step-by-step explanation:

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