Answer:
The area of outer Δ = 40 cm²
Explanation:
∵ Area of the small triangle = 10 cm²
If we join the center of the circle with the 3 vertices of the inner Δ
These 3 segments are the radii of the circle
Now the inner triangle has 3 isosceles Δ their sides are r , r and s1 with vertex angle 120° ⇒ (360° ÷ 3 = 120°)
Where r is the radius of the circle and s1 is the side of the inner triangle
By using cosine rule
(s1)² = r² + r² - 2r²cos120 = r² + r² - 2r² (-0.5) = r² + r² + r² = 3r²
∴ s1 = r√3
∵ The radius of the circle ⊥ to the side of the outer Δ because the side of the outer Δ is a tangent to the circle
If we join a vertex of the outer Δ with the center of the circle
We will have a right angle triangle of two legs r and half s2 with angle 60° (120° ÷ 2 )between them ⇒ s2 is the side of outer Δ
∴ tan 60° = 1/2 (s2) ÷ r ⇒ √3 = 1/2 (s2) ÷ r = (s2)/2r
∴ s2 = 2r√3
∴ s2 : s1 = 2r√3 ÷ r√3 = 2 : 1
∴ The side of the outer Δ is double the side of the inner Δ
By using similarity ratio
A2/A1 = (s2/s1)² ⇒ A2 and A1 are the areas of outer and inner triangles
∴ A2 : A1 = (2/1)² = 4/1
∴ A2 = 4 A1
∴ A2 = 4 × 10 = 40 cm²