The outer and inner triangles are both equilateral and the circle touches all three sides of the outer triangle. if the area of the inner triangle is 10 square centimeters, what is the area of the outer triangle

The area of outer Δ = 40 cm²

Step-by-step 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

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day