The image above consists of an isosceles triangle, △ A B C where the height: h = 4 units ; A B = B C ; and ∠ A B C = 1 2 0 ∘ . △ A B C lies on top of the square A C D E and a circle of radius r is inscribed in the square A C D E . Determine the area of the pentagon formed from combining the triangle A B C and the square A C D E minus the area of the inscribed circle.
Give your answer to one decimal place.
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Δ A C B i s 3 0 − 3 0 − 1 2 0 . ∴ A C = 2 ∗ h ∗ C o t 3 0 = 2 ∗ 4 ∗ 3 . = 8 3 . A r e a s P e n t a g o n − c i r c l e = a r e a s Δ A C B + □ A E D C − c i r c l e , A r e a s P e n t a g o n − c i r c l e = 2 1 ∗ 8 3 ∗ 4 + ( 8 3 ) 2 − π ∗ 4 8 3 ) 2 A r e a s P e n t a g o n − c i r c l e = 1 6 3 + 1 9 2 − 4 8 ∗ π = 6 8 . 9 .
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△ A B C can be split into two congruent right triangles with legs h and r and an angle of 6 0 ∘ . To determine r : t a n ( 6 0 ∘ ) = r / 4 ⇒ r = 4 t a n ( 6 0 ∘ ) = 4 3 u n i t s The areas of each shape can be calculated as follows: A r e a o f △ A B C = 2 1 × h × 2 r = 2 4 ( 8 3 ) = 1 6 3 u n i t s 2 A r e a o f □ A C D E = ( 2 r ) 2 = ( 8 3 ) 2 = 6 4 ( 3 ) = 1 9 2 u n i t s 2 A r e a o f ◯ = π r 2 = π ( 4 3 ) 2 = 4 8 π u n i t s 2
A n s w e r : 1 9 2 + 1 6 3 − 4 8 π ≈ 6 8 . 9 u n i t s 2