Area: Pentagon with a circular hole

Geometry Level 4

The image above consists of an isosceles triangle, A B C \triangle ABC where the height: h = 4 units h = 4 \text{ units} ; A B = B C \overline{AB}=\overline{BC} ; and A B C = 12 0 \angle ABC = 120^{\circ} . A B C \triangle ABC lies on top of the square A C D E ACDE and a circle of radius r r is inscribed in the square A C D E ACDE . Determine the area of the pentagon formed from combining the triangle A B C ABC and the square A C D E ACDE minus the area of the inscribed circle.

Give your answer to one decimal place.


The answer is 68.9.

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2 solutions

David Hontz
May 22, 2016

A B C \triangle ABC can be split into two congruent right triangles with legs h h and r r and an angle of 6 0 60^{\circ} . To determine r r : t a n ( 6 0 ) = r / 4 r = 4 t a n ( 6 0 ) = 4 3 u n i t s tan(60^{\circ})=r/4 \Rightarrow r=4tan(60^{\circ}) =4\sqrt{3} \space units The areas of each shape can be calculated as follows: A r e a o f A B C = 1 2 × h × 2 r = 4 ( 8 3 ) 2 = 16 3 u n i t s 2 Area \space of \space \triangle ABC = \frac{1}{2} \times h \times 2r = \frac{4(8\sqrt{3})}{2} = 16\sqrt{3} \space units^2 A r e a o f A C D E = ( 2 r ) 2 = ( 8 3 ) 2 = 64 ( 3 ) = 192 u n i t s 2 Area \space of \space \square ACDE = (2r)^2 = (8\sqrt{3})^2 = 64(3) = 192 \space units^2 A r e a o f = π r 2 = π ( 4 3 ) 2 = 48 π u n i t s 2 Area \space of \space\bigcirc = \pi r^2 = \pi (4\sqrt{3})^2 = 48\pi \space units^2

A n s w e r : 192 + 16 3 48 π 68.9 u n i t s 2 Answer: 192 + 16\sqrt{3} - 48\pi \approx \boxed{68.9 \space units^2}

Δ A C B i s 30 30 120. A C = 2 h C o t 30 = 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 = 1 2 8 3 4 + ( 8 3 ) 2 π 8 3 ) 2 4 A r e a s P e n t a g o n c i r c l e = 16 3 + 192 48 π = 68.9. \Delta~ACB~is~30-30-120.\\ \therefore~AC=2*h*Cot30=2*4*\sqrt3.=8\sqrt3.\\ Areas~Pentagon~ -circle~=areas~\Delta~ACB+\square~AEDC~-~circle,\\ Areas~Pentagon~ -circle~=\frac 1 2*8\sqrt3*4 + (8\sqrt3)^2-\pi*\dfrac{8\sqrt3)^2} 4\\ Areas~Pentagon~ -circle~=16\sqrt3+192 - 48*\pi=\Large~~~~\color{#D61F06}{68.9}.

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