The following figure is made of six semicircles with the same diameter, and of four congruent equilateral triangles represented by AEG, EGF, BEF, and GFC. If AE = 3cm, then the area of this figure is, in cm2, equal to:
A) [tex] 9(3\pi + 4\sqrt{3} ) [/tex]
B) [tex] 4(3\pi + 4\sqrt{3} ) [/tex]
C) [tex] \frac{9}{4}(3\pi + 4\sqrt{3}) [/tex]
D) [tex] \frac{4}{9}(3\pi + 4\sqrt{3}) [/tex]
Answer: "C."
Why is it letter "C"?
No matter what I try, I cannot obtain this result. Does anyone know the correct procedure for this?

The following figure is made of six semicircles with the same diameter and of four congruent equilateral triangles represented by AEG EGF BEF and GFC If AE 3cm class=

Respuesta :

gmany

6 semicircles = 3 circles

The formula of the area of a equilateral triangle:

[tex]A_\triangle=\dfrac{a^2\sqrt3}{4}[/tex]

We have a = |AE| = 3cm. Substitute:

[tex]A_\triangle=\dfrac{3^2\sqrt3}{4}=\dfrac{9\sqrt3}{4}\ cm^2[/tex]

The formula of the area of a circle:

[tex]A_O=\pi r^2[/tex]

We have 2r = |AE| =3cm → r = 1.5cm. Substitute:

[tex]A_O=\pi\cdot1.5^2=2.25\pi\ cm^2[/tex]

The area of the figure:

[tex] A=4A_\triangle+3A_O\\\\A=4\cdot\dfrac{9\sqrt3}{4}+3\cdot2.25\pi=9\sqrt3+6.75\pi=6\dfrac{3}{4}\pi+9\sqrt3\\\\=\dfrac{27}{4}\pi+\dfrac{36\sqrt3}{4}=\dfrac{9}{4}\cdot3\pi+\dfrac{9}{4}\cdot4\sqrt3=\dfrac{9}{4}(3\pi+4\sqrt3)\ cm^2 [/tex]

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