Question

Plsssss help.

Answer 1

Answer:

a)  $AE=4\sqrt{3}$

b) $A_s=16(4\sqrt{3}-\pi)\ cm^2$

Step-by-step explanation:

Lines in a Circle

(a)

Points AEO form a right triangle, which hypotenuse is the radius of the larger circle (8 cm) and one of its legs is the radius of the smaller circle (4 cm).

Thus, the distance AE is calculated by using the Pythagora's Theorem:

$AE^2=8^2-4^2$

$AE^2=64-16=48$

$AE=\sqrt{48}$

Since 48=16*3:

$AE=\sqrt{16}\sqrt{3}$

$\boxed{AE=4\sqrt{3}}$

(b)

The shaded area is the area of the rectangle ABCD minus the area of the smaller circle:

$A_s=A_r-A_c$

The rectangle has a length of twice AE:

$L=8\sqrt{3}\ cm$

And a width equal to the diameter of the smaller circle:

$W=8\ cm$

The area of the rectangle is:

$A_r=8\sqrt{3}\ cm*8\ cm$

$A_r=64\sqrt{3}\ cm^2$

The area of the smaller circle is:

$A_c=\pi\ r^2$

$A_c=\pi\ 4^2\ cm^2$

$A_c=16\pi\ cm^2$

Thus, the shaded area is:

$A_s=(64\sqrt{3}-16\pi)\ cm^2$

Factoring:

$\boxed{A_s=16(4\sqrt{3}-\pi)\ cm^2}$