[tex]\sf \bf {\boxed {\mathbb {GIVEN:}}}[/tex]
Radius of the circle "[tex]r[/tex]" = 5.5 yd
[tex]\sf \bf {\boxed {\mathbb {TO\:FIND:}}}[/tex]
The area of the given figure.
[tex]\sf \bf {\boxed {\mathbb {SOLUTION :}}}[/tex]
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf {b. \:47.5\:sq\:yd.}}}}}}[/tex]
[tex]\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:EXPLANATION:}}}[/tex]
We know that,
[tex]\sf\pink{Area\:of\:a\:semi-circle}[/tex] = [tex]\frac{\pi \: {r}^{2} }{2} [/tex]
Using 3.14 for π, we have
[tex] = \frac{3.14 \times ( {5.5 \: yd})^{2} }{2} \\[/tex]
[tex] = \frac{3.14 \times 5.5 \times 5.5 \: {yd}^{2} }{2}\\ [/tex]
[tex] = \frac{94.985 \: {yd}^{2} }{2} \\[/tex]
[tex] = 47.5 \: {yd}^{2} \\[/tex]
Using [tex]\frac{22}{7} [/tex] for π,
[tex] = \frac{22 \times 5.5 \times 5.5 \: {yd}^{2} }{7 \times 2} \\[/tex]
[tex] = \frac{665.5 \: {yd}^{2} }{14}\\ [/tex]
[tex] = 47.5 \: {yd}^{2} \\[/tex]
Therefore, the area of the semi-circle is [tex]47.5\: yd²[/tex].
[tex]\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♨}}}}}[/tex]
