Explanation
In the first case, we will find the area of the triangle. This is given as;
[tex]\begin{gathered} \text{Area of}\triangle=ab\sin \theta \\ =5\times5\sin 40^0 \\ =16.0697 \end{gathered}[/tex]
To get the area of the semicircle we will require the diameter of the semicircle, which is the unknown side of the triangle.
Let the unknown side be x
Therefore, we can use the cosine rule.
[tex]\begin{gathered} x=\sqrt[]{a^2+b^2-2ab\cos \theta} \\ x=\sqrt[]{5^2+5^2-2\times5\times5\cos 40^0} \\ x=\sqrt[]{25^{}+25^{}-50\cos 40^0} \\ x=\sqrt[]{50^{}-50\cos 40^0} \\ x=3.4202 \end{gathered}[/tex]
x represents the diameter of the semicircle. We can then have the radius as half of x.
[tex]r=\frac{x}{2}=\frac{3.4202}{2}=1.7101[/tex]
The area of the semicircle is given as
[tex]\text{area of a semicircle =}\frac{1}{2}(\pi r^2)=\frac{1}{2}(\pi\times1.7101\times1.7101)=4.5937^{}[/tex]
Therefore, the area of the shape is the sum of the triangle and the semi-circle.
[tex]\begin{gathered} \text{Area of the shape =16.06}97+4.5937 \\ =20.67 \end{gathered}[/tex]
Answer:
[tex]20.67[/tex]
