Answer:
![4.35$ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/c1ub8ku8ksps5f0nlnyvf007ic3551v3l9.png)
Explanation:
For the equilateral triangle in a semicircle of radius 4cm.
Side Lengths =4cm
Angles =60 degrees
![\text{Area of a Triangle}=(1)/(2)ab\sin \theta](https://img.qammunity.org/2021/formulas/mathematics/high-school/wkn8yq72y75wr4occct83n2kwdv8a2cfs0.png)
Therefore, the area of one equilateral triangle
![=(1)/(2)*4*4*\sin 60^\circ](https://img.qammunity.org/2021/formulas/mathematics/high-school/jnc91647lxh7y35gxxj9ayk7s8gpdnj62t.png)
Area of the three equilateral triangle
![=3*(1)/(2)*4*4*\sin 60^\circ\\=20.7846$ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/cpwkawvsfnpjvhsyt5fehrwj1sytpfj9f6.png)
Area of the semicircle
![=0.5 * \pir^2\\=0.5 * \pi * 4^2\\=25.1327$ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/p5n68tdnqemvscaml7x64xdfzrbft3ckg4.png)
Therefore, the area left over =Area of Semicircle -Area of Triangles
=25.1327-20.7846
=
![4.35$ cm^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/c1ub8ku8ksps5f0nlnyvf007ic3551v3l9.png)