We have been given net of a pyramid. We are asked to find the total surface area of the pyramid.
The surface area of the pyramid will be area of base plus area of 4 triangular sides.
We have been given that base of pyramid is square, so area of base will be square of base side.
![\text{Area of base}=(\text{10 in})^2=100\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2drz4bt039djcg0ixwqkbhz7n47g1p40rm.png)
Let us find area of one triangular side.
![\text{Area of triangle}=(1)/(2)* \text{Base}*\text{Height}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/hcfwn26jdcy1pke2987ubvsa26fhi3qzy3.png)
![\text{Area of triangle}=(1)/(2)* \text{10 in}*\text{6 in}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ufh51li4xpqo7sdeqoee1adclrfd4nszis.png)
![\text{Area of triangle}=\text{5 in}*\text{6 in}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/36710puhcr7i7hubcbghv8knys1k1cjbh4.png)
![\text{Area of triangle}=30\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/uly2q8ajd90bakjjvl7xrb1pufa6wsy0vc.png)
Now we will multiply area of one triangular face by 4 to find area of 4 triangular faces.
![\text{Area of 4 triangular faces}=4* 30\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/25zp6j1prdspo8mgzt2r9c4jpvgf29oh22.png)
![\text{Area of 4 triangular faces}=120\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ds78ceaowfpws519m24rezxl1wkzeeay46.png)
Total surface area of pyramid would be area of base plus area of 4 triangular faces.
![\text{Total surface area of pyramid}=100\text{ in}^2+120\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/fquf2v3wrf2vxm55ssjh4y6elssrc4ojt6.png)
![\text{Total surface area of pyramid}=220\text{ in}^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/5ol5mqd2o8hikv08gjw7gih98dns3kvcxq.png)
Therefore, 220 square inches of copper sheet are needed to make one pyramid.