Answer:
The slant height of the pyramid refers to the height of one face-triangle. Now, the cube has dimensions of 2 x 2 x 2, which means the height, base and length of the pyramid are equal, that's why is a squared pyramid.
Notice that the slant height forms a right triangle with two faces of the cube, and the pyramid intercepts the cube at the middlepoint of the plane.
First, we need to find the half-length of a diagonal, which is a leg of the right triangle formed by the slant height.
![d^(2)=2^(2) +2^(2)\\ d=√(4+4)= √(8)\\ d=2√(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/z3qowzqp0tiper4cmnaecd3jlwrbxi66ap.png)
The part related with the slant height is
![d_(half) =(2√(2) )/(2)=√(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/84bxcqdto54sm8zx9vw1w48mysnq2mnoxo.png)
So, the slant height is
![h_(slant)= \sqrt{(√(2) )^(2) +2^(2) }=√(2+4)=√(6)](https://img.qammunity.org/2021/formulas/mathematics/high-school/jsjymeko888c0tjt6zcod9h4kl3j6mo9ex.png)
Therefore, the slant height is the square root of 6 yards.
The surface area of a squared pyramid is
![S_(area)=2(2)(√(6))+(2)^(2) =4√(6)+4 \approx 13.8yd^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/74m7eusx964v27ujo246clins5jnzf63jl.png)
The five faces of the cube have a surface area
![S_(cube)=5(2)^(2) =20yd^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/awgeynw3t0au90jolul4a7lm0cvth53gm8.png)
Therefore, the composite surface area of the figure is
![S_(total)=13.8+20=33.8yd^(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/4nprbshmdlzyvp0rfrn9pjhxxul0odyvt6.png)