Answer:
Part 1) ![LA=(80+16\sqrt{13})\ in^{2}](https://tex.z-dn.net/?f=LA%3D%2880%2B16%5Csqrt%7B13%7D%29%5C%20in%5E%7B2%7D)
Part 2) ![TA=(104+16\sqrt{13})\ in^{2}](https://tex.z-dn.net/?f=TA%3D%28104%2B16%5Csqrt%7B13%7D%29%5C%20in%5E%7B2%7D)
Step-by-step explanation:
Part 1) Find the lateral area of the prism
we know that
The lateral area of the prism is equal to
![LA=Ph](https://tex.z-dn.net/?f=LA%3DPh)
where
P is the perimeter of the base
h is the height of the prism
Applying the Pythagoras Theorem
<u>Find the hypotenuse of the triangle</u>
![c^{2}=4^{2}+6^{2}\\ \\c^{2}=52\\ \\c=2\sqrt{13}\ in](https://tex.z-dn.net/?f=c%5E%7B2%7D%3D4%5E%7B2%7D%2B6%5E%7B2%7D%5C%5C%20%5C%5Cc%5E%7B2%7D%3D52%5C%5C%20%5C%5Cc%3D2%5Csqrt%7B13%7D%5C%20in)
<u>Find the perimeter of triangle</u>
![P=4+6+2\sqrt{13}=(10+2\sqrt{13})\ in](https://tex.z-dn.net/?f=P%3D4%2B6%2B2%5Csqrt%7B13%7D%3D%2810%2B2%5Csqrt%7B13%7D%29%5C%20in)
<u>Find the lateral area</u>
![LA=Ph](https://tex.z-dn.net/?f=LA%3DPh)
we have
![P=(10+2\sqrt{13})\ in](https://tex.z-dn.net/?f=P%3D%2810%2B2%5Csqrt%7B13%7D%29%5C%20in)
![h=8\ in](https://tex.z-dn.net/?f=h%3D8%5C%20in)
substitutes
![LA=(10+2\sqrt{13})*8=(80+16\sqrt{13})\ in^{2}](https://tex.z-dn.net/?f=LA%3D%2810%2B2%5Csqrt%7B13%7D%29%2A8%3D%2880%2B16%5Csqrt%7B13%7D%29%5C%20in%5E%7B2%7D)
Part 2) Find the total area of the prism
we know that
The total area of the prism is equal to
![TA=LA+2B](https://tex.z-dn.net/?f=TA%3DLA%2B2B)
where
LA is the lateral area of the prism
B is the area of the base of the prism
<u>Find the area of the base B</u>
The area of the base is equal to the area of the triangle
![B=\frac{1}{2}bh](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B1%7D%7B2%7Dbh)
substitute
![B=\frac{1}{2}(6)(4)=12\ in^{2}](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B1%7D%7B2%7D%286%29%284%29%3D12%5C%20in%5E%7B2%7D)
Find the total area of the prism
![TA=LA+2B](https://tex.z-dn.net/?f=TA%3DLA%2B2B)
we have
![B=12\ in^{2}](https://tex.z-dn.net/?f=B%3D12%5C%20in%5E%7B2%7D)
![LA=(80+16\sqrt{13})\ in^{2}](https://tex.z-dn.net/?f=LA%3D%2880%2B16%5Csqrt%7B13%7D%29%5C%20in%5E%7B2%7D)
substitute
![TA=(80+16\sqrt{13})+2(12)=(104+16\sqrt{13})\ in^{2}](https://tex.z-dn.net/?f=TA%3D%2880%2B16%5Csqrt%7B13%7D%29%2B2%2812%29%3D%28104%2B16%5Csqrt%7B13%7D%29%5C%20in%5E%7B2%7D)