Question

Can someone help me solve this?

Answer 1

So...the tent, seems to be the first picture below

the second picture, is just a break-down of the pyramid

$\bf \textit{area of a square}=A=a^2=4.9\implies a=\sqrt{4.9} \\\\\\ \textit{slant height will be}=c^2=\left( \frac{\sqrt{4.9}}{2} \right)^2+1^2 \\\\\\ c^2=\cfrac{4.9}{4}+1\implies c^2=\cfrac{89}{40}\implies c=\sqrt{\cfrac{89}{40}}$

so... that's the value of "a" and the "c" or "slant height", namely the missing slanted side on that second picture

we need the slant height, in order to get the area of that triangular face, and the base of that triangle is, just "a"

since the area of a triangle is 1/2 bh, and the base is "a" and the altitude or height here is the slant height "c", then the area of that triangular face is $\bf \left( \cfrac{1}{2} \right)\left( \sqrt{4.9}\right)\left( \sqrt{\cfrac{89}{40}} \right)$

and the area of the lateral sides for the prism, are " a * 1.5"  or $\bf \sqrt{4.9} \cdot 1.5$

so, you have 4 triangles, and 4 rectangles to be added, so.. .let's do that, to get the canvas then

$\bf \begin{array}{clclll} 4\left[ \left( \cfrac{1}{2} \right)\left( \sqrt{4.9}\right)\left( \sqrt{\cfrac{89}{40}} \right) \right]&+&4[(\sqrt{4.9})(1.5)]\\ \uparrow &&\uparrow \\ \textit{4 triangles}&&\textit{4 rectangles} \end{array}$

that'd be the area of the canvas, or namely, the surface area

notice, we are not including the base of the pyramid, because that's inside the tent, and excluding the base of the prism, because that'd be the ground, and the tent may not include that, well, depends on the tent, in this case, I think a tent this big may not, other smaller ones do though