Slant height of the pyramid
to find s we can use pythagoras on half triangle
[tex]a^2+b^2=h^2[/tex]
where a and b are legs and h the hypotenuse
replacing
[tex]\begin{gathered} 7^2+3.5^2=S^2 \\ 49+12.25=S^2 \\ S^2=61.25 \\ \\ S=\frac{7\sqrt[]{5}}{2}\approx7.83 \end{gathered}[/tex]
lenght of the slant height is 7.83inches
Surface area
we find the erea of each side of the pyramid, square(base) and 4 triangles
area of the square
[tex]\begin{gathered} A=l\times l \\ A=7\times7 \\ A=49 \end{gathered}[/tex]
area of one triangle
[tex]\begin{gathered} A=\frac{b\times h}{2} \\ \\ A=\frac{7\times7}{2} \\ \\ A=\frac{49}{2} \end{gathered}[/tex]
now sum the base and four triangles
[tex]\begin{gathered} S=49+\frac{49}{2}+\frac{49}{2}+\frac{49}{2}+\frac{49}{2} \\ \\ S=147 \end{gathered}[/tex]
surface area of the pyramid is 147 square inches
Volume
we use the formula of the volume of the pyramid
[tex]V=\frac{A_b\times h}{3}[/tex]
where Ab is the area of the base (square=49in) and h is the height
replacing
[tex]\begin{gathered} V=\frac{49\times7}{3} \\ \\ V=\frac{343}{3}\approx114.33 \end{gathered}[/tex]
Volume of the pyramid is 114.33 cubic inches
