1: Refer to the attached image. We can split the composite figure in easy figures: triangles ABC and CDE have a base AC=4 and height of 3. Their area is thus
[tex]A_{ABC}=A_{CDE}=\dfrac{4\cdot 3}{2}=6[/tex]
Rectangle AEFH has sides AE=8 and AH=3. So, it has area
[tex]A_{AEFH}=8\cdot 3=24[/tex]
Finally, triangle FGH has a base HG=2 and height HF=8. So, its area is
[tex]A_{FGH}=\dfrac{2\cdot 8}{2}=8[/tex]
So, the total area is
[tex]A_{ABC}+A_{CDE}+A_{AEFH}+A_{FGH}=6+6+24+8=44[/tex]
2:
The base radius is 3, so the base area is
[tex]\pi 3^2 = 9\pi[/tex]
The lateral area is the product between the height and the base circumference:
[tex]9\cdot 6\pi=54\pi[/tex]
So, the total area is twice the base area plus the lateral area:
[tex]54\pi+18\pi=62\pi\approx 72\cdot 3.14=226.08[/tex]
3:
The surface area of a sphere is
[tex]S=4\pir^2[/tex]
Solving for r, we have
[tex]r=\sqrt{\dfrac{S}{4\pi}}=\sqrt{\dfrac{196\pi}{4\pi}}=\sqrt{49}=7[/tex]
The volume of a sphere is
[tex]\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi 7^3=\dfrac{1372}{3}\pi[/tex]
