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
![A_{ABC}=A_{CDE}=\dfrac{4\cdot 3}{2}=6](/tpl/images/0375/9162/953b6.png)
Rectangle AEFH has sides AE=8 and AH=3. So, it has area
![A_{AEFH}=8\cdot 3=24](/tpl/images/0375/9162/5bef8.png)
Finally, triangle FGH has a base HG=2 and height HF=8. So, its area is
![A_{FGH}=\dfrac{2\cdot 8}{2}=8](/tpl/images/0375/9162/b9f5f.png)
So, the total area is
![A_{ABC}+A_{CDE}+A_{AEFH}+A_{FGH}=6+6+24+8=44](/tpl/images/0375/9162/9ec98.png)
2:
The base radius is 3, so the base area is
![\pi 3^2 = 9\pi](/tpl/images/0375/9162/77f34.png)
The lateral area is the product between the height and the base circumference:
![9\cdot 6\pi=54\pi](/tpl/images/0375/9162/b3959.png)
So, the total area is twice the base area plus the lateral area:
![54\pi+18\pi=62\pi\approx 72\cdot 3.14=226.08](/tpl/images/0375/9162/39b0b.png)
3:
The surface area of a sphere is
![S=4\pir^2](/tpl/images/0375/9162/cd53b.png)
Solving for r, we have
![r=\sqrt{\dfrac{S}{4\pi}}=\sqrt{\dfrac{196\pi}{4\pi}}=\sqrt{49}=7](/tpl/images/0375/9162/908ef.png)
The volume of a sphere is
![\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi 7^3=\dfrac{1372}{3}\pi](/tpl/images/0375/9162/e3e5b.png)
![Asap you 1. find the area of the composite figure. 40 units2 38.5 units2 39.75 units2 44 units2 #2:](/tpl/images/0375/9162/5c316.jpg)