Don't get surface area questions

I posted this last night but I have since gotten answers for one of the questions but still do not get two at all. (Headbang)

1. "Find the total surface area of the following figure.

Hint: The surface area formula for the cone includes the base of the cone. Substract this value."

Image: http://www.omegamath.com/scripts/geometry/Gtest/image924.gif

I __do not__ get this at all. Someone please just respond with clear directions on how I measure the surface area for this particular image. An answer would be appreciated too of course, it would help in figuring out how I do this!

2. "Find the total surface area of the following figure. The back of the figure is a rectangle. Disregard the bottom of the figure."

Image: http://www.omegamath.com/scripts/geometry/Gtest/image925.gif

This one seems to be the same way, so same deal as #1.

Thank you in advance!

Re: Don't get surface area questions

1.) The lateral surface area of a cone is:

$\displaystyle A_1=\pi rl$

where $\displaystyle r$ is the radius and $\displaystyle l$ is the lateral height.

The surface area of a hemisphere is:

$\displaystyle A_2=\frac{2}{3}\pi r^3$

Now add these the get the total area:

$\displaystyle A=A_1+A_2=\pi rl+\frac{2}{3}\pi r^3=\frac{\pi r}{3}\left(3l+2r^2 \right)$

Now use $\displaystyle r=2\text{ cm}$ and $\displaystyle l=5\text{ cm}$ to find the surface area of the given solid.

2.) The back of the figure is a rectangle having a base $\displaystyle b$ of 16 in and a height $\displaystyle h$ of 7 in.

The area of a rectangle is $\displaystyle A_R=bh$

The two sides are trapezoids having big base $\displaystyle B$ of 7 in and little base $\displaystyle b$ of 3 in and height $\displaystyle h$ of 3 in.

The area of a trapezoid is $\displaystyle A_T=\frac{h}{2}(B+b)$

The front is two rectangles. The vertical rectangle has a base of 16 in and a height of 3 in, and the slanted rectangle has a base of 16 in and a height which is the hypotenuse of a right triangle having legs of 3 in and 4 in. Using the Pythagorean theorem, we find this is:

$\displaystyle \sqrt{3^2+4^2}=5$

So, put this all together to find the requested surface area.

Re: Don't get surface area questions

I am always fond of mathematics and find it really interesting. Some parts of it like trigonometry always troubled me with the formulas. Even integral calculus was really hard to understand. You are maintaining a nice math forum here.for details

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Gasper Millan