# Thread: a few algebra 2 and geometry word problems

1. ## a few algebra 2 and geometry word problems

here are some word problems that if you could help would be great! you don't have to do all of them either if you don't want to or are unable too.

a) frame of uniform width surrounds a 16 cm by 20 cm picture so that the total area of the framed picture is 464 cm sq. what is the width of the frame?

b) find coordinates of all points that are 10 units from the origin and 17 units from the point (0, -9)

c) each lateral edge of a square pyramid is twice as long as the altitude of the pyramid. the slant height of the pyramid is 5 cm. find the lenght of the altitude and a base edge.

d) two cones have the same volume. one of them has a radius of 3 cm less than the other and a height twice that of the first. what are the radii of the two cones?

2. Hello,

For the first one, let x be the width of the frame. As it is uniform, you don't have to care about its variation.

If you draw the thing, you may have a rectangle into another one (the frame).

You know the area of the big rectangle, which is 464.

The dimensions of this rectangle (the big one) is 16+2x and 20+2x (you add twice the width to the normal dimensions).

Hence the area of the rectangle is (16+2x)(20+2x) and is equal to 464.
So you have to solve for x in :

(16+2x)(20+2x)=464

3. $\left( {20 + 2x} \right)\left( {16 + 2x} \right) = 464$

$320 + 72x + 4x^2 = 464$

$4x^2 + 72x - 144 = 0$

$4\left( {x^2 + 18x - 36} \right) = 0$

$x^2 + 18x - 36 = 0$

$x = \frac{{ - \left( {18} \right) \pm \sqrt {\left( {18} \right)^2 - \left( 4 \right)\left( 1 \right)\left( { - 36} \right)} }}
{{2\left( 1 \right)}} \Rightarrow \frac{{ - 18 \pm 6\sqrt {13} }}
{2} \Rightarrow - 9 \pm 3\sqrt {13} \Rightarrow x \approx 1.81665$

For b, you have to find the intersection of the two circles centered at the given points with the given radius, which translates to solving this system of equations:

$x^2 + y^2 = 10^2$

$x^2 + \left( {y - \left( { - 9} \right)} \right)^2 = 17^2$