1. ## 6 geometry questions

I have what I think are the correct answers to all of these but the last two. Can someone help a poor mom out please?

2. Originally Posted by sanee66
I have what I think are the correct answers to all of these but the last two. Can someone help a poor mom out please?
Hello,

I've attached a sketch which will show you how to do the construction problem.

To the last problem:

Calculate the centres of the circles by completing the square:

$x^2+y^2 - 8y = 12 ~ \Longrightarrow ~ x^2 + y^2 - 8y +16 = 12 + 16 ~\Longrightarrow~x^2+(y-4)^2=28$

Thus you get the coordinates of the first centre as: $C_1(0, 4)$

$x^2-4x+y^2=16~\Longrightarrow~x^2-4x+4+y^2=16+4~\Longrightarrow~(x-2)^2+y^2=20$

Thus you get the coordinates of the second centre as: $C_2(2, 0)$

Now use Pythagorean theorem to calculate the distance:

$d = \sqrt{4^2+2^2}=\sqrt{20} = 2\cdot \sqrt{5} \approx 4.47$

3. Hello, Sanee!

I have what I think are the correct answers to all of these but the last two.

In the fifth problem: .OQ is a radius (12) perpendicular to the tangent $QR$.

. . Since $OS \parallel QR$, then: . $OS \perp\ OQ$ . . . and $OS = 12$.

From $S$ draw perpendicular $ST$ to $QR$.

. . Then quadrilateral $OQTS$ is a square: . $QT= 12$

$\Delta RTS$ is an isosceles right triangle: . $ST \:=\:TR = 12$

. . Therefore: . $QR \:=\:QT + TR \:=\:12 + 12 \:=\:24$

4. ## area of a parallegram in a rectangle

Please take a look and see if I have the correct answer. Thanks for all you have done.

5. Hello, sanee!

For the area of a parallelogram, you need its base $(GD = 11)$
. . and its height $(DB)$

$\Delta DBF$ is an isosceles right triangle, so $DB = FB$.

Let $x = DB = FB$
. . Then we have: . $x^2 + x^2 \:=\:(8\sqrt{2})^2\quad\Rightarrow\quad 2x^2 \:=\:128\quad\Rightarrow\quad x \:=\:8$

Now you can find the area . . .

6. Thank you Soroban for looking at this!
So the answer would then be 88? Base of 11 times the height of 8?

7. Hello, Sanee!

Yes . . . you've got it!