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:
$\displaystyle x^2+y^2 - 8y = 12 ~ \Longrightarrow ~ x^2 + y^2 - 8y +16 = 12 + 16 ~\Longrightarrow~x^2+(y-4)^2=28$
Thus you get the coordinates of the first centre as: $\displaystyle C_1(0, 4)$
$\displaystyle x^2-4x+y^2=16~\Longrightarrow~x^2-4x+4+y^2=16+4~\Longrightarrow~(x-2)^2+y^2=20$
Thus you get the coordinates of the second centre as: $\displaystyle C_2(2, 0)$
Now use Pythagorean theorem to calculate the distance:
$\displaystyle d = \sqrt{4^2+2^2}=\sqrt{20} = 2\cdot \sqrt{5} \approx 4.47$
Hello, Sanee!
Your four answers are correct.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 $\displaystyle QR$.
. . Since $\displaystyle OS \parallel QR$, then: .$\displaystyle OS \perp\ OQ$ . . . and $\displaystyle OS = 12$.
From $\displaystyle S$ draw perpendicular $\displaystyle ST$ to $\displaystyle QR$.
. . Then quadrilateral $\displaystyle OQTS$ is a square: .$\displaystyle QT= 12$
$\displaystyle \Delta RTS$ is an isosceles right triangle: .$\displaystyle ST \:=\:TR = 12$
. . Therefore: .$\displaystyle QR \:=\:QT + TR \:=\:12 + 12 \:=\:24$
Hello, sanee!
Sorry, your answer is incorrect . . .
For the area of a parallelogram, you need its base $\displaystyle (GD = 11)$
. . and its height $\displaystyle (DB)$
$\displaystyle \Delta DBF$ is an isosceles right triangle, so $\displaystyle DB = FB$.
Let $\displaystyle x = DB = FB$
. . Then we have: .$\displaystyle x^2 + x^2 \:=\:(8\sqrt{2})^2\quad\Rightarrow\quad 2x^2 \:=\:128\quad\Rightarrow\quad x \:=\:8$
Now you can find the area . . .