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Okay. Thank you!! I got it right.
Is the answer 9sqrt(5)?
Thank you!! I got 'em. Can you help me with another prob plz:
In the diagram below, $WXYZ$ is a trapezoid such that $\overline{WX}\parallel \overline{ZY}$ and $\overline{WY}\perp\overline{ZY}$. If $YZ = 12$, $\tan Z = 1.5$, and $\tan X = 2$, then what is $XY$?
^Those were correct. Thanks!! Do you have any ideas for other prob I posted?
Thank you!!! I got it. Next Prob:
In right triangle $ABC$ with $\angle B = 90^\circ$, we have $\sin A = 2\cos A$. What is $\cos A$?
What is a compuhigh problem? Never heard of it...
Problem 1: A rectangular box $P$ is inscribed in a sphere of radius $r$. The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$?
Problem 2: The equation of the line passing through $(1,8)$ and $(5,6)$ can be expressed in the form
\[\frac{x}{a} + \frac{y}{b} = 1.\]
Find $a$.
Problem 3: Let $P = (5,1)$, and let $Q$ be the reflection of $P$ over the line $y = \frac{1}{2} x + 2$. Find the coordinates of $Q$.
Problem 4: The vertices of a triangle are the points of intersection of the line $y = -x-1$, the line $x=2$, and $y = \frac{1}{5}x+\frac{13}{5}$. Find an equation of the circle passing through all three vertices.
Problem 5: In triangle $PQR$, we have $\angle P = 90^\circ$, $QR = 15$, and $\tan R = 5\cos Q$. What is $PQ$?
Problem 6: Two circles of radius 1 are externally tangent at $Q$. Let $\overline{PQ}$ and $\overline{QR}$ be diameters of the two circles. From $P$ a tangent is drawn to the circle with diameter $\overline{QR}$, and from $R$ a parallel tangent is drawn to the circle with diameter $\overline{PQ}$. Find the distance between these two tangent lines.
Problem 1: Let $ABCDEFGH$ be a rectangular prism, as shown, where $AB = 2$, $AD = 3$, and $AE = 5$. Find the volume of pyramid $ACFH$.
I don't understand that solution - I posted my questions at that thread
Hi Bob;
LB extends to D because HBD = 90° and LBA = BDH = x (AB being parallel to CD); similarly, LA extends to C (HAC = 90° and BAL = HCA = y). Therefore BAL = DHB = HCA = y, and LBA = AHC = BDH = x.
Triangles LBA, BDH and AHC are congruent (similar angles, LA = BH and LB = AH). Therefore their hypotenuses are equal to each other (CH = HD = AB), proving that H is CD's midpoint.
We already know that GH, AG and GB are radii of circle G, and therefore GH (ie, the distance between the midpoints of AB and CD) = AB/2.
https://onedrive.live.com/download?resid=C20C46B976D069EE!3932&authkey=!AHpqkk_oTZr-SGg&v=3&ithint=photo%2cjpg
... But how to you know HBD=90????
Two circles are externally tangent at point $P$, as shown. Segment $\overline{CPD}$ is parallel to common external tangent $\overline{AB}$. Prove that the distance between the midpoints of $\overline{AB}$ and $\overline{CD}$ is $AB/2$.
Here is the diagram: http://latex.artofproblemsolving.com/f/0/f/f0fccd0e6e157a2063fd2c451a6812eb1a4fa7f9.png
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