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Mon 23 Oct, 2006 11:44 am
Hi,
I've been working through these two problems for the past five days and am completely stuck. If you could offer any help, equations, hints, I'd APPRECIATE it!
Thanks.
-Mason
Question #1: The base edges of a triangular pyramid are 25 inches, 39 inches, and 50 inches long, and all three lateral faces are 24 inches in height. Find the height of this pyramid.
What I did: I drew an accurate diagram but honestly have no idea how to proceed from here. Please offer any help and advice, the more thorough, the better!! I also hate it when people assume that when someone asks for help, it's because he or she is lazy and doesn't want to do the work. This is NOT the case for me!! I spent a lot of time trying to figure this out!
Question #2: Consider the dilation whose center is (4,4) and whose magnification factor is 3. Apply this transformation to the points on the circle whose equation is x(squared) + 2x + (ysquared) - 4y + 1= 0.
a. Determine an equation that describes the configuration of image points.
b. The image and original figures have two common external tangents, which meet at the dilation center. Find the size of the angle formed by these lines and write an equation for each.
c. If the center of dilation is moved to (3, 5), determine equations for the two common external tangents?
What I did: I converted the equation x(squared) + 2x + (ysquared) - 4y + 1= 0 to a regular circle equation:
(x+1)(squared) + (y-2)(squared)=4
thus, the center is (-1,2) and the circle has a radius of two.
I'm not sure what to do now! Please help!
Thank you SO much in advance!!!!
mwatson
Welcome to A2K.
#1. Suggestion
....isn't the vertex of the tetrahedron situated above the intersection of the perpendicular bisectors of the sides of the base triangle ? If so... are the distances to this intersection from the mid points in the ratio of the base sides ?.....then manipulating pythagoras equations on separate vertical triangles with same height h and hypotenuse 24 should yield the ratio which would then give the height h by re-substitution.
(You need to check this out.)
(LATER EDIT. The apex will in fact lie vertically above the centre of the inscribed circle of the base tiangle i.e. where the bisectors of the angles intersect...not the bisectors of the sides as I thought above. above. You therefore need to find the radius which is in fact the base of a vertical right angled triangle height h hypotenuse 24)
http://courses.ncssm.edu/goebel/statecon/Topics/triangl/TRIANGL.htm
#2. Again a suggestion...
Since the centre of the original circle is (-1,2) can we not find the centre of the new circle by the translation (X - (-1), Y- 2)/3 where (X,Y) is the centre of dilation ? I am basing this on the fact that equation of a circle is (x-a)sqrd +(y-b)sqrd=Rsqd, where (a,b) is the centre and R the radius.
Presumably the original radius also dilates to R/3 I am assuming "dilation" involves
reduction to 1/3, but the term might mean a
magnification of 3. You are more likely to know than me!
(LATER EDIT I've just found out it does mean magnification therefore the above should be corrected for multiplication by 3 instead of division)
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