Rectangle PQRS is inscribed in rectangle ABCD, as shown. If DR = 3, RP = 13, and PA = 8, compute the area of rectangle ABCD.
Credits – Southern California ARML, Oct. 2009, Individual round
Posted in Math
Tags: Math Problems
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44+11 rad 80
I did it in my head really quickly
no, the answer is prettier than that >>”
remember, people, rectangles have four right angles…
Is there really nobody out there who can solve this?
[spoiler]AB and CD are parallel… and RP is a transversal… by transversals, guess what? OMG, RP cuts ABCD in half! That means BP = RD! How amazing!
Now I wonder how I can find AD?? Hmm, doesn’t that look rather like a Pythagorean triple?[/spoiler]
Huge hints, idk why I’m not just directly posting a solution ><
You get a pat on the back ^_^
[spoiler] 121? [/spoiler]
Omg I multiplyed wrong
[spoiler] 132 [/spoiler]
wait then whats the 13 for?
alternate solution? ::
draw an imaginary line from R to line PA so that it is || to DA. Lets call the point where the imaginary line intersects PA X. As you can see, PO is 5 since the difference of PA and RD is PX. Using the pythagorean theorem with RP as the hypotenuse and PX as one of the sides, RX is 12 (5-12-13 triangle). RX is equal to DA and square it and get 144. whats wrong with my reasoning?
I assume you meant PX is 5, not PO. Anyways, note the problem states ABCD is a rectangle. Not a square. ^^”
Note the dimensions are 12 (as you correctly found) and 11 (the sum of AP=8 and PB=RD=3).
Anyways this post is over one year old… I commend you for bringing it back from the dead lol
And it’s really fun to see people actually visiting my blog… XD i wonder where people are getting linked from
Oh and I found it strange that you commented on a math problem post. And on top of that, a math problem post from over a year ago. Why didn’t you comment on some of K’s and my (I just had an entertaining discourse about the proper grammar of the previous construction) contemplations on life?
Or my awesome, epic, obsessive anime posts from mid-sophomore year? XD
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