<p>Can someone help me solve this please? I got this from Dr. Chungs Math book, and although there's a solution I do not know how they reached the numbers given in the solution. (does that make sense?)</p>
<p>Anyways,</p>
<p>In Triangle ABC above, AB // PQ // RS and the ratio of the lengths, AQ:QS:SC = 2:2:3. If the area of the quadrilateral PRSQ is 48, what is the area of Triangle ABC?</p>
<p>I understand it can be hard without a picture, so I uploaded one onto tinypic.com
<a href="http://i45.tinypic.com/zvy0jl.jpg%5B/url%5D">http://i45.tinypic.com/zvy0jl.jpg</a></p>
<p>Please help! And Thank you!</p>
<p>I’m assuming ABC is a right triangle?</p>
<p>Let AQ = 2x, QS = 2x, SC = 3x. Triangles ABC, QPC, and SRC are all similar. We know that AC:QC:SC = 7:5:3 (done by combining all the above ratios), so therefore AB:QP:SR = 7:5:3.</p>
<p>Let AB = 7y, QP = 5y, and SR = 3y. PQRS is a trapezoid, so we use the trapezoid’s area:</p>
<p>[PQRS] = 48 = (1/2)(2x)(5y+3y) → 96 = 2x(8y) → xy = 6.</p>
<p>The area of ABC is given by</p>
<p>[ABC] = (1/2)AB<em>AC = (1/2)(7y)(7x) = (49/2)xy. Since we know xy = 6, [ABC] = (49/2)(6) = 49</em>3 = 147.</p>
<p>Thank you for the quick reply, however the answer is 147. I was just wondering if you could explain this work then?</p>
<p>Since the area of PRSQ is 48, (25k-9k)=16k=48, or k=3. Therefore, the area of triangle ABC is 49K=49*3=147.</p>
<p>Where do you get 25k and 9k? O.o</p>
<p>Yes I had originally put 392, but I quickly realized I put xy = 16 instead of xy = 6. So 147.</p>
<p>That solution works as well. It’s obtained by taking the area of triangle QPC and subtracting the area of SRC.</p>
<p>Note that the corresponding sides of triangles ABC, QPC, and SRC are in the ratio 7:5:3. That means the areas of those triangles are in the ratio (7^2) : (5^2) : (3^2), or 49:25:9. That’s where those numbers come from. :)</p>
<p>Ahhh! Now I see it, thank you so much! It never clicked in my head, that because they are all similar triangles, the ratios should all be similar, at least for two sides.</p>
<p>Thank you</p>
<p>Now that’s a hard problem - I would consider it Level 6 which means it is a bit harder than anything that would appear on an actual SAT. Let me give an algebraic solution</p>
<p>Start by choosing values for AB, PQ and RS. The easiest choice would be 2, 2, and 3. Now I suggest you draw the 3 triangles in the picture separately: triangles ABC, QPC, and SRC. Note that the bases of these triangles have lengths 7, 5 and 3 respectively., and they are all similar. Now, let RS have length b1, PQ have length b2, and BA have length b3. By similarity of the last 2 triangles we have 3/b1 = 5/b2, or equivalently 3b2=5b1 (by cross multiplying). Also, using the area of a trapezoid we have 1/2 (b1 + b2)(2) = 48 (QS is the height of the trapezoid). So b1 + b2 = 48. We now have the following system of equations:</p>
<p>5b1=3b2
b1+b2=48</p>
<p>Multiply the second equation by 3 and substitute</p>
<p>3b1+3b2=48(3)
3b1+5b1=48(3)
8b1=48(3)
b1=48(3)/8=6(3)=18</p>
<p>Now using similarity of triangles ABC and PSC we have b3/7=18/3. Cross multiplying gives 3b3=18(7), or b3=18(7)/3 = 6(7)=42. </p>
<p>Thus the area of the big triangle is 1/2(7)(42)=21(7)=147.</p>
<p>I just did that very quickly, so hopefully there are no computational errors, but I’m sure someone will correct me if there are. :)</p>
<p>I will look at it again later and see if I can find a quick, sneaky way to do it.</p>
<p>It’s an interesting problem, but unless you’re consistently scoring over a 750 on practice tests I wouldn’t recommend spending a lot of time on problems this difficult.</p>
<p>Holy Crapola! Well then, I’m glad that this is a question rarely on the SAT’s but I’m glad I’ve prepared myself for that chance. I took the June SAT, and I’m hoping that I scored 730+, I believe I only missed 2 questions, both were careless mistakes. But I’m just wondering, where did you find the leveling system that you use? I’ve never heard of giving math questions, levels?</p>
<p>They sometimes assign level 1-5 for SAT problems, with 5 being the hardest.</p>
<p>This is probably comparable to a level 5 problem. It’s definitely solvable using elementary techniques, but takes a little time.</p>
<p>If you want hard geometry problems, look at AMC and AIME problems.</p>
<p>Ok - I thought of a quicker way!</p>
<p>Since the ratios of the sides of the 3 similar triangles are 7:5:3, the ratios of the areas of the triangles are 49:25:9.</p>
<p>So first let’s find the area of the small triangle - call it x.</p>
<p>25/(48+x)=9/x. Cross multiplying gives 25x = 9x+48(9). So 16x=48(9), and x = 48(9)/16=3*9=27</p>
<p>Now let’s find the area of the large triangle - call it y.</p>
<p>49/y=9/27. Cross multiplying gives 9y=49(27). So y=49(27)/9=49(3)=147.</p>
<p>Oh goodness, I haven’t looked at those in years… but thank you for the help! I feel very stupid that I wasn’t able to complete that question yet I scored a 3 on the AIME :/</p>
<p>@keunglh I’ve felt the same way. I scored an 11 on 2010 AIME, only to score 4 on 2011 AIME (I was sick that day, and it was extremely hard compared to previous years).</p>
<p>@DrSteve, that’s just a variant of the solution keunglh posted (#3). It’s a good solution, though.</p>
<p>Wow you guys are amazingly intelligent, and I must say I now know who to turn to for math help.</p>
<p>@rspence</p>
<p>Aha - sometimes I post before I look!</p>
