Hardest SAT Math Question Ever….

Test Preparation SAT Preparation
1

<p>Well not hard in the sense that its the toughest question ever perse, but just that the answer seems to be VERY strange. </p>

<p>My younger sister is taking her SAT for the first time in a months time and she gave me the following question (about triangles inscribed in circles);</p>

<p>[URL=<a href="http://img183.imageshack.us/i/satquestion.png/%5D%5BIMG%5Dhttp://img183.imageshack.us/img183/7572/satquestion.png%5B/IMG%5D%5B/URL"&gt;http://img183.imageshack.us/i/satquestion.png/]

http://img183.imageshack.us/img183/7572/satquestion.png

[/URL</a>]</p>

<p>I am certain the answer is D (area of circle - area of triangle), but she insists the book states its C. I've double-checked this and the only conclusion I have come to is that the book is wrong, but I just want to be certain before I tell her of this (incase it turns out I am incorrect and thus teaching her the wrong thing etc).</p>

<p>Thanks for the help CC.</p>

2

<p>Answer E (which is cut off from the picture annoyingly), is the following;</p>

<p>E = 12.5pi - 48</p>

3

<p>The answer is definitely D.</p>

4

<p>It should be D. The area of the semicircle is (0.5)pi(5)^2, while the area of the triangle is (1/2)(6)(8). So this is 12.5pi - 24. You’re right.</p>

5

<p>Well you find the diameter using Pythagorean theorem, then find the area of semi-circle minus the area of the triangle, which is 12.5pi - 24. So you are right and book is wrong.</p>

6

<p>I’m pretty sure the answer is D also.</p>

7

<p>Those book have typos like that a lot…its absolutely D.</p>

8

<p>The answer is D. Tell her not to worry about it.</p>

9

<p>That was a really easy math question.</p>

10

<p>^Agreed. That questions was definitely easy. The hardest math questions on the SAT are much, much harder than this-mind boggling.</p>

11

<p>SAT doesn’t have hard math questions.</p>

12

<p>Answer is definitely D.
By the way, this is from Gruber’s Math Workbook, which is notoriously known for all its mistakes and typos in answers.
I’m sure this was never an actual SAT question.</p>

13

<p>There’s really nothing quirky or unusual about this problem, and it’s certainly one that could EASILY be on an actual SAT. What skills are we using here?
Pythagoras: 2 sides are 6 and 8, what’s the 3rd side. Seen it a million time on real SATs.
Semicircle area, triangle area, shaded area determined by whole area minus part not shaded. All very standard concepts for the SAT.</p>

14

<p>I’d like to say, not hard indeed. In fact, yes, D is correct.
You should get the area of that semicicle first; it is 12.5 π
Then the triangle area is 24.
Finally use the bigger area, 12.5π minus 24
So you get it!</p>

15

<p>I’m guessing he thinks its hard because his books tells him the wrong answer… chill out people lol</p>

16

<p>Definitely an easy problem.</p>

<p>They practically give you the area of the triangle and you know the formula for the area of the circle. And since you already discovered that the diameter of the circle was 10, 1/2A = 1/2(pi)(25) = 12.5pi.</p>

<p>12.5pi-24.</p>

<p>The reason she might be getting 25pi - 48 is because she is mistaking the area of the triangle to be A = (b)(h) and then this misstep allows her to say that A = pi*(r^2), which is where she got the 25 pi from.</p>

17

<p>Failing math education is the reason why American teenagers find problems like this difficult.</p>

18

<p>You find the area of the triangle, which is 24. Than you find the radius, which is 5. Then you find the area of the semicircle which is (pi(r^2))\2, which is 12.5pi. than you subtract that from 24, so the answer is 12.5pi-24.</p>

<p>Your sister is wrong. If she gives you information, be very suspicious.</p>

19

<p>Why are you guys explaining this… its been covered at least ten times. Do you just want to show us that you know how to do it?</p>

20

<p>The thread has already been resolved… we thoroughly covered every part of the original posters problem at least 6 times. Are you just trying to up your post count?</p>