Math Questions

Test Preparation SAT Preparation
1

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<p>In the figure above, the circle with center O is inscribed in square ABCD. What is the area of the shaded portion of the circle?</p>

<p>π/4
π/2
π
3π/2
2π</p>

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<p>The flag shown above is made of overlapping equilateral triangles ADF and BCE. Because ribbon is to be sewn around the entire outer edge, it is necessary to know the perimeter of the flag. If CD, DE, and EF each have length 10 inches, what is the length, in inches, of the perimeter shown in bold?</p>

<p>^^ grid-in</p>

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<p>Figure not drawn to scale.</p>

<p>The pyramid shown above has altitude h and a square base of side m. The four edges meet at the vertex of the pyramid, each have length e. If e=m, what is the value of h in terms of m?</p>

<p>m/√2</p>

<p>m√3/2</p>

<p>m</p>

<p>2m/√3</p>

<p>m√2</p>

2

<p>q3) If e = m, then each of the four side triangles of the pyramid is equilateral
Hence each angle is 60 degrees.
and tan 60 = h / (m/2)
sqrt(3) = 2h / m
h = sqrt(3)m / 2</p>

3

<p>q1) The selected triangle is obviously one fourth of the square, and by symmetry, encloses 1/4th of the circle too.
So the shaded area is 1/4th of the circle’s total area.
You haven’t explained what ‘n’ signifies</p>

4

<p>I believe that “n” is actually the symbol for pi.</p>

5

<p>q2)
CD + DE = DE + EF
Therefore, the bases of the 2 triangles are equal, and since they are equilateral, all their sides are equal (to 20 inches).
Now,
When 2 congruent equilateral triangles overlap, the triangle formed is also equilateral.
So (let the unknown point be X)
XD = DE = EX = 10</p>

<p>So the perimeter is = CB + BX + XA + AF + CD + DE + EF
20 + 20 + 10 + 10 + 10 + 10 + 10
= 90 inches</p>

6

<p>well, I don’t know the radius of the circle. But assuming it’s a unit circle, the area required is pi.r^2 / 4 = pi/4</p>

7

<p>Okay, so CE and EF is where you are getting 20?</p>

8

<p>Uh, no. CB and AF are 20</p>

9

<p>I haven’t used CE and EF in the addition but yes they are 20 each too</p>

10

<p>Sorry, this isn’t “gelling”. How is CB 20?</p>

11

<p>The triangles are equilateral, which means the sides are equal in length.
SO EB= BC = CE
And CE = CD + DE = 10 + 10 = 20
so EB = BC = CE = 20 each</p>

12

<p>Do you do the same for the other triangle? Like AD = DF = AF = 20?</p>

<p>Thanks for all your help</p>

13

<p>yes, but if you add all that, you have to subtract XD and XE ( X is the unnamed point)</p>

14

<p>Spidey, you’re the best. What did you get on the SAT Math?</p>

<p>I’m applying to Colgate too :smiley: How do you like the campus?</p>

15

<p>I just arrived in the US yesterday and will head to Colgate tomorrow.
I had 800 in math :)</p>