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In the figure above the chord s is length 6 and the circle has a radiu [#permalink]
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Tough question both per se both to figure out a simpler way to deal with

Attachment:
GRE geometry figure.jpg
GRE geometry figure.jpg [ 13.06 KiB | Viewed 211 times ]


we know that the radius is 7 and therefore the diameter is 14. At the same time the diameter is also the diagonal of the rectangle.

S the chord is 6. Also the base of the triangle is the radius of the circle and it is 7

Now we can use the Heron's formula to calculate the height

\(Area=\frac{7+7+6}{2}=\frac{20}{2}=10\)

\(Area=\sqrt{10(10-7)(10-7)(10-6)}=\sqrt{10*3*3*4}=18.9\)

\(18.9=\frac{1}{2}*7*h\)

\(h=5.4\)

Double the height of the triangle and you have the length of the rectangle \(5.4*2=10.8\) (this is a huge estimation by myself)

Now we have the diagonal 14 and the length 10.8 and we can calculate the width

\(w=\sqrt{14^2-10.8^2}=\sqrt{196-116.64}=\sqrt{79.36}=8.9\)

Now the area of rectangle is \(8.9*10.8=96\)

However, I worked by estimation. So clearly the numbers are inferior than that and because the last two answer are pretty close (95,96) I personally during exam I would be pretty sure the answer is C

sorry it is the only way to make it more simple to reach the solution
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Re: In the figure above the chord s is length 6 and the circle has a radiu [#permalink]
Thanks for the explanation Carcass, but does such lengthy questions generally show up in GRE?
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Re: In the figure above the chord s is length 6 and the circle has a radiu [#permalink]
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That's a good question.

I think no because too many steps are involved.

However, it is still good for practice to reinforce your mental muscles.

Regards
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