Here,
Let refer to the diagram

ABC is equilateral \(\triangle\)

AD is the height of the \(\triangle\) = 8

For the \(\triangle\) ADC

\(\angle CAD = 30\)
\(\angle ADC = 90\)
\(\angle DCA = 60\)

It is a 30 -60-90 \(\triangle \) and the sides are distributed in the ratio \(1\) : \(\sqrt{3}\) : \(2\)

since, the height AD = 8, hence the sides are distributed as

\(\frac{8}{(\sqrt{3})}\) : \(8\) : \(\frac{16}{(\sqrt{3})}\)

i.e one side of the equilateral \(\triangle\) = \(\frac{16}{(\sqrt{3})}\)

Area of equilateral \(\triangle\) = \(\frac{(\sqrt 3 * {side}^2)}{4}\) =\(\frac{\sqrt 3 * ({\frac {16}{\sqrt3}})^2}{4}\) = 37 (approx)

Now, Area of the circle of radius '7' = \(\pi * {radius}^2 \) = \(\pi *49\) = 154

Therefore, Area of the shaded region = 154 - 37 = 117

QTY A > QTY B
_________________

I believe this problem was a good one to exemplify estimation. We know it cannot be D because there are no variables. We also can roughly estimate the circle's area as 150 and thus the shaded region is ~150- the triangle's area. Knowing that the base will have a root in it because of the 8 height in the equilateral triangle quickly gets rid of C as well. From here we will need to calculate if we want to get all the way, but good knowledge of the 30-60-90 triangle will show that 64/root3(~1.7) is less than 45... 64/1.5 would be like 2/3 of 63 and we are dividing by more than that.

the area of the circle is pai or 3.1416 * r^2 (7^2) which is(3.1416*49) 153 ,now subtract with the area of the triangle which is root 3/ 4* 8^2(1.732*8^2/4) 27.712
the result is 126 which is greater than 105

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