Find the approximate area using Simpson's Rule

Question

Find the approximate area of the shaded figure shown using Simpson's rule. Each of the equidistant parallel chords is measured from the base to a point on the curve. All units are expressed in km.

So I tried this problems and they said the answer is $77km^2$ I got $137.76km^2$ I don't understand where I was wrong

$S.Rule=\frac{1}{3}d[(y^1+y^6)+4(y^3+y^5)+2(y^2+y^4)]$

$S.Rule=\frac{1}{3}2.89$$[(10+9)+4(22)+2(18)]$ I started from the right and got my interval by finding the base using soh cah toa by using $30^\circ$ and $10$ and by dividing by $6$ to get the interval $\frac{10\sqrt{3}}{6}$

Answer 1

There are 7 points (usually we need odd number of points for the Simpson's rule). The sum should be

$$\frac{10\cot30^{\circ}}{3} \times \frac{1}{6} (y_0 + 4 y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6)$$

And the question actually asks for the shaded area, while the Simpson's rule give the total area of the shaded region and the right-angled triangle. So the formula is

$$\frac{10\cot30^{\circ}}{3} \times \frac{1}{6} (y_0 + 4 y_1 + 2y_2 + 4y_3 + 2y_4 + 4y_5 + y_6) - \frac{1}{2} \times 10 \times 10\cot30^{\circ}$$

The numerical answer is 76.980035892.

Answer 2

After solving multiple times and with the help of the comments

$S.Rule=\frac{1}{3}2.89$$[(10)+4(31)+2(18)]$ Starting from the left I divided my base to 6 and got $2.89$ as my interval. Finally I got $S.Rule=\frac{1}{3}2.89$$(170)$ - $86.62$ the area of the right triangle and finally got $77.14km^2$