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Difference between revisions of "2019 Mock AMC 10B Problems/Problem 12"

(Created page with "Observe that the triangle is a 30-60-90 triangle because of the 10-10sqrt(3) ratio. Let the center of the circle be O, the upper left point of the triangle be A, the right ang...")
 
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Observe that the triangle is a 30-60-90 triangle because of the 10-10sqrt(3) ratio. Let the center of the circle be O, the upper left point of the triangle be A, the right angle be C, the point on the right be B, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be I. Then OI and OB are the radii of the semicircle = 5sqrt(3). Because ABC = 30 degrees, then BIO = 30 degrees (radii are the two legs of the triangle -> isosceles triangle). That means that COI is 60 degrees and IOB is 120 degrees.
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We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a <math>30</math> - <math>60</math> - <math>90</math> triangle because of the <math>10</math> to <math>10\sqrt{3}</math> ratio. Let the center of the circle be <math>O</math>, the upper left point of the triangle be <math>A</math>, the right angle be <math>C</math>, the point on the right be <math>B</math>, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be <math>I</math>. Then <math>OI</math> and <math>OB</math> are the radii of the semicircle <math>= 5\sqrt{3}</math>. Because <math>\angle ABC = 30</math> degrees, then <math>\angle BIO = 30</math> degrees (radii are the two legs of the triangle <math>\rightarrow</math> isosceles triangle). That means that <math>\angle COI</math> is <math>60</math> degrees and <math>\angle IOB</math> is <math>120</math> degrees.
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We know that the area of the 60-degree sector is 1/6*(5sqrt(3)^2)pi = 25pi/2. We also can draw the altitude of the remaining area we need to find, which forms a 30-60-90 triangle, giving us the height of the triangle is 15/2. Therefore, the area of that triangular portion = 5*sqrt(3)*15pi/4 = 75sqrt(3)/4. Therefore, the total area = 75sqrt(3)/4 + 25pi/2 = (75*sqrt(3)+50pi)/4. Our answer = 75 + 3+ 50 + 4 = 132.
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To find the area of the overlappingregion, we can find the area of the 60-degree sector and the area of the remaining region, which is a triangle. We know that the area of the 60-degree sector is <math>\frac{1}{6} \cdot (5\sqrt{3})^2 \pi = \frac{25}{2} \pi</math>. We also can draw the altitude of the remaining area we need to find, which forms a <math>30</math> - <math>60</math> - <math>90</math> triangle, giving us the height of the triangle is <math>\frac{15}{2}</math>. Therefore, the area of that triangular portion = <math>5\sqrt{3} \cdot \frac{15}{2} \pi \cdot \frac{1}{2} = \frac{75\sqrt{3}}{4}</math>. Therefore, the total area <math>= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}</math>. Our answer <math>= 75 + 3+ 50 + 4 = \boxed{132}</math>.

Revision as of 20:20, 1 July 2023

We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a $30$ - $60$ - $90$ triangle because of the $10$ to $10\sqrt{3}$ ratio. Let the center of the circle be $O$, the upper left point of the triangle be $A$, the right angle be $C$, the point on the right be $B$, and the intersection between the hypotenuse of the triangle and the curve of the semicircle be $I$. Then $OI$ and $OB$ are the radii of the semicircle $= 5\sqrt{3}$. Because $\angle ABC = 30$ degrees, then $\angle BIO = 30$ degrees (radii are the two legs of the triangle $\rightarrow$ isosceles triangle). That means that $\angle COI$ is $60$ degrees and $\angle IOB$ is $120$ degrees. To find the area of the overlappingregion, we can find the area of the 60-degree sector and the area of the remaining region, which is a triangle. We know that the area of the 60-degree sector is $\frac{1}{6} \cdot (5\sqrt{3})^2 \pi = \frac{25}{2} \pi$. We also can draw the altitude of the remaining area we need to find, which forms a $30$ - $60$ - $90$ triangle, giving us the height of the triangle is $\frac{15}{2}$. Therefore, the area of that triangular portion = $5\sqrt{3} \cdot \frac{15}{2} \pi \cdot \frac{1}{2} = \frac{75\sqrt{3}}{4}$. Therefore, the total area $= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}$. Our answer $= 75 + 3+ 50 + 4 = \boxed{132}$.

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