# 3. Radius of incenter, C. 4. The centroid is the point that corresponds to the center of gravity in a triangle. B

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1 1. triangle that contains one side that has the same length as the diameter of its circumscribing circle must be a right triangle, which cannot be acute, obtuse, or equilateral Radius of incenter, 4. The centroid is the point that corresponds to the center of gravity in a triangle. 5. Triangle is a triangle with hypoteneuse 6, as stated. Using our knowledge of ratios of sides in triangles, we know that and. Triangle is also a triangle which shares a side with triangle,. Using ratios, we find that.. The area of rhombus is twice the area of two equilaterial triangles with sides of length. rea ( ) 6. In a regular 7-pointed star, there are 7 exterior acute angles and 7 interior obtuse angles, labeled and I respectively in the diagram below. ach exterior angle is equal to half of the subtended arc length in the circumscribed circle. ( ). I, the value of (360-I) is equal to half of the two subtended arc lengths along the circumscribed circle. ( ). The sum is equal to ( ).

2 I 7. Solving the length of using angle bisector ratios, we find that, which exceeds the sum of the other two sides. This is not a possible triangle. 8. y drawing perpendicular lines from the sides of the hexagon to the triangle, as drawn below, we find that the side of the triangle is equal to the sum of 6 and two bases of triangles with hypotenuse of length 3. Side of triangle 9. s in the diagram below, if you label one half the base of the triangle x, we find that the diagonal of the triangle has length. Setting this equal to, we can solve for x. ( ) =. ( ) ( ) The side of the triangles

3 x x x 10. ecause, we know that. lso, since, we know that. Using the information that, we can solve. The sum of arcs and is 120º. Therefore, the remaining arcs sum to 240º, which is equal to twice the sum of ( ). The height of the triangle is ( ) and the width is. Thus, the area is ( )( ) ( ) ( ) 12. The radius of the circumscribed circle is as follows: ( )( )( ) 13. The uler line contains the centroid, circumcenter, and orthocenter. 14. The radius of the inscribed circle in a triangle is ( ). Using Heron s formula, we can solve for the area of the triangle.. Thus, the radius of the inscribed circle is and the diameter is.

4 15. Since segment is a midsegment, we know that the length of. rawing a perpendicular line from point to point along (such that segment intersects segment at point ), we find that triangle and are triangles and triangle and are triangles. iven segment has length 4, we know segment and has length iven segment has length, we know that segment has length. 16. is supplementary to, therefore,. Within triangle, using the information known about the other two angles, we can solve ( ). Using the property of vertical angles, we know as well. is supplementary to, therefore. Within triangle, based on the other two angles, we can solve. is supplementary to, therefore. Within triangle, based on the other two angles, we can solve. is supplementary to, therefore. Within triangle I, based on the other two angles, we can solve ( ) I H J

5 17. ecause the circumcenter lies along, we know that is the diameter of the circle as well as the hypotenuse of the right triangle with right angle at. ecause is a right angle, we know Using law of Sines, we can solve for, the length of the diameter of the circle.. The circumference of the circle is. 18. Quadrilateral is composed of two right triangles that share the same hypotenuse, diagonal, of length Using Heron s formula, we can solve for the area of the triangle. ( )( )( ) ( )( )( ) The altitude to each side can be found by ( ). Therefore, the three altitudes have lengths and the sum is. 20. iven two sides the sine of an angle that is not the interior angle, we do not have enough information to solve for. 21. right isosceles triangle with hypotenuse 8 has two legs of length triangle with hypotenuse 8 has legs of length 4 and. Thus, the perimeter of the quadrilateral whose sides are composed of the legs of the triangles is 22. The shortest distance from the base to vertex is the altitude which is length of the leg of a right triangle with hypotenuse of length of 12 and other leg of length 7. Using Pythagorean Theorem, we can solve the shortest distance possible from vertex to to be. The maximum length from vertex to is along the edge of the triangle, which has legs of length 12. is the only value outside of the range. 23. Looking at the ratios of the sides, we see that is a median as it intersects at its midpoint. Point is the centroid of the triangle since it intercepts the median two-thirds of the way from the vertex. Since goes through point, we know that is also a median. ecause the medians of a triangle separate the triangle into 6 triangles of equal area, we know that triangle is a sixth of the area of triangle. rea of triangle. Since and is parallel to, we know that and therefore, triangle is ½ the area of triangle. rea of triangle. Within triangle, and therefore, triangle is ½ the area of triangle. rea of.

6 H 24. The regular hexagon can be split up into 6 regular triangles, such as drawn below. The altitude of the inscribed smaller triangle can be evaluated by subtracting the altitude of the big triangle from the radius of the circle to give. Using ratios of triangles, we can solve the length of the side of the inscribed triangle to be ( ). The area of the small equilateral triangle is ( ). The area of the regular hexagon is ( ). The ratios of ( ) the area of the triangle to the hexagon is as follows: ( ( ) ) ( ). 25.Since the triangles are both right triangles, we know the hypotenuse coincides with the diameter of the circumscribed triangle. Thus, the circumcenter is the midpoint of the hypotenuse. iven this information, we can solve for the lengths and. We also know using the property of similar triangles and given the fact that = (otherwise, the angles would sum to 90 degrees). Using the Law of osines, we can solve for ( ) ( )( ). It is also possible to simply recognize that triangle is proportional to triangle.

7 26. The equation of a circle is, where a, b, and c are real numbers. Thus, we can plug in the three pairs and solve for a, b, and c. Solving, we get. ompleting the square, we get ( ) ( ). H 27. Triangles with the following vertices are pictured: H, H, H, H,,,,,,,,,,,,, H, H,,,, These 10 triangles with the following vertices are isosceles:,,,,,, H, H, H, H. Triangle is an isosceles triangle with vertex as the altitude drawn from point perpendicularly bisects the base. There are no equilateral triangles. Thus, there are 22-10=12 scalene triangles. 29. raw an altitude down from within isosceles triangle to center H of the square and call the intersection of and I. ecause I bisects, ( ). In triangle I, is a right angle since is the altitude from the vertex of an isosceles triangle. iven

8 that and, we can use the Pythagorean theorem to solve and thus,. is equal to the length of half the diagonal of square. and thus,. Using the formula, where is the angle between sides and, we can determine the area of pentagon. Pentagon is composed of triangles and. ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ). I H 30. since. Point I is the centroid of the triangle and is two-thirds the distance from each vertex. Since and, and. Triangle has sides of length 6, 6, and 4. Using Herons s formula, The medians divide a triangle into 6 triangles of equal area. Thus, the area of triangle I is equal to the areal of triangle I. Since is one third the length of, we know that the area of triangle I is one-third the area of triangle I.. I H

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