How Long is the Triangle's Base? - July 15-19, 1996
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In an isosceles triangle, the perpendicular bisector of one leg passes through the midpoint of the base. If the length of this leg is 10, how long is the base?
Quite a range of ages submitted full solutions this week, all the way from Jennifer Kaplan (Grade 6), Thomas Kuo (Grade 7), and Lori Parcel (Grade 10), to Greg Pack, a "Geometry teacher practicing for the fall," and Bob Overkamp of Odyssey Ultraware Inc.Note: Some people's drawings had to be straightened out a bit because they used tabs - different computers interpret tabs differently, so it's important to use only the space bar when lining up your asterisk drawings.
Jennifer Kaplan Grade: 6 School: Castilleja First I drew the isosceles triangle and labeled the angles A, B, and C. Since the triangle was isosceles angle CAB = angle BCA. Then I drew the perpendicular bisector of one leg and made it pass through the midpoint of the base. I labled the point at the bisector of the leg D. I drew a line from angle ABC to the midpoint of the base (E). I found that the two triangles that the line made were congruent because the sides were the same, and since the triangles were congruent, their angles must be the same - therefore the two angles around the midpoint must be equal, and both angles must be 90 degees. Also the line was the angle bisector of angle ABC because the two top angles were the same. Since the line from point D is perpendicular to AB, both angles on either side of point D equal 90 degrees. Then I looked at the two triangles that the bisector of line AB made. I knew that for two triangles to be congruent a side, angle and side must be equal. The two triangles have a common side of five, a common angle of 90, and share a side (the bisector of side AB). Therefore, since the two triangles are congruent, AE equals EB. Since the hypotenuse of triangle AEB equals 10, each of the sides must be the square root of 50, (using the Pythagoreon theorem, the sum of each side of a triangle squared = the hypotonuse squared). Since 10 squared equals 100, and both sides are equal, each side must be 5-the-square-root-of-2. Since one side of the triangle equals half the base, the base of the triangle must equal 10-the-square-root-of-2!
From: Thomas S. Kuo School: Murray Middle School, Ridgecrest, California Grade: 7th A E B * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * D * * * * * * * * * * * C Triangle ABC is an isosceles triangle. (AB = BC, AC is the base) AB = 10. ED is perpendicular to AB. AE = EB. AD = DC. The base is 10 x sqrt(2). (1) Find BC. Because the triangle is an isosceles and AB and BC are the legs, BC = AB = 10 (2) Find ED. Because of the theorem that if a line's endpoints are at the middle of two of the sides of a triangle that the line is half the length of the third and parallel side, ED = 1/2 (BC) = 5 (3) Find AD. Because of the theorem that in a right triangle if the legs are equal, the hypotenuse is equal to that length times sqrt(2), AD = 5 x sqrt(2). (4) Find AC. Because AD is a half of AC, AC = 10 x sqrt(2).
Lori Parcel Grade: 10 School: Center Grove High School Draw a line to connect the midpoint of the base with the angle opposite. Since this is an isosceles triangle, it will form a right angle with the base resulting in two right triangles (the first being formed by the perpendicular bisector of the leg, half of base, and half of leg). Then use the following to determine the base of the larger right triangle (half the base of the isosceles triangle): part of hyp closest to base of larg. rt tri ------------------------------------------- = x x -------------------------- entire hyp of lrg. rt. tri 5 (10) = x squared 50 = x squared 5 radical 2 = x Thus, half of the base of the isosceles triangle is 5 radical 2, so the entire base is 10 radical 2.
Organization: Odyssey Ultraware Inc. The midpoint of a segment is equidistant from the endpoints of the segment, and a point on a perpendicular bisector of a segment is equidistant from the endpoints of the segment. Since "In an isosceles triangle, the perpendicular bisector of one leg passes through the midpoint of the base", the midpoint of the base of this triangle is equidistant from the three vertices of the triangle; in fact, it is the circumcenter of the triangle, and the base of the isosceles triangle is a diameter of the circumcircle of the triangle. An angle inscribed in a circle measures half as much as the arc it intercepts, so the angle opposite the base of our isosceles triangle is a right angle. If "the length of this leg is 10", then the length of the base is 10 times the square root of 2, in the same scale of measurement.
Gregory W. Pack, Geometry teacher practicing for the fall Pensacola Catholic High School Pensacola, FL The drawing below uses isosceles triangle AJF with congruent legs AF and AJ to illustrate the problem below. Segment IL, the perpendicular bisector of leg AJ, intersects base FJ at its midpoint I. AJ = 10. Find: How long is base FJ?The altitude AI of isosceles triangle AJF intersects base FJ perpendicularly at its midpoint I. This forms right triangle AIJ with hypotenuse AJ. Since segment IL is perpendicular to hypotenuse AJ and intersects vertex I of right angle AIJ, it is an altitude of right triangle AIJ. The altitude to the hypotenuse of right triangle AIJ forms two triangles that are similar to triangle AIJ and to each other. Hence, the altitude to the hypotenuse AJ of right triangle AIJ intersects it so that the length of leg IJ is the geometric mean between the length of its adjacent segment LJ of the hypotenuse and the length of the entire hypotenuse AJ. Since AJ = 10 and L is the midpoint of hypotenuse AJ, LJ = 5. Solving for the length of leg IJ of right triangle AIJ using the geometric mean: (IJ/5) = (10/IJ) IJ^2 = 50 IJ = sqrt(50) = 5*sqrt(2) Since I is the midpoint of base FJ, FI = IJ = 5*sqrt(2) Since I is between F and J, FJ = FI + IJ Hence, FJ = 5*sqrt(2) + 5*sqrt(2) = 10*sqrt(2) Therefore, the length of base AJ is 10*sqrt(2) or approx 14.14213562373.
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