Difference between revisions of "2002 AMC 12A Problems/Problem 23"
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<math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math> | <math>\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | |||
<asy> | <asy> | ||
unitsize(0.25 cm); | unitsize(0.25 cm); | ||
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Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. | Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have <math>[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math>. | ||
− | + | ==Solution 3== | |
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = MC</math>. Also, by the angle bisector theorem, <math>\frac {AB}{BC} = \frac{9}{7}</math>. Thus, let <math>AB = 9x</math> and <math>BC = 7x</math>. In addition, <math>BM = 3.5x</math>. | Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, <math>BD = DC = 7</math> and <math>BM = MC</math>. Also, by the angle bisector theorem, <math>\frac {AB}{BC} = \frac{9}{7}</math>. Thus, let <math>AB = 9x</math> and <math>BC = 7x</math>. In addition, <math>BM = 3.5x</math>. | ||
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Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math> | Thus, <math>AB = 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}</math> | ||
− | + | ==Solution 4== | |
Note that because the perpendicular bisector and angle bisector meet at side <math>AC</math> and <math>CD = BD</math> as triangle <math>BDC</math> is isosceles, so <math>BD = 7</math>. By the angle bisector theorem, we can express <math>AB</math> and <math>BC</math> as <math>9x</math> and <math>7x</math> respectively. We try to find <math>x</math> through Stewart's Theorem. So | Note that because the perpendicular bisector and angle bisector meet at side <math>AC</math> and <math>CD = BD</math> as triangle <math>BDC</math> is isosceles, so <math>BD = 7</math>. By the angle bisector theorem, we can express <math>AB</math> and <math>BC</math> as <math>9x</math> and <math>7x</math> respectively. We try to find <math>x</math> through Stewart's Theorem. So |
Revision as of 07:37, 26 August 2022
Contents
[hide]Problem
In triangle , side and the perpendicular bisector of meet in point , and bisects . If and , what is the area of triangle ?
Solution 1
Looking at the triangle , we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let , so that from given and the previous deducted. Then because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means and are similar, so .
Then by using Heron's Formula on (with sides ), we have .
Solution 3
Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, and . Also, by the angle bisector theorem, . Thus, let and . In addition, .
Thus, . Additionally, using the Law of Cosines and the fact that ,
Substituting and simplifying, we get
Thus, . We now know all sides of . Using Heron's Formula on ,
Solution 4
Note that because the perpendicular bisector and angle bisector meet at side and as triangle is isosceles, so . By the angle bisector theorem, we can express and as and respectively. We try to find through Stewart's Theorem. So
We plug this to find that the sides of are . By Heron's formula, the area is . ~skyscraper
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.