Difference between revisions of "2000 AMC 12 Problems/Problem 19"

Revision as of 18:21, 6 January 2015

Problem

In triangle $ABC$ , $AB = 13$ , $BC = 14$ , $AC = 15$ . Let $D$ denote the midpoint of $\overline{BC}$ and let $E$ denote the intersection of $\overline{BC}$ with the bisector of angle $BAC$ . Which of the following is closest to the area of the triangle $ADE$ ?

$\text {(A)}\ 2 \qquad \text {(B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4$

Solution

The area of $[[triangle]] ADE = 1.95$ which rounded equals = $2\ \mathrm{(A)}$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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.

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 == Solution ==
-By the [[Angle Bisector Theorem]], <math>\frac{13}{BE} = \frac{15}{14 - BE} \Longrightarrow BE = 6.5</math>. Since <math>BD = 7</math>, then <math>DE = 0.5</math>.
+The area of <math>[[triangle]] ADE = 1.95</math> which rounded equals = <math>2\ \mathrm{(A)}</math>.
-By [[Heron's Formula]], <math>[ABC] = \sqrt{21(6)(7)(8)} = 84</math>, so the height of <math>\triangle ABC</math> from <math>A</math> is <math>h = \frac{2 \cdot 84}{14} = 12</math>. Notice that the heights of <math>\triangle ABC</math> and <math>\triangle ADE</math> are the same, so <math>[ADE] = \frac{1}{2}(12)(0.5) = 3\ \mathrm{(C)}</math>.
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Difference between revisions of "2000 AMC 12 Problems/Problem 19"

Revision as of 18:21, 6 January 2015

Problem

Solution

See also