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| ==Solution 1== | | ==Solution 1== |
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− | Essentially, we see that if we draw a line from point A to imaginary point D, that line would apply to both triangles. Let us say that <math>x</math> is the lenmeter of <math>\triangle{ABD}</math> would be <math>\overline{AD} + 4 + x</math>, while the perimeter of <math>\triangle{ACD}</math> would be <math>\overline{AD} + 3 + (5 - x)</math>. Notice that we can find out <math>x</math> from these two equations. We can find out that <math>x = 2</math>, and because the height of the triangles is the same, the ratio of the areas is <math>2:3</math>, so that means that the area of <math>\triangle ABD = \frac{2 \cdot 6}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
| + | We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math>. |
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− | ==Solution 2== | + | ==Video Solutions== |
− | | + | https://youtu.be/itz3JyoZQYg |
− | We know that the perimeters of the two small triangles are <math>3+CD+AD</math> and <math>4+BD+AD</math>. Setting both equal and using <math>BD+CD = 5</math>, we have <math>BD = 2</math> and <math>CD = 3</math>. Now, we simply have to find the area of <math>\triangle ABD</math>. Since <math>\frac{BD}{CD} = \frac{2}{3}</math>, we must have <math>\frac{[ABD]}{[ACD]} = 2/3</math>. Combining this with the fact that <math>[ABC] = [ABD] + [ACD] = \frac{3\cdot4}{2} = 6</math>, we get <math>[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} \cdot 6 = \boxed{\textbf{(D) } \frac{12}{5}}</math>
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− | ==Solution 3==
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− | Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} \cdot 4 \cdot \frac{6}{5} = \frac{12}{5} \implies \boxed{\textbf{(D) } \frac{12}{5}}</math>
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− | ==Solution 4==
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− | Using any preferred method, realize <math>BD = 2</math>. Since we are given a 3-4-5 right triangle, we know the value of <math>\sin(\angle ABC) = \frac{3}{5}</math>. Since we are given <math>AB = 4</math>, apply the Sine Area Formula to get <math>\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{3}{5} = \boxed{\textbf{(D) } \frac{12}{5}}</math>.
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| ==See Also== | | ==See Also== |
Latest revision as of 19:14, 5 July 2024