Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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(→Solution 5: Trig: Changed "area of triangle $ABC$ is equal to $\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}$ to "area of triangle $ADE$...") |
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==Solution 5: Trig == | ==Solution 5: Trig == | ||
− | We try to find the area of quadrilateral <math>FDBG</math> by subtracting the area outside the quadrilateral but inside triangle <math>ABC</math>. Note that the area of <math>\triangle | + | We try to find the area of quadrilateral <math>FDBG</math> by subtracting the area outside the quadrilateral but inside triangle <math>ABC</math>. Note that the area of <math>\triangle ADE</math> is equal to <math>\frac{1}{2} \cdot 25 \cdot 5 \cdot \sin{A}</math> and the area of triangle <math>ABC</math> is equal to <math>\frac{1}{2} \cdot 50 \cdot 10 \cdot \sin A</math>. The ratio <math>\frac{[ADE]}{[ABC]}</math> is thus equal to <math>\frac{1}{4}</math> and the area of triangle <math>ADE</math> is <math>\frac{1}{4} \cdot 120 = 30</math>. Let side <math>BC</math> be equal to <math>6x</math>, then <math>BG = 5x, GC = x</math> by the angle bisector theorem. Similarly, we find the area of triangle <math>AGC</math> to be <math>\frac{1}{2} \cdot 10 \cdot x \cdot \sin C</math> and the area of triangle <math>ABC</math> to be <math>\frac{1}{2} \cdot 6x \cdot 10 \cdot \sin C</math>. A ratio between these two triangles yields <math>\frac{[ACG]}{[ABC]} = \frac{x}{6x} = \frac{1}{6}</math>, so <math>[AGC] = 20</math>. Now we just need to find the area of triangle <math>AFE</math> and subtract it from the combined areas of <math>[ADE]</math> and <math>[ACG]</math>, since we count it twice. Note that the angle bisector theorem also applies for <math>\triangle ADE</math> and <math>\frac{AE}{AD} = \frac{1}{5}</math>, so thus <math>\frac{EF}{ED} = \frac{1}{6}</math> and we find <math>[AFE] = \frac{1}{6} \cdot 30 = 5</math>, and the area outside <math>FDBG</math> must be <math> [ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45</math>, and we finally find <math>[FDBG] = [ABC] - 45 = 120 -45 = \boxed{75}</math>, and we are done. ~skyscraper |
==See Also== | ==See Also== |
Revision as of 01:00, 29 December 2019
Contents
Problem
Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ?
Solution 1
Let , , , and the length of the perpendicular to through be . By angle bisector theorem, we have that where . Therefore substituting we have that . By similar triangles, we have that , and the height of this trapezoid is . Then, we have that . We wish to compute , and we have that it is by substituting.
Solution 2
For this problem, we have because of SAS and . Therefore, is a quarter of the area of , which is . Subsequently, we can compute the area of quadrilateral to be . Using the angle bisector theorem in the same fashion as the previous problem, we get that is times the length of . We want the larger piece, as described by the problem. Because the heights are identical, one area is times the other, and .
Solution 3
The area of to the area of is by Law of Sines. So the area of is . Since is the midsegment of , so is the midsegment of . So the area of to the area of is , so the area of is , by similar triangles. Therefore the area of quad is
Solution 4
The area of quadrilateral is the area of minus the area of . Notice, , so , and since , the area of . Given that the area of is , using on side yields . Using the Angle Bisector Theorem, , so the height of . Therefore our answer is
Solution 5: Trig
We try to find the area of quadrilateral by subtracting the area outside the quadrilateral but inside triangle . Note that the area of is equal to and the area of triangle is equal to . The ratio is thus equal to and the area of triangle is . Let side be equal to , then by the angle bisector theorem. Similarly, we find the area of triangle to be and the area of triangle to be . A ratio between these two triangles yields , so . Now we just need to find the area of triangle and subtract it from the combined areas of and , since we count it twice. Note that the angle bisector theorem also applies for and , so thus and we find , and the area outside must be , and we finally find , and we are done. ~skyscraper
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
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