Difference between revisions of "2020 AMC 10A Problems/Problem 20"
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== Solution == | == Solution == | ||
It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}</math> | It's crucial to draw a good diagram for this one. Since <math>AC=20</math> and <math>CD=30</math>, we get <math>[ACD]=300</math>. Now we need to find <math>[ABC]</math> to get the area of the whole quadrilateral. Drop an altitude from <math>B</math> to <math>AC</math> and call the point of intersection <math>F</math>. Let <math>FE=x</math>. Since <math>AE=5</math>, then <math>AF=5-x</math>. By dropping this altitude, we can also see two similar triangles, <math>BFE</math> and <math>DCE</math>. Since <math>EC</math> is <math>20-5=15</math>, and <math>DC=30</math>, we get that <math>BF=2x</math>. Now, if we redraw another diagram just of <math>ABC</math>, we get that <math>(2x)^2=(5-x)(15+x)</math>. Now expanding, simplifying, and dividing by the GCF, we get <math>x^2+2x-15=0</math>. This factors to <math>(x+5)(x-3)</math>. Since lengths cannot be negative, <math>x=3</math>. Since <math>x=3</math>, <math>[ABC]=60</math>. So <math>[ABCD]=[ACD]+[ABC]=300+60=\boxed {D)360}</math> | ||
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+ | (I'm very sorry if you're a visual learner - Ultraman) | ||
==Video Solution== | ==Video Solution== |
Revision as of 23:59, 31 January 2020
Contents
Problem
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find to get the area of the whole quadrilateral. Drop an altitude from to and call the point of intersection . Let . Since , then . By dropping this altitude, we can also see two similar triangles, and . Since is , and , we get that . Now, if we redraw another diagram just of , we get that . Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
(I'm very sorry if you're a visual learner - Ultraman)
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.