Difference between revisions of "2022 MMATHS Individual Round Problems/Problem 2"
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Triangle <math>ABC</math> has <math>AB = 3, BC = 4,</math> and <math>CA = 5</math>. Points <math>D, E, F, G, H, </math> and <math>I</math> are the reflections of <math>A</math> over <math>B</math>, <math>B</math> over <math>A</math>, <math>B</math> over <math>C</math>, <math>C</math> over <math>B</math>, <math>C</math> over <math>A</math>, and <math>A</math> over <math>C</math>, respectively. Find the area of hexagon <math>EFIDGH</math>. | Triangle <math>ABC</math> has <math>AB = 3, BC = 4,</math> and <math>CA = 5</math>. Points <math>D, E, F, G, H, </math> and <math>I</math> are the reflections of <math>A</math> over <math>B</math>, <math>B</math> over <math>A</math>, <math>B</math> over <math>C</math>, <math>C</math> over <math>B</math>, <math>C</math> over <math>A</math>, and <math>A</math> over <math>C</math>, respectively. Find the area of hexagon <math>EFIDGH</math>. | ||
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+ | ==Solution 1== | ||
+ | The area of rectangle <math>BEHG</math> is <math>6 \cdot 4 = 24</math>. The area of triangle <math>BEF = \frac {8 \cdot 6}{2} = 24</math>. The area of rectangle <math>BDIF</math> is <math>3 \cdot = 24</math>. The area of triangle <math>DBG</math> is <math>\frac {4 \cdot 3}{2} = 6</math>. Add them all together to get <math>\boxed {78}</math>. | ||
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+ | ~Arcticturn |
Latest revision as of 20:13, 18 December 2022
Problem
Triangle has and . Points and are the reflections of over , over , over , over , over , and over , respectively. Find the area of hexagon .
Solution 1
The area of rectangle is . The area of triangle . The area of rectangle is . The area of triangle is . Add them all together to get .
~Arcticturn