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  • .... A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theorem. ...ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,
    5 KB (885 words) - 22:03, 5 October 2024
  • ...idean [[geometry]], and are the simplest possible polygon. In [[physics]], triangles are noted for their durability, since they have only three [[vertex|vertice Triangles are split into six categories; three by their [[angle]]s and three by their
    4 KB (631 words) - 20:16, 8 October 2024
  • ...sides of length <math>1</math> and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the [[Category: Introductory Geometry Problems]]
    4 KB (691 words) - 17:38, 19 September 2021
  • Draw extra lines to create similar triangles! (Hint: Draw <math>AD</math> on all three figures. Draw another line as wel == Problems ==
    5 KB (859 words) - 15:11, 8 December 2024
  • ...hing this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is <ma ...seed is <math>\frac{33(ay+yx)}{11xy}.</math> Finally note that by similar triangles <math>\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.</math> Therefore the r
    4 KB (709 words) - 00:50, 10 January 2022
  • (Similar to Solution 1) Note that triangles <math>\triangle AFE</math> and <math>\triangle AZE</math> share the same hy
    6 KB (958 words) - 22:29, 28 September 2023
  • {{AMC10 Problems|year=2006|ab=A}} [[2006 AMC 10A Problems/Problem 1|Solution]]
    13 KB (2,028 words) - 15:32, 22 March 2022
  • ...riangle AEO_2 \sim \triangle AFC</math>. Using the first pair of [[similar triangles]], we write the proportion: [[Category:Introductory Geometry Problems]]
    5 KB (811 words) - 10:44, 30 November 2024
  • ...s to come to the same conclusion using different [[right triangle|45-45-90 triangles]]. Drawing lines as shown above and piecing together the triangles, we see that <math>ABCD</math> is made up of <math>12</math> squares congru
    6 KB (1,106 words) - 09:20, 4 November 2024
  • How many non-[[similar]] triangles have angles whose degree measures are distinct positive integers in [[arith [[Category:Introductory Geometry Problems]]
    2 KB (259 words) - 02:10, 22 June 2023
  • ...<math>O_3</math> to <math>\overline{AB}</math> be <math>T</math>. From the similar [[right triangle]]s <math>\triangle HO_1T_1 \sim \triangle HO_2T_2 \sim \tr ...=10-4=6</math>, and <math>O_1O_2=14</math>. But <math>O_1O_3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}
    4 KB (693 words) - 12:03, 28 December 2021
  • === Solution 3 (similar triangles)=== ...et <math>a = OE</math>, <math>b = OF</math>. Applying Stewart's Theorem on triangles <math>AOB</math> twice, first using <math>E</math> as the base point and th
    13 KB (2,080 words) - 12:14, 23 July 2024
  • Similarly, using the 5-12-13 triangles, we easily see that <math>12+5i</math> represents the bisection of the angl ...rom this tangent point to the origin is <math>\sqrt{69}.</math> By similar triangles, the slope of this line is then <math>\frac{\sqrt{69}}{5\sqrt{3}}.</math> W
    12 KB (2,001 words) - 19:26, 23 July 2024
  • == Solution 2 (Similar Triangles)== ...the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>
    14 KB (2,340 words) - 15:38, 21 August 2024
  • ...riangle]]s <math>\triangle CDA</math> and <math>\triangle CEB</math> are [[similar]] [[right triangle]]s. By the Pythagorean Theorem <math>CD=8\cdot\sqrt{6}< [[Category:Intermediate Geometry Problems]]
    4 KB (729 words) - 00:00, 27 November 2022
  • ...{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, ...w the formula for the distance from a point to a line, you can use similar triangles to get the ratio:
    5 KB (836 words) - 06:53, 15 October 2023
  • ...frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac{70}{3}</m ...so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 2
    9 KB (1,500 words) - 19:06, 8 October 2024
  • ...we can find the third side of the triangle using [[Stewart's Theorem]] or similar approaches. We get <math>AC = \sqrt{56}</math>. Draw additional lines as indicated. Note that since triangles <math>AQP</math> and <math>BPR</math> are isosceles, the altitudes are also
    14 KB (2,351 words) - 20:06, 8 December 2024
  • ...to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity <math>K = \dfrac{ab\sin C}{2}</math> to show that the ...f each of the three triangles would all be equal. The areas of each of the triangles are all perfect squares, so we could assume <math>\dfrac{base}{height} = \d
    4 KB (726 words) - 12:39, 13 August 2023
  • ...frac{1}{1985} = 1</math>, where <math>X</math> is the area of the of the 4 triangles. ...c{n-1}{\sqrt{1985}}</math>. Notice that <math>\triangle CEL</math> is also similar to <math>\triangle CDF</math> by <math>AA</math> similarity. Thus, <math>\f
    3 KB (484 words) - 20:40, 2 March 2020

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