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- ...r</math> to one of the sides creating an <math>r,r, r\sqrt{2}</math> right triangle. This means that we have <math>r + r\sqrt{2} = 8\sqrt{2}</math> so <math>r ...rom <math>O</math> to <math>AB</math>, which forms a <math>45-45-90</math> triangle. The length of the perpendicular is <math>\frac{r}{\sqrt{2}}</math>. Note a4 KB (707 words) - 10:11, 16 September 2021
- ...he center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment <math>x</math>. Then by the [[Pythagorean Theorem]]: ...ound by partitioning the leg into three sections and using <math>45-45-90 \triangle</math>s to see that the leg is <math>100\sqrt{2} + 200 + 100\sqrt{2} = 200(3 KB (496 words) - 12:02, 5 August 2019
- In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> ...h>, it also follows that <math>AQ=PC</math>. Now apply the Law of Sines to triangle <math>PBC</math> to find <cmath>\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{9 KB (1,505 words) - 22:17, 31 December 2024
- The [[perimeter]] of triangle <math>APM</math> is <math>152</math>, and the angle <math>PAM</math> is a [ ...</math> at <math>B</math>. Then note <math>\triangle OPB</math> and <math>\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by [[power of a4 KB (658 words) - 18:15, 19 December 2021
- ...>, <math>CA=15</math>, and that the distance from <math>O</math> to <math>\triangle ABC</math> is <math>\frac{m\sqrt{n}}k</math>, where <math>m</math>, <math>n ...riangles <math>\triangle OAD</math>, <math>\triangle OBD</math> and <math>\triangle OCD</math> we get:4 KB (628 words) - 15:27, 20 January 2025
- ...gle OAB</math> is isoceles with base <math>AB</math>. The height of <math>\triangle OAB</math> from <math>O</math> to <math>AB</math> is <math>\sqrt {\overline ...e PAB</math> is also isoceles with base <math>AB</math>. The height of the triangle from <math>P</math> to <math>AB</math> is <math>\sqrt {\overline{PB}^2 - (\2 KB (256 words) - 00:45, 26 June 2016
- In triangle <math>ABC</math>, <math>AB=125</math>, <math>AC=117</math> and <math>BC=120 ...cular to <math>{BM}</math>, <math>\triangle BCP</math> must be an isoceles triangle, so <math>BP=BC=120</math>, and <math>M</math> is the midpoint of <math>{CP9 KB (1,523 words) - 11:08, 15 July 2024
- ..., <math>AD_2E_3</math>, and <math>AD_2E_4</math>, each congruent to <math>\triangle ABC</math>, ...\theta</math>, where <math>\theta</math> is the apex angle in the isoceles triangle with sides <math>\sqrt{111}</math>, <math>\sqrt{111}</math>, and <math>\sqr13 KB (2,052 words) - 17:02, 5 February 2024
- Let <math>ABC</math> be an acute triangle with <math>AB>AC</math>. Let <math>\Gamma</math> be its circumcircle, <math ...<math>M \longleftrightarrow Q`</math>. Of course <math>\triangle HFM \sim \triangle HQ'A</math> so <math>\angle HQ'A = 90</math>. Hence, <math>Q' = Q</math>. S2 KB (422 words) - 14:02, 1 June 2024
- ...h>1</math>. Any <math>3</math> of the <math>12</math> vertices determine a triangle. Find the number of these triangles that are isosceles (including equilater ...we can call one of the vertices <math>A</math> for distinction. Either the triangle can have sides <math>1, 1, \sqrt{3}</math> with 6 cases or <math>\sqrt{3},6 KB (1,023 words) - 09:38, 21 November 2023
- In acute triangle <math>ABC,</math> <math>\ell</math> is the bisector of <math>\angle BAC</ma ...</math> is isoceles with base <math>EF</math>. Using <math>30-60-90</math> triangle ratio, we find <math>\theta = 30^\circ</math>3 KB (463 words) - 14:49, 12 July 2023
- Thus by SAS, <math>\triangle MO_{2}M_{2}</math> is congruent to <math>\triangle MO_{1}M_{1}</math> ...} = MM_{2}</math>, hence <math>\triangle MM_{1}M_{2}</math> is an isoceles triangle.3 KB (429 words) - 15:38, 9 December 2018
- ...math>DEF</math> and <math>DEG</math> is at most equal with the area of the triangle <math>ABC</math>. When does the equality hold? ...>DF</math> and <math>EG</math>, so <math>AP</math> is an angle bisector of triangle <math>ADE</math>.3 KB (565 words) - 23:59, 8 January 2023
- ...angle ABC</math> that is tangent to side <math>\overline{BC}</math> of the triangle and tangent to the extensions of the other two sides. Let <math>E</math>, < So we recall that it is well known that triangle <math>AEF</math> is similar to <math>ABC</math>. This motivates reflecting10 KB (1,536 words) - 19:27, 12 April 2021
- ...side of line <math>AB</math> so that <math>\triangle ABC</math> and <math>\triangle BAD</math> are congruent with <math>AB = 9</math>, <math>BC=AD=10</math>, a Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is th9 KB (1,508 words) - 13:02, 7 September 2024
- ...whose areas are <math>2, 3, 4,</math> and <math>5</math> starting with the triangle with base <math>\overline{CD}</math> and moving clockwise as shown in the d ...e height of the bottom triangle, <math>h_2</math> be the height of the top triangle.3 KB (484 words) - 11:58, 10 September 2023
- ...e radian measure of the smallest angle in a <math>7{-}24{-}25</math> right triangle. In terms of <math>\alpha</math>, what is <math>\beta</math>? ...measure of angle COP is <math>{2}{\alpha}</math>. Triangle COP is a right triangle with angle OCP being the same as angle ACD (<math>\beta</math>), angle COP7 KB (1,069 words) - 20:22, 28 December 2024