Create the page "Heron" on this wiki! See also the search results found.

• 2003 AIME II Problems/Problem 6
  ...area of <math>\Delta ABC</math> either like the first solution or by using Heron’s Formula. Then, draw the medians from <math>G</math> to each of <math>A, <math>[ABC]</math> can be calculated as 84 using Heron's formula or other methods. Since a <math>180^{\circ}</math> rotation is eq
  5 KB (787 words) - 17:38, 30 July 2022
• 2000 AIME II Problems/Problem 12
  By [[Heron's Formula]] the area of <math>\triangle ABC</math> is (alternatively, a <ma
  3 KB (532 words) - 13:14, 22 August 2020
• Incircle
  *The formula above can be simplified with Heron's Formula, yielding <math>r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math> *The [[area]] of the [[triangle]] by [[Heron's Formula]] is <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>.
  2 KB (384 words) - 18:38, 9 March 2023
• Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
  ...of the squares is <math> 4^{2}+13^{2}+15^{2}=410. </math> Now after using Heron’s Formula, we have that the area of the triangle is 24. Thus, the total a ...) \Longrightarrow \cos{\alpha}=\frac{253}{13 \cdot 25}. </math> Now, using Heron’s Formula, we see that the area of the triangle is 204, so <math> \frac{1
  9 KB (1,364 words) - 15:59, 21 July 2006
• University of South Carolina High School Math Contest/1993 Exam/Problem 29
  One simple solution is using [[area]] formulas: by [[Heron's formula]], a [[triangle]] with sides of length 2, 3 and 4 has area <math>
  2 KB (219 words) - 09:57, 31 August 2012
• Hero's Formula
  #REDIRECT [[Heron's Formula]]
  29 bytes (3 words) - 13:27, 7 January 2008
• 2007 AIME II Problems/Problem 9
  ...we can find that <math>EF = \sqrt {63^2 + 280^2} = 287</math>. We then use Heron's formula to get:
  5 KB (818 words) - 11:05, 7 June 2022
• Mock AIME 3 Pre 2005 Problems/Problem 1
  ...triangle are <math>10</math>, <math>3+r</math>, and <math>7+r</math>. From Heron's Formula, <math>84=\sqrt{(10+r)(r)(7)(3)}</math>, or <math>84*84=r(10+r)*2
  795 bytes (129 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 14
  Using Heron's formula,
  3 KB (563 words) - 02:05, 25 November 2023
• 2007 AIME II Problems/Problem 15
  First, apply [[Heron's formula]] to find that <math>[ABC] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = Consider a 13-14-15 triangle. <math>A=84.</math> [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
  11 KB (2,099 words) - 17:51, 4 January 2024
• 1961 IMO Problems/Problem 2
  By Heron's formula, we have and the RHS becomes <math>4\sqrt{3}\sqrt{(x+y+z)xyz}</math> If we use Heron's formula.
  5 KB (860 words) - 13:12, 13 February 2024
• Circumradius
  ...w the inradius, you can find the area of the triangle by [[Heron's Formula|Heron’s Formula]]: Which follows from the Heron's Formula and <math>R=\frac{abc}{4A}</math>.
  4 KB (729 words) - 16:52, 19 February 2024
• 2006 AIME II Problems/Problem 15
  ...of the triangle is <math>s = \frac{8A + 10A + 12A}{2} = 15A</math> so by [[Heron's formula]] we have <cmath>A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^
  4 KB (725 words) - 17:18, 27 June 2021
• 2000 AMC 12 Problems/Problem 19
  ...DE, DE</math>, to be <math>\frac{1}{2}</math> units long. We can now use [[Heron's Formula]] on <math>ABC</math>. Let's find the area of <math>\Delta ABC</math> by Heron,
  3 KB (547 words) - 17:37, 17 February 2024
• 2007 AMC 10B Problems/Problem 11
  ...1}{2}Bh = \frac {abc}{4R}</math> (or we could use <math>s = 4</math> and [[Heron's formula]]),
  5 KB (851 words) - 22:02, 26 July 2021
• 2008 AMC 12B Problems/Problem 25
  ...side of length <math>8</math> in a <math>5-7-8</math> triangle, and using Heron's, the area of such a triangle is <math>\sqrt{10(5)(3)(2)} = 10 \sqrt{3} =
  12 KB (2,015 words) - 20:54, 9 October 2022
• 2002 AMC 10A Problems/Problem 13
  By [[Heron's formula]], the area is <math>150</math>, hence the shortest altitude's le
  3 KB (395 words) - 13:22, 8 November 2021
• 2009 AIME I Problems/Problem 12
  ...Now we can compute the area of <math>\triangle ABI</math> in two ways: by heron's formula and by inradius times semiperimeter, which yields ...ath>, <math>y + z</math> and <math>x + z</math>, the square of its area by Heron's formula is <math>(x+y+z)xyz</math>.
  12 KB (1,970 words) - 22:53, 22 January 2024
• 2002 AMC 12A Problems/Problem 23
  Then by using [[Heron's Formula]] on <math>ABD</math> (with sides <math>12,7,9</math>), we have < ...= 12</math>. We now know all sides of <math> \triangle ABD</math>. Using [[Heron's Formula]] on <math>\triangle ABD</math>, <math>\sqrt{(14)(2)(7)(5)} = 14\
  6 KB (899 words) - 01:41, 5 July 2023
• 2010 AIME II Problems/Problem 12
  ...gles must be <math>s-8x,s-8x,16x</math> and <math>s-7x,s-7x,14x</math>. By Heron's Formula, we have ...l side lengths. Plugging <math>8x</math> and <math>7x</math> directly into Heron's gives <math>s=338</math>, but for this to be true, the second triangle wo
  2 KB (386 words) - 12:54, 21 November 2023

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)