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• Heron's Formula
  '''Heron's Formula''' (sometimes called Hero's formula) is a [[mathematical formula * [http://www.scriptspedia.org/Heron%27s_Formula Heron's formula implementations in C++, Java and PHP]
  4 KB (675 words) - 00:05, 22 January 2024
• Heron's formula
  #REDIRECT [[Heron's Formula]]
  29 bytes (3 words) - 13:55, 22 December 2007

Page text matches

• Heron's Formula
  '''Heron's Formula''' (sometimes called Hero's formula) is a [[mathematical formula * [http://www.scriptspedia.org/Heron%27s_Formula Heron's formula implementations in C++, Java and PHP]
  4 KB (675 words) - 00:05, 22 January 2024
• Brahmagupta's Formula
  Brahmagupta's formula reduces to [[Heron's formula]] by setting the side length <math>{d}=0</math>.
  3 KB (465 words) - 18:31, 3 July 2023
• Triangle
  ...s(s-a)(s-b)(s-c)}</math>, where <math>s</math> is the [[semiperimeter]] ([[Heron's Formula]]).
  4 KB (628 words) - 17:17, 17 May 2018
• Semiperimeter
  Two other well-known examples of formulas involving the semiperimeter are [[Heron's formula]] and [[Brahmagupta's formula]].
  641 bytes (97 words) - 00:28, 31 December 2020
• Area
  * [[Heron's formula]]: <math>K=\sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>a, b</math === Other formulas <math>K = f(a,b,c)</math> equivalent to Heron's ===
  6 KB (1,181 words) - 22:37, 22 January 2023
• Trigonometry
  ...thagorean Theorem]] and is used to prove several famous results, such as [[Heron's Formula]] and [[Stewart's Theorem]]. However, it sees limited applicabili
  8 KB (1,217 words) - 20:15, 7 September 2023
• 2005 AIME I Problems/Problem 15
  ...ath>m = 4\sqrt{2}</math>, and thus <math>AB = 26</math>. You can now use [[Heron's Formula]] to finish. The answer is <math>24 \sqrt{14}</math>, or <math>\b Finally, you can use [[Heron's Formula]] to get that the area is <math>24\sqrt{14}</math>, giving an ans
  5 KB (906 words) - 23:15, 6 January 2024
• 2005 AIME II Problems/Problem 14
  From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*
  13 KB (2,129 words) - 18:56, 1 January 2024
• 1985 AIME Problems/Problem 9
  This triangle has [[semiperimeter]] <math>\frac{2 + 3 + 4}{2}</math> so by [[Heron's formula]] it has [[area]] <math>K = \sqrt{\frac92 \cdot \frac52 \cdot \fr
  5 KB (763 words) - 16:20, 28 September 2019
• 1986 AIME Problems/Problem 2
  ...th side-lengths <math>2\sqrt5,2\sqrt6,</math> and <math>2\sqrt7,</math> by Heron's Formula, the area is the square root of the original expression.
  3 KB (460 words) - 00:44, 5 February 2022
• 1989 AIME Problems/Problem 15
  === Solution 2 (Mass Points, Stewart's Theorem, Heron's Formula) === ...se and the <math>h_{\triangle ABC} = 2h_{\triangle BCP}</math>. Applying [[Heron's formula]] on triangle <math>BCP</math> with sides <math>15</math>, <math>
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1992 AIME Problems/Problem 13
  ...>, so the area is <math>\frac14\sqrt {(81^2 - 81x^2)(81x^2 - 1)}</math> by Heron's formula. By AM-GM, <math>\sqrt {(81^2 - 81x^2)(81x^2 - 1)}\le\frac {81^2 ...e after letting the two sides equal <math>40x</math> and <math>41x</math>. Heron's gives
  4 KB (703 words) - 02:40, 29 December 2023
• 1995 AIME Problems/Problem 14
  ...minor arc <math>\stackrel{\frown}{BC}</math>. The former can be found by [[Heron's formula]] to be <math>[BCE] = \sqrt{60(60-48)(60-42)(60-30)} = 360\sqrt{3
  3 KB (484 words) - 13:11, 14 January 2023
• 1999 AIME Problems/Problem 15
  Now see that by Heron's, <cmath>[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2
  7 KB (1,169 words) - 15:28, 13 May 2024
• 1999 AIME Problems/Problem 14
  ...th>[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84</math>, by [[Heron's formula]].
  7 KB (1,184 words) - 13:25, 22 December 2022
• 1999 AIME Problems/Problem 12
  ...x \cdot 2}{2} = 50 + x</math>, we get <math>(21)(50 + x) = A</math>. By [[Heron's formula]], <math>A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}</
  3 KB (472 words) - 15:59, 25 February 2022
• 2001 AIME I Problems/Problem 12
  ...</math> and <math>\sqrt{4^{2}+6^{2}}</math>, so using the expanded form of heron's formula, <cmath>\begin{align*}[ABC]&=\sqrt{\dfrac{2(a^{2}b^{2}+b^{2}c^{2}
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2001 AIME I Problems/Problem 7
  ...ABC</math> is <math>s = \frac{20 + 21 + 22}{2} = \frac{63}{2}</math>. By [[Heron's formula]], the area of the whole triangle is <math>A = \sqrt{s(s-a)(s-b)(
  9 KB (1,540 words) - 08:31, 1 December 2022
• 2002 AIME I Problems/Problem 13
  ...asy to get that <math>\sin \angle AEP = \frac{\sqrt{55}}{8}</math> (equate Heron's and <math>\frac{1}{2}ab\sin C</math> to find this). Now note that <math>\ \end{matrix}\right|=\frac{16}{81}.</cmath>By Heron's Formula, we have <math>[ABC]=\frac{81\sqrt{55}}{2}</math> which immediate
  6 KB (974 words) - 13:01, 29 September 2023
• 2002 AIME I Problems/Problem 10
  ...Now we have all segments of triangles AGF and ADC. Joy! It's time for some Heron's Formula. This gives area 10.95 for triangle AGF and 158.68 for triangle A
  4 KB (643 words) - 22:44, 8 August 2023
• 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
• 2010 AMC 10B Problems/Problem 7
  An alternative way to find the area of the triangle is by using Heron's formula, <math>A=\sqrt{(s)(s-a)(s-b)(s-c)}</math> where <math>s</math> is
  2 KB (318 words) - 09:00, 1 September 2022
• 2011 AIME I Problems/Problem 4
  ...find <math>r</math>, we can use the formula <math>rs = [ABC]</math> and by Heron's, <math>[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\
  6 KB (1,068 words) - 18:52, 2 August 2023
• 2011 AIME I Problems/Problem 8
  ...lue of <math>h</math> is thus <math>\frac{2K}{57},</math> and note that by Heron's formula the area of <math>\triangle ABC</math> is <math>20\sqrt{221}</mat Note that the area is given by Heron's formula and it is <math>20\sqrt{221}</math>. Let <math>h_i</math> denote
  6 KB (1,077 words) - 21:47, 12 April 2022
• 2011 AIME II Problems/Problem 10
  .../math>, <math>20</math>, and <math>24</math>, we can compute its area with Heron's formula:
  11 KB (1,720 words) - 03:12, 18 December 2023
• 1996 AHSME Problems/Problem 27
  ...math> and <math>B</math>). We can now find the area of the triangle using Heron's formula:
  7 KB (1,046 words) - 11:42, 30 September 2023
• 2011 AMC 8 Problems/Problem 16
  ==Solution 2 (Using Heron's Formula)== Using Heron's formula, we can calculate the area of the two triangles. The formula stat
  2 KB (371 words) - 16:51, 21 January 2024
• 2012 AMC 12A Problems/Problem 16
  ...triangle is <math>\frac{abc}{4A}</math> and that the area of a triangle by Heron's formula is <math>\sqrt{(S)(S-a)(S-b)(S-c)}</math> with <math>S</math> as
  9 KB (1,496 words) - 02:40, 2 October 2022
• 2012 AMC 12A Problems/Problem 18
  By Heron's formula for the area of a triangle we have that the area of triangle <mat
  4 KB (717 words) - 19:07, 28 July 2021
• Mock AIME I 2012 Problems/Problem 3
  Use Heron's formula to find <math>A=[MNO]=\frac{33}{4}\sqrt{195}</math>. Also note fr
  1 KB (208 words) - 17:31, 7 April 2012
• 2012 AMC 12B Problems/Problem 20
  By [https://en.wikipedia.org/wiki/Heron%27s_formula Heron's Formula] <math>S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdo
  4 KB (670 words) - 07:14, 27 December 2022
• 2013 AMC 10B Problems/Problem 16
  ...edians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its s
  5 KB (761 words) - 19:33, 11 January 2024
• 2013 AIME I Problems/Problem 7
  ...+ 64}</math>, and <math>\sqrt{(x/2)^2 + 36}</math>. Therefore, we can use Heron's formula to set up an equation for the area of the triangle. ...)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2</math>. Therefore, when we square Heron's formula, we find
  6 KB (934 words) - 20:06, 24 January 2021
• 2013 AIME I Problems/Problem 9
  ...but it only requires simple understanding of areas, similar triangles, and Heron's formula. I'll just put the strategy here because I am too lazy to calcula 2. use heron's formula to find the areas of those two triangles. remember that it is sqr
  9 KB (1,530 words) - 17:12, 18 April 2024
• 2013 AIME I Problems/Problem 13
  ...peats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90. Using Heron's Formula we can get the area of the triangle <math>\Delta AB_0C_0 = 90</ma
  7 KB (1,085 words) - 22:48, 17 July 2023
• 2013 AIME II Problems/Problem 13
  ...[CAE]=3[EDB]=3[AEB]=\frac{3}{8}[ABC]</math>, because of base-ratios. Using Heron's Formula on <math>\triangle EDB</math>, as it is simplest, we see that <ma Note to writter: Couldn't we just use Heron's formula for <math>[CEB]</math> after <math>x</math> is solved then notici
  13 KB (2,116 words) - 23:24, 21 March 2024
• 2013 USAMO Problems/Problem 6
  ...= 4[ABC] = \sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}</math> by Heron's Formula (where <math>[ABC]</math> is the area of triangle <math>ABC</math
  4 KB (691 words) - 18:29, 10 May 2023
• 2014 AIME II Problems/Problem 3
  By Heron's Formula, the area of each isosceles triangle is <math>\sqrt{(30)(12)(12)(
  3 KB (454 words) - 22:00, 24 January 2024
• 2014 AMC 10A Problems/Problem 14
  ==Solution 4(Heron's Formula)== ...0)=20</math> since they have the same <math>x</math> coordinate. Now using Heron's formula, we have
  6 KB (1,001 words) - 13:07, 25 July 2022
• 2014 AMC 12A Problems/Problem 20
  ...lication of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As <math>A</math> lies on both <math>AB</math> a
  4 KB (665 words) - 04:35, 22 January 2024
• 2014 AMC 10B Problems/Problem 21
  The area of <math>\Delta AED</math> is by Heron's, <math>4\sqrt{9(4)(3)(2)}=24\sqrt{6}</math>. This makes the length of the
  4 KB (652 words) - 09:18, 23 September 2021
• 1964 AHSME Problems/Problem 35
  ...the origin and point <math>B</math>, logically, as <math>(15,0)</math>. By Heron's Formula, the area of this triangle is <math>84.</math> Thus the height pe
  6 KB (892 words) - 00:02, 12 July 2023
• Mock AIME 5 2005-2006 Problems/Problem 12
  By Heron's theorem , We get
  2 KB (294 words) - 16:24, 24 August 2022
• 1993 UNCO Math Contest II Problems/Problem 10
  Heron's Formula states that in a triangle with sides <math>a, b, c</math> and <ma
  2 KB (246 words) - 23:57, 3 June 2022
• 1986 AHSME Problems/Problem 11
  Let's start by finding <math>AH</math>. By Heron's Formula, <math>s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21- P.S.: Although this is torturous, this is a good practice of Heron's formula and Stewart's theorem though.
  2 KB (290 words) - 11:41, 1 December 2022
• 2015 AMC 10A Problems/Problem 21
  ...th>ABE</math>. The area <math>A</math> of triangle <math>ABE</math> is, by Heron's Formula, given by
  4 KB (574 words) - 07:18, 16 August 2023
• 2015 AMC 12A Problems/Problem 20
  ...semiperimeter is <math>9</math>, and <math>a = 9 - \frac12b</math>. Using Heron's formula, <math>\sqrt{9\left(\frac{b}{2}\right)\left(\frac{b}{2}\right)(9- Using Heron's, we get <math>\sqrt{(9)(4)^2(1)} = \sqrt{(\frac{2a+b}{2})\left(\frac{b}{2
  5 KB (912 words) - 22:32, 7 June 2021
• 2015 AIME II Problems/Problem 7
  ...so the area of <math>PQRS</math> is half the area of the triangle. Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>, ...for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>AC</ma
  7 KB (1,180 words) - 14:08, 14 February 2023
• 2015 AMC 8 Problems/Problem 6
  ...h>\triangle ABC</math> is <math>\frac{29+29+42}{2}=50</math>. Next, we use Heron's Formula to find that the area of the triangle is just <math>\sqrt{50(50-2
  1 KB (189 words) - 15:55, 21 January 2024
• 2015 AMC 8 Problems/Problem 19
  ===Solution 6 (Heron's Formula, Not Recommended)=== ...engths of the sides by using the [[Pythagorean Theorem]]. Then, we apply [[Heron's Formula]] to find the area.
  4 KB (596 words) - 00:07, 17 May 2023
• 2016 AMC 12A Problems/Problem 15
  ...+ a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = ==Solution 5 (Heron’s)==
  8 KB (1,255 words) - 09:05, 5 September 2022
• 2016 AMC 12B Problems/Problem 17
  ...angle by [https://artofproblemsolving.com/wiki/index.php/Heron%27s_Formula Heron's Formula]: Using Heron's Formula, we see that <math>II' \cdot 12 = \sqrt{12 \cdot 3 \cdot 4 \cdot
  7 KB (873 words) - 20:12, 23 October 2023
• 2017 AMC 12A Problems/Problem 16
  [[Heron's Formula]] states that the area of an triangle with sides <math>a</math> < Use Heron's Formula to obtain
  13 KB (1,982 words) - 17:12, 20 December 2022
• 2017 AMC 10B Problems/Problem 19
  ...th>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula,
  13 KB (2,008 words) - 23:42, 17 July 2023
• 2017 AMC 12B Problems/Problem 15
  ...th>. We can evaluate the area of triangle <math>ABC</math> by simply using Heron's formula,
  9 KB (1,416 words) - 14:30, 23 November 2023
• 2017 AIME I Problems/Problem 4
  ==Solution 3 (Heron's Formula)== Using Heron's Formula, the area of <math>\triangle CPM</math> can be written as
  7 KB (1,177 words) - 15:55, 5 January 2024
• 2016 IMO Problems/Problem 3
  ...ction on <math>k</math>, with the base case of <math>k=3</math> settled by Heron's formula: If <math>a,b,c</math> are the side lengths of the triangle, then
  2 KB (437 words) - 01:36, 19 November 2023
• 2017 AMC 8 Problems/Problem 18
  ...triangle. So the area of <math>\triangle BCD</math> is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides hav
  2 KB (337 words) - 17:37, 21 January 2024
• 1963 AHSME Problems/Problem 35
  By [[Heron's Formula]], the area of the triangle is <math>\sqrt{24 \cdot 3(24-b)(b-3)}
  1 KB (217 words) - 17:58, 7 June 2018
• 2018 AIME I Problems/Problem 13
  ...6\sqrt{21}</math>. Solving for <math>r_1</math> and <math>r_2</math> using Heron's in <math>ABX</math> and <math>ACX</math>, we get that <math>r_1=3\sqrt{21
  13 KB (2,200 words) - 21:36, 6 January 2024
• 2018 AIME II Problems/Problem 12
  .... Since they share base <math>AC</math>, their areas are equal. We can use Heron's formula. To not have any fractions, let <math>AC=2x.</math> ...ly, we get <math>x^2=32.</math> Plugging this <math>x</math> back into the Heron's formula, we get that the area of <math>ABC</math>(or <math>ADC</math>) is
  18 KB (2,912 words) - 13:12, 24 January 2024
• Mathematical formula
  ...example, a formula for the area of a triangle may be found in the use of [[Heron's Formula]].
  285 bytes (45 words) - 22:38, 30 June 2018
• 2008 iTest Problems/Problem 37
  Using [[Heron's Formula]], the area of the triangle is
  1 KB (173 words) - 02:17, 3 July 2018
• 1998 JBMO Problems/Problem 2
  <math>[ACD]</math> can be found by [[Heron's formula]].
  3 KB (554 words) - 08:31, 2 July 2020
• 2006 iTest Problems/Problem 11
  ...of the triangle is <math>\tfrac12 \cdot (13 + 30 + 37) = 40</math>. By [[Heron's Formula]], the area of the triangle is <math>\sqrt{40 \cdot 27 \cdot 10 \
  2 KB (232 words) - 20:49, 21 January 2020
• 2006 iTest Problems/Problem 29
  ...A}{15}</math>. Since we know the lengths of all three sides, we can use [[Heron's Formula]] to solve for <math>A</math>.
  2 KB (288 words) - 12:12, 4 December 2018
• 2019 AMC 10A Problems/Problem 7
  ...er vertices are at <math>(4, 6)</math> and <math>(6, 4)</math>. Then apply Heron's Formula: the semi-perimeter will be <math>s = \sqrt{2} + \sqrt{20}</math> ==Solution 12 (Heron's Formula) ==
  7 KB (1,079 words) - 22:24, 10 November 2023
• 2019 AMC 10B Problems/Problem 10
  Heron's formula states that for real numbers <math>x</math>, <math>y</math>, <mat
  3 KB (535 words) - 10:27, 24 June 2023
• 2019 AMC 12B Problems/Problem 18
  ...d <math>QR = \sqrt{1^2 + \left(-1\right)^2 + 2^2} = \sqrt{6}</math>. Using Heron's formula, or by dropping an altitude from <math>P</math> to find the heigh
  5 KB (799 words) - 19:30, 12 November 2022
• 2019 AMC 12B Problems/Problem 25
  ...e given its side length to find the area of <math>\triangle ABD</math> and Heron's formula to find the area of <math>\triangle BCD</math>.
  8 KB (1,301 words) - 15:03, 15 September 2023
• 2019 AIME I Problems/Problem 3
  ...es <math>AF = \sqrt{20}</math> and <math>CB = \sqrt{10}</math>. Now, using Heron's Formula, we find <math>\triangle{APF} = 10</math> and <math>\triangle{CQB
  6 KB (901 words) - 09:38, 10 May 2024
• 2019 AIME II Problems/Problem 1
  ...<math>\triangle ABC-</math> the area of <math>\triangle BCE</math>. Using Heron's formula, we compute the area of <math>\triangle ABC=36</math>. Using the This means <math>CE=17-BE=17-\frac{51}{10}=\frac{119}{10}</math>. Next, apply Heron's formula to get the area of <math>\triangle BCE</math>, which equals <math
  11 KB (1,794 words) - 15:32, 14 January 2024
• 2019 AIME I Problems/Problem 11
  ...rom it: <cmath>a=2y, b=x+y, s=x+2y</cmath> It is known (easily proved with Heron's and <math>a=rs</math>) that <cmath>r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sq
  21 KB (3,915 words) - 19:55, 10 October 2023
• 2020 AIME I Problems/Problem 13
  By Heron's, we can calculate the circumradius <math>R = 8/\sqrt{7}</math>, and by La ...nd <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s
  35 KB (5,215 words) - 23:08, 29 October 2023
• 2019 AMC 8 Problems/Problem 21
  Warning: Do not use the distance formula for the base then use Heron's formula. It will take you half of the time you have left!
  4 KB (607 words) - 07:23, 17 January 2024
• 2020 AMC 12A Problems/Problem 24
  ...h>. Let <math>s</math> be the side of the equilateral triangle, we use the Heron's formula: ...ralized for any point in a general triangle (although that requires use of Heron's, and potentially Law of Sines and Cosines).
  16 KB (2,509 words) - 17:49, 8 February 2024
• 2021 AIME I Problems/Problem 9
  ==Solution 8 (Heron's Formula)== For convenience, let <math>AB = 5x</math>. By Heron's formula on <math>\triangle ABD</math>, we have sides <math>5x,6x,9x</math
  23 KB (3,640 words) - 18:16, 25 January 2024
• 2020 CIME II Problems/Problem 7
  ...BC]}{73}.</cmath> The area of triangle <math>ABC</math> can be found using Heron's formula. It is just <cmath>\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{438 \cdot 98 \cd
  2 KB (395 words) - 21:41, 5 September 2020
• 2021 AMC 12B Problems/Problem 11
  ==Solution 4 (Heron's Formula, Pythagorean Theorem, Similar Triangles)== Let the brackets denote areas. By Heron's Formula, we have
  7 KB (1,184 words) - 07:55, 1 October 2021
• 2021 JMC 10 Problems/Problem 21
  ...o <math>\triangle AO_1 O_2</math>'s perimeter is <math>2R</math>. Thus, by Heron's Formula <math>[\triangle{AO_1 O_2}]=\sqrt{Rr_1 r_2 (R-r_1 -r_2)} =\tfrac{
  2 KB (377 words) - 15:56, 1 April 2021
• 2022 AIME II Problems/Problem 11
  By Heron's formula and the inradius area formula, ...equired side lengths; we can find the area of <math>\Delta AMD</math> with Heron's formula. Doing so yields <math>\dfrac72\sqrt5</math>. We could also bash
  14 KB (2,254 words) - 18:26, 8 February 2024
• 2021 Fall AMC 12B Problems/Problem 24
  Denote <math>B=(0, 0)</math> and <math>C=(24, 0)</math>. Note that by Heron's formula the area of <math>\triangle ABC</math> is <math>\frac{165\sqrt{7}
  11 KB (1,733 words) - 11:11, 23 November 2023
• 2021 WSMO Speed Round Problems/Problem 7
  ...c}=\frac{\sqrt{12\cdot42\cdot13\cdot14}}{42}=2\sqrt{13}.</cmath> Now, from Heron's formula, we find that the area of triangle <math>ABC</math> is <cmath>\sq
  2 KB (296 words) - 11:59, 23 December 2021
• 2022 AMC 10A Problems/Problem 13
  Using Heron's formula, <cmath>[ABC] = \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\lef
  7 KB (1,170 words) - 08:04, 19 May 2024
• 2024 AIME I Problems/Problem 14
  We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as <math>A</math>, is <math>A = 6 ...th> must the volume. Each face has the same area by SSS congruence, and by Heron's it is <math>\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b)
  13 KB (2,042 words) - 09:34, 3 April 2024
• Mock AIME 6 2006-2007 Problems/Problem 9
  minimum possible area of triangle <math>ABC</math> using Heron's formula is <math>ABC</math> is:
  2 KB (370 words) - 13:48, 26 November 2023
• 1992 OIM Problems/Problem 3
  Using Heron's formula: And since we use Heron's formula, if the triangle was not possible it would have given us imaginar
  4 KB (814 words) - 09:42, 23 December 2023