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- ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==7 KB (1,198 words) - 23:28, 19 September 2024
- ==Circumference and Area== ...circumference]] (distance around a circle) is <math>2 \pi r</math> and the area is <math>\pi r^2</math>. Both formulas involve the mathematical constant [9 KB (1,585 words) - 12:46, 2 September 2024
- Pi is the [[ratio]] of the [[circumference]] ([[perimeter]]) of a given [[circle]] to its [[d The number pi often shows up in problems in [[number theory]], particularly [[algebraic number theory]]. For example8 KB (1,469 words) - 20:11, 16 September 2022
- == Problems == <math>\ln a</math> can also be defined as the area under the curve <math>y=\frac{1}{x}</math> between 1 and a, or <math>\int^a4 KB (680 words) - 11:54, 16 October 2023
- Right triangles are very useful in [[geometry]] and for finding the [[area]]s of [[polygon]]s. The most important relationship for right triangles is ...h>30^{\circ}-60^{\circ}-90^{\circ}</math> triangle, which has sides in the ratio of <math>x:x\sqrt3:2x</math>. This triangle is analogous to an equilateral3 KB (499 words) - 22:41, 11 June 2022
- {{AIME Problems|year=2006|n=I}} [[2006 AIME I Problems/Problem 1|Solution]]7 KB (1,173 words) - 02:31, 4 January 2023
- We will use <math>[...]</math> to denote volume (four letters), area (three letters) or length (two letters). We have the volume ratio <math>\frac {[TSBC]}{[TABC]} = \frac {[TS]}{[TA]} = \frac {4}{5}</math>.6 KB (980 words) - 20:45, 31 March 2020
- ...B </math> is 11/5. Find the ratio of shaded region <math> D </math> to the area of shaded region <math> A. </math> Since the area of the triangle is equal to <math>\frac{1}{2}bh</math>,4 KB (709 words) - 00:50, 10 January 2022
- {{AMC12 Problems|year=2006|ab=B}} [[2006 AMC 12B Problems/Problem 1|Solution]]13 KB (2,058 words) - 11:36, 4 July 2023
- {{AMC12 Problems|year=2006|ab=A}} [[2006 AMC 12A Problems/Problem 1|Solution]]15 KB (2,223 words) - 12:43, 28 December 2020
- {{AMC12 Problems|year=2005|ab=A}} [[2005 AMC 12A Problems/Problem 1|Solution]]13 KB (1,965 words) - 21:18, 7 September 2024
- {{AMC12 Problems|year=2003|ab=A}} [[2003 AMC 12A Problems/Problem 1|Solution]]13 KB (1,955 words) - 20:06, 19 August 2023
- {{AMC12 Problems|year=2002|ab=A}} [[2002 AMC 12A Problems/Problem 1|Solution]]12 KB (1,792 words) - 12:06, 19 February 2020
- {{AMC12 Problems|year=2000|ab=}} [[2000 AMC 12 Problems/Problem 1|Solution]]13 KB (1,948 words) - 09:35, 16 June 2024
- {{AMC12 Problems|year=2003|ab=B}} [[2003 AMC 12B Problems/Problem 1|Solution]]13 KB (1,987 words) - 17:53, 10 December 2022
- What is the ratio of the area of <math>S_2</math> to the area of <math>S_1</math>? <math>S_1</math> is a circle with a radius of <math>7</math>. So, the area of <math>S_1</math> is <math>49\pi </math>.2 KB (280 words) - 16:35, 17 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
- ...th>. In addition, <math>AH=AC=2</math>, and <math>AD=3</math>. What is the area of quadrilateral <math>WXYZ</math> shown in the figure? ...em]] we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{16 KB (1,106 words) - 09:20, 4 November 2024
- From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use simil ==Solution 4 (Ratio Lemma and Angle Bisector Theorem)==14 KB (2,340 words) - 15:38, 21 August 2024
- ...s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both e ...rac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[s5 KB (839 words) - 21:12, 16 December 2015