{"id":3054,"date":"2021-07-17T14:55:41","date_gmt":"2021-07-17T14:55:41","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3054"},"modified":"2021-12-18T22:24:47","modified_gmt":"2021-12-18T16:54:47","slug":"what-is-the-formula-for-triangle-area","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-formula-for-triangle-area\/","title":{"rendered":"Formula for Triangle Area"},"content":{"rendered":"
Earlier we find Area of Triangle<\/strong> by using the formula –<\/p>\n Area of Triangle = \\(\\sqrt{s(s-a)(s-b)(s-c)}\\)<\/p>\n where s = \\(a+b+c\\over 2\\)<\/p>\n and a, b, c are the sides of the triangle.<\/p>\n<\/blockquote>\n we have used this heron’s formula to find the area of a triangle when the lengths of its sides are given.<\/p>\n Here, we will learn what is the formula for triangle area in terms of coordinates of its vertices.<\/p>\n The area of triangle, the coordinates of whose vertices are A(\\(x_1,y_1\\)), B(\\(x_2,y_2\\)) and C(\\(x_3,y_3\\)) is given by –<\/p>\n Area of Triangle ABC = \\(1\\over 2\\) |[\\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\)]|<\/p>\n<\/blockquote>\n\n\n Example : <\/span>Find the area of a triangle whose vertices are A(3, 2), B(11, 8) and C(8, 12).<\/p>\n Solution : <\/span>Let A = (\\(x_1, y_1\\)) = (3, 2), B = (\\(x_2, y_2\\)) = (11, 8) and C = (\\(x_3, y_3\\)) = (8, 12) be the given points. Then, Remarks :<\/strong><\/p>\n (i) If the area of triangle joining three points is zero, then the points are collinear.<\/p>\n (ii) If altitude of any equilateral triangle is P, then its area = \\(P^2\\over {\\sqrt{3}}\\). If ‘a’ be the side of equilateral triangle, then its area = (\\(a^2\\sqrt{3}\\over 4\\))<\/p>\n<\/blockquote>\n\n\n Example : <\/span>Prove that the area of triangle whose vertices are (t, t-2), (t+2, t+2) and (t+3, t) is independent of t.<\/p>\n Solution : <\/span>Let A = (\\(x_1, y_1\\)) = (t, t-2), B = (\\(x_2, y_2\\)) = (t+2, t+2) and C = (\\(x_3, y_3\\)) = (t+3, t) be the vertices of given triangle. Then,\n
Formula for Triangle Area<\/h2>\n
\n
\nArea of Triangle ABC = \\(1\\over 2\\) |[\\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\)]|
\n\\(\\implies\\) A = |[\\(3(8-12)+11(12-2)+8(2-8)\\)]|
\n\\(\\implies\\) A = |(-12+110-48)| = 25 sq. units<\/p>\n\n\n\n
\nArea of Triangle ABC = \\(1\\over 2\\) |[\\(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\\)]|
\n\\(\\implies\\) A = |[\\(t(t+2-t)+(t+2)(t-t+2)+(t+3)(t-2-t-2)\\)]|
\n\\(\\implies\\) A = |(2t+2t+4-4t-12)| = |-4| = 4 sq. units
\nClearly, area of triangle ABC is independent of t.<\/p>\n\n\n\n