{"id":654,"date":"2022-12-15T10:00:34","date_gmt":"2022-12-15T04:30:34","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=654"},"modified":"2022-12-16T09:26:34","modified_gmt":"2022-12-16T03:56:34","slug":"areas-of-two-similar-triangles","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/","title":{"rendered":"Areas Of Two Similar Triangles"},"content":{"rendered":"<h2><strong>Areas Of Two <a href=\"https:\/\/www.aplustopper.com\/similarity-of-triangles\/\">Similar Triangles<\/a><\/strong><\/h2>\n<p><strong>Theorem 1: \u00a0<\/strong>The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides.<br \/>\n<strong>Given:<\/strong> Two triangles ABC and DEF such that \u2206ABC ~ \u2206DEF.<br \/>\n<strong>To Prove: <\/strong>\u00a0\\(\\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{A{{B}^{2}}}{D{{E}^{2}}}=\\frac{B{{C}^{2}}}{E{{F}^{2}}}=\\frac{A{{C}^{2}}}{D{{F}^{2}}}\\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png\" alt=\"areas-of-two-similar-triangles-1\" width=\"266\" height=\"136\" \/><br \/>\nConstruction: Draw AL \u22a5\u00a0BC and DM \u22a5\u00a0EF.<br \/>\n<strong>Proof:<\/strong> Since similar triangles are equiangular and their corresponding sides are proportional. Therefore,<br \/>\n\u2206ABC ~ \u2206DEF<br \/>\n\u21d2 \u00a0\u2220A = \u2220D, \u00a0\u00a0\u2220B = \u2220E, \u00a0\u00a0\u2220C = \u2220F \u00a0\u00a0and \u00a0 \\(\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF} \\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\nThus, in \u2206ALB and \u2206DME, we have<br \/>\n\u21d2 \u2220ALB = \u2220DME \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[Each equal to 90\u00ba]<br \/>\nand, \u2220B = \u2220E \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From (i)]<br \/>\nSo, by AA-criterion of similarity, we have<br \/>\n\u2206ALB ~ \u2206DME<br \/>\n\\(\\Rightarrow \\frac{AL}{DM}=\\frac{AB}{DE} \\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(ii)<br \/>\nFrom (i) and (ii), we get<br \/>\n\\(\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF}=\\frac{AL}{DM} \u00a0\\)<br \/>\nNow,<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{\\frac{1}{2}(BC\\times AL)}{\\frac{1}{2}(EF\\times DM)}\\)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{BC}{EF}\\times \\frac{AL}{DM} \\)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{BC}{EF}\\times \\frac{BC}{EF}\\text{\u00a0\u00a0\u00a0 }\\left[ From\\ (iii),\\ \\frac{BC}{EF}=\\frac{AL}{DM} \\right]\\)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{B{{C}^{2}}}{E{{F}^{2}}} \\)<br \/>\n\\(But\\frac{BC}{EF}=\\frac{AB}{DE}=\\frac{AC}{DF} \\)<br \/>\n\\(\\Rightarrow \\frac{B{{C}^{2}}}{E{{F}^{2}}}=\\frac{A{{B}^{2}}}{D{{E}^{2}}}=\\frac{A{{C}^{2}}}{D{{F}^{2}}} \\)<br \/>\n\\(Hence,\\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{A{{B}^{2}}}{D{{E}^{2}}}=\\frac{B{{C}^{2}}}{E{{F}^{2}}}=\\frac{A{{C}^{2}}}{D{{F}^{2}}} \\)<\/p>\n<p><strong>Theorem 2: \u00a0<\/strong>If the areas of two similar triangles are equal, then the triangles are congruent i.e. equal and similar triangles are congruent.<br \/>\n<strong>Given:<\/strong> Two triangles ABC and DEF such that<br \/>\n\u2206ABC ~ \u2206DEF and Area (\u2206ABC) = Area (\u2206DEF).<br \/>\n<strong>To Prove:<\/strong> We have,<br \/>\n\u2206ABC \u2245\u00a0\u2206DEF<br \/>\n<strong>Proof:<\/strong>\u00a0\u2206ABC ~ \u2206DEF<br \/>\n\u2220A = \u2220D, \u00a0\u00a0\u2220B = \u2220E, \u00a0\u00a0\u2220C = \u2220F \u00a0\u00a0and \u00a0 \\(\\frac{AB}{DE}=\\frac{BC}{EF}=\\frac{AC}{DF} \\)<br \/>\nIn order to prove that \u2206ABC \u2245\u00a0\u2206DEF, it is sufficient to show that AB = DE, BC = EF and AC = DF.<br \/>\nNow, Area (\u2206ABC) = Area (\u2206DEF)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=1\\)<br \/>\n\\(\\Rightarrow \\frac{A{{B}^{2}}}{D{{E}^{2}}}=\\frac{B{{C}^{2}}}{E{{F}^{2}}}=\\frac{A{{C}^{2}}}{D{{F}^{2}}}=1\\text{\u00a0\u00a0\u00a0\u00a0\u00a0 }\\left[ \\because \\ \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{A{{B}^{2}}}{D{{E}^{2}}}=\\frac{B{{C}^{2}}}{E{{F}^{2}}}=\\frac{A{{C}^{2}}}{D{{F}^{2}}} \\right] \\)<br \/>\n\u21d2 \u00a0AB<sup>2<\/sup>\u00a0= DE<sup>2<\/sup>, \u00a0 \u00a0BC<sup>2<\/sup>\u00a0= EF<sup>2 \u00a0<\/sup>\u00a0and \u00a0 AC<sup>2<\/sup> = DF<sup>2<\/sup><br \/>\n\u21d2 \u00a0AB = DE, \u00a0 BC = EF \u00a0 and \u00a0 AC = DF<br \/>\nHence, \u2206ABC \u2245\u00a0\u2206DEF.<\/p>\n<p><b>Read More:<\/b><\/p>\n<ul>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/angle-sum-property-of-triangle\/\"><span style=\"font-weight: 400;\">Angle Sum Property of a Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/median-altitude-of-triangle\/\"><span style=\"font-weight: 400;\">Median and Altitude of a Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/find-angle-of-isosceles-triangle\/\"><span style=\"font-weight: 400;\">The Angle of An Isosceles Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/area-of-a-triangle\/\"><span style=\"font-weight: 400;\">Area of A Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/congruent-triangles\/\"><span style=\"font-weight: 400;\">To Prove Triangles Are Congruent<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/similarity-of-triangles\/\"><span style=\"font-weight: 400;\">Criteria For Similarity of Triangles<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/construction-of-triangles-with-compass-and-protractor\/\"><span style=\"font-weight: 400;\">Construction of an Equilateral Triangle<\/span><\/a><\/li>\n<li style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/classification-of-triangles\/\"><span style=\"font-weight: 400;\">Classification of Triangles<\/span><\/a><\/li>\n<\/ul>\n<h2><strong>Areas Of Two Similar Triangles With Examples<\/strong><\/h2>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>The areas of two similar triangles \u2206ABC and \u2206PQR are 25 cm<sup>2<\/sup>\u00a0and 49 cm<sup>2<\/sup>\u00a0respectively. If QR = 9.8 cm,\u00a0find BC.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>It is being given that \u2206ABC ~ \u2206PQR,<br \/>\nar (\u2206ABC) = 25 cm<sup>2<\/sup> and ar (\u2206PQR) = 49 cm<sup>2<\/sup>.<br \/>\nWe know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8477\/28425476953_35492c9878_o.png\" alt=\"areas-of-two-similar-triangles-2\" width=\"308\" height=\"155\" \/><br \/>\n\\(\\therefore \\text{\u00a0 }\\frac{ar\\ (\\Delta ABC)}{ar\\ (\\Delta PQR)}=\\frac{B{{C}^{2}}}{Q{{R}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{25}{49}=\\frac{{{x}^{2}}}{{{(9.8)}^{2}}} \\)<br \/>\n\\(\\Rightarrow {{x}^{2}}=\\left( \\frac{25}{49}\\times 9.8\\times 9.8 \\right) \\)<br \/>\n\\(\\Rightarrow x=\\sqrt{\\frac{25}{49}\\times 9.8\\times 9.8}=\\left( \\frac{5}{7}\\times 9.8 \\right)=\\left( 5\\times 1.4 \\right)=7 \\)<br \/>\nHence BC = 7 cm.<\/p>\n<p><strong>Example 2: \u00a0 \u00a0<\/strong>In two similar triangles ABC and PQR, if their corresponding altitudes AD and PS are in the ratio 4 : 9, find the ratio of the areas of \u2206ABC and \u2206PQR.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Since the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.<br \/>\n\\(\\therefore \\text{\u00a0 }\\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta PQR)}=\\frac{A{{D}^{2}}}{P{{S}^{2}}}\\)<br \/>\n\\(\\Rightarrow \\frac{Area\\,(\\Delta ABC)}{Area\\ (\\Delta PQR)}={{\\left( \\frac{4}{9} \\right)}^{2}}=\\frac{16}{81} \\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 AD : PS = 4 : 9]<br \/>\nHence, Area (\u2206ABC) : Area (\u2206PQR) = 16 : 81<\/p>\n<p><strong>Example 3: \u00a0 \u00a0<\/strong>If \u2206ABC is similar to \u2206DEF such that \u2206DEF = 64 cm<sup>2<\/sup>, DE = 5.1 cm and area of \u2206ABC = 9 cm<sup>2<\/sup>. Determine the area of AB.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Since the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.<br \/>\n\\(\\therefore \\text{\u00a0 }\\frac{Area\\ (\\Delta ABC)}{Area\\,\\,(\\Delta DEF)}=\\frac{A{{B}^{2}}}{D{{E}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{9}{64}=\\frac{A{{B}^{2}}}{{{(5.1)}^{2}}} \\)<br \/>\n\\(\\Rightarrow AB=\\sqrt{3.65} \\)<br \/>\n\u21d2 \u00a0AB = 1.912 cm<\/p>\n<p><strong>Example 4: \u00a0 \u00a0<\/strong>If \u2206ABC ~ \u2206DEF such that area of \u2206ABC is 16cm<sup>2<\/sup> and the area of \u2206DEF is 25cm<sup>2<\/sup> and<br \/>\nBC = 2.3 cm. Find the length of EF.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0We have,<br \/>\n\\(\\frac{\\text{Area}\\ \\text{(}\\Delta \\text{ABC})}{Area\\ (\\Delta DEF)}=\\frac{B{{C}^{2}}}{E{{F}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{16}{25}=\\frac{{{(2.3)}^{2}}}{E{{F}^{2}}} \\)<br \/>\n\\(\\Rightarrow EF=\\sqrt{8.265}~~=\\text{ }2.875\\text{ }cm\\)<\/p>\n<p><strong>Example 5: \u00a0 \u00a0<\/strong>In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD and AB = 2 \u00d7 CD. If the area of \u2206AOB = 84 cm<sup>2<\/sup>. Find the area of \u2206COD.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0In \u2206AOB and \u2206COD, we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c2.staticflickr.com\/8\/7477\/29043025625_397464fa4f_o.png\" alt=\"areas-of-two-similar-triangles-3\" width=\"188\" height=\"129\" \/><br \/>\n\u2220OAB = \u2220OCD \u00a0 \u00a0 \u00a0 \u00a0 (alt. int. \u2220s)<br \/>\n\u2220OBA = \u2220ODC \u00a0 \u00a0 \u00a0 \u00a0 (alt. int. \u2220s)<br \/>\n\u2234\u00a0\u2206AOB ~ \u2206COD \u00a0 \u00a0 \u00a0[By AA-similarity]<br \/>\n\\(\\Rightarrow \\frac{ar\\ (\\Delta AOB)}{ar\\ (\\Delta COD)}=\\frac{A{{B}^{2}}}{C{{D}^{2}}}=\\frac{{{(2CD)}^{2}}}{C{{D}^{2}}} \\) \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 AB = 2 \u00d7 CD]<br \/>\n\\(\\Rightarrow \\frac{4\\times C{{D}^{2}}}{C{{D}^{2}}}=4 \\)<br \/>\n\u21d2 \u00a0ar (\u2206COD) = 1\/4 \u00d7 ar (\u2206AOB)<br \/>\n\\(\\Rightarrow \\left( \\frac{1}{4}\\times 84 \\right)c{{m}^{2}}=21c{{m}^{2}} \\)<br \/>\nHence, the area of \u2206COD is 21 cm<sup>2<\/sup>.<\/p>\n<p><strong>Example 6: \u00a0 \u00a0<\/strong>Prove that the area of the triangle BCE described on one side BC of a square ABCD as base is one half the area of the similar triangle ACF described on the diagonal AC as base.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0ABCD is a square. \u2206BCE is described on side BC is similar to \u2206ACF described on diagonal AC.<br \/>\nSince ABCD is a square. Therefore,<br \/>\nAB = BC = CD = DA and, AC = \u221a2 BC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0[\u2235 Diagonal = \u221a2 (Side)]<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/c2.staticflickr.com\/8\/7578\/28937675172_72ffde7b49_o.png\" alt=\"areas-of-two-similar-triangles-4\" width=\"231\" height=\"150\" \/><br \/>\nNow, \u2206BCE ~ \u2206ACF<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta BCE)}{Area\\ (\\Delta ACF)}=\\frac{B{{C}^{2}}}{A{{C}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta BCE)}{Area\\ (\\Delta ACF)}=\\frac{B{{C}^{2}}}{{{(\\sqrt{2}BC)}^{2}}}=\\frac{1}{2} \\)<br \/>\n\u21d2 \u00a0Area (\u2206BCE) = \\(\\frac { 1 }{ 2 }\\) \u00a0Area (\u2206ACF)<\/p>\n<p><strong>Example 7: \u00a0 \u00a0<\/strong>D, E, F are the mid-point of the sides BC, CA and AB respectively of a \uf044ABC. Determine the ratio of the\u00a0areas of \u2206DEF and \u2206ABC.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Since D and E are the mid-points of the sides BC and AB respectively of \u2206ABC. Therefore,<br \/>\nDE || BA<br \/>\nDE || FA \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(i)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8243\/29043025515_c94f53dba3_o.png\" alt=\"areas-of-two-similar-triangles-5\" width=\"145\" height=\"148\" \/><br \/>\nSince D and F are mid-points of the sides BC and AB respectively of \u2206ABC. Therefore,<br \/>\nDF || CA \u00a0\u00a0\u21d2 \u00a0\u00a0DF || AE<br \/>\nFrom (i), and (ii), we conclude that AFDE is a parallelogram.<br \/>\nSimilarly, BDEF is a parallelogram.<br \/>\nNow, in \u2206DEF and \u2206ABC, we have<br \/>\n\u2220FDE = \u2220A \u00a0 \u00a0\u00a0[Opposite angles of parallelogram AFDE]<br \/>\nand, \u2220DEF = \u2220B \u00a0 \u00a0 \u00a0\u00a0[Opposite angles of parallelogram BDEF]<br \/>\nSo, by AA-similarity criterion, we have<br \/>\n\u2206DEF ~ \u2206ABC<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta DEF)}{Area\\ (\\Delta ABC)}=\\frac{D{{E}^{2}}}{A{{B}^{2}}}=\\frac{{{(1\/2AB)}^{2}}}{A{{B}^{2}}}=\\frac{1}{4}\\text{ \u00a0 \u00a0}\\left[ \\because \\ DE\\ =\\frac{1}{2}AB \\right] \\)<br \/>\nHence, Area (DDEF) : Area (DABC) = 1 : 4.<\/p>\n<p><strong>Example 8: \u00a0 \u00a0<\/strong>D and E are points on the sides AB and AC respectively of a \u2206ABC such that DE || BC and divides \u2206ABC into two parts, equal in area. Find .<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0We have,<br \/>\nArea (\u2206ADE) = Area (trapezium BCED)<br \/>\n\u21d2 \u00a0Area (\u2206ADE) + Area (\u2206ADE)<br \/>\n= Area (trapezium BCED) + Area (\u2206ADE)<br \/>\n\u21d2\u00a02 Area (\u2206ADE) = Area (\u2206ABC)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8027\/28756819460_b0a8e3b907_o.png\" alt=\"areas-of-two-similar-triangles-6\" width=\"154\" height=\"136\" \/><br \/>\nIn \u2206ADE and \u2206ABC, we have<br \/>\n\u2220ADE = \u2220B \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[\u2235\u00a0 DE || BC \u2234\u00a0\u00a0\u2220ADE = \u2220B (Corresponding angles)]<br \/>\nand, \u2220A = \u2220A \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [Common]<br \/>\n\u2234\u00a0\u00a0\u2206ADE ~ \u2206ABC<br \/>\n\\( \\Rightarrow \\frac{Area\\ (\\Delta ADE)}{Area\\ (\\Delta ABC)}=\\frac{A{{D}^{2}}}{A{{B}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{Area\\ (\\Delta ADE)}{2\\,Area\\,(\\Delta ADE)}=\\frac{A{{D}^{2}}}{A{{B}^{2}}} \\)<br \/>\n\\(\\Rightarrow \\frac{1}{2}={{\\left( \\frac{AD}{AB} \\right)}^{2}}\\Rightarrow \\frac{AD}{AB}=\\frac{1}{\\sqrt{2}} \\)<br \/>\n\u21d2 AB = \u221a2 AD \u00a0AB = \u221a2 (AB \u2013 BD)<br \/>\n\u21d2 (\u221a2 \u2013 1) AB = \u221a2 BD<br \/>\n\\(\\Rightarrow \\frac{BD}{AB}=\\frac{\\sqrt{2}-1}{\\sqrt{2}}=\\frac{2-\\sqrt{2}}{2} \\)<\/p>\n<p><strong>Example 9: \u00a0 \u00a0<\/strong>Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Let \u2206ABC and \u2206DEF be the given triangles such that AB = AC and DE = DF, \u2220A = \u2220D.<br \/>\nand \u00a0\\( \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{16}{25} \u00a0\\) \u00a0 \u00a0&#8230;&#8230;.(i)<br \/>\nDraw AL \u22a5\u00a0BC and DM \u22a5\u00a0EF.<br \/>\nNow, AB = AC, DE = DF<br \/>\n\\( \\Rightarrow \\frac{AB}{AC}=1\\text{\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0 }\\frac{DE}{DF}=1 \\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8388\/28425476623_70e1f88105_o.png\" alt=\"areas-of-two-similar-triangles-7\" width=\"290\" height=\"147\" \/><br \/>\n\\( \\Rightarrow \\frac{AB}{AC}=\\frac{DE}{DF}\\text{\u00a0\u00a0 }\\Rightarrow \\text{\u00a0 }\\frac{AB}{DE}=\\frac{AC}{DF} \u00a0\\)<br \/>\nThus, in triangles ABC and DEF, we have<br \/>\n\\( \\frac{AB}{DE}=\\frac{AC}{DF}\\text{\u00a0\u00a0 and\u00a0\u00a0 } \\) \u00a0 \u00a0 \u00a0\u00a0and \u2220A = \u2220D \u00a0 \u00a0[Given]<br \/>\nSo, by SAS-similarity criterion, we have<br \/>\n\u2206ABC ~ \u2206DEF<br \/>\n\\( \\Rightarrow \\frac{Area\\ (\\Delta ABC)}{Area\\ (\\Delta DEF)}=\\frac{A{{L}^{2}}}{D{{M}^{2}}} \u00a0\\)<br \/>\n\\( \\Rightarrow \\frac{16}{25}=\\frac{A{{L}^{2}}}{D{{M}^{2}}} \\) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0[Using (i)]<br \/>\n\\( \\frac{AL}{DM}=\\frac{4}{5} \u00a0\\)<br \/>\nAL : DM = 4 : 5<\/p>\n<p><strong>Example 10: \u00a0 \u00a0<\/strong>In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of \u2206ADE and the trapezium BCED.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8155\/28756819350_c9a1b6132c_o.png\" alt=\"areas-of-two-similar-triangles-8\" width=\"156\" height=\"135\" \/><br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 \u2206ADE ~ \u2206ABC.<br \/>\n\\( \\therefore \\frac{ar(\\Delta ADE)}{ar(\\Delta ABC)}=\\frac{D{{E}^{2}}}{B{{C}^{2}}}={{\\left( \\frac{DE}{BC} \\right)}^{2}}={{\\left( \\frac{3}{5} \\right)}^{2}}=\\frac{9}{25} \u00a0\\)<br \/>\nLet ar (\u2206ADE) = 9x sq units<br \/>\nThen, ar (\u2206ABC) = 25x sq units<br \/>\nar (trap. BCED) = ar (\u2206ABC) \u2013 ar (\u2206ADE)<br \/>\n= (25x \u2013 9x) = (16x) sq units<br \/>\n\\( \\therefore \\frac{ar(\\Delta ADE)}{ar(trap.BCED)}=\\frac{9x}{16x}=\\frac{9}{16} \\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Areas Of Two Similar Triangles Theorem 1: \u00a0The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides. Given: Two triangles ABC and DEF such that \u2206ABC ~ \u2206DEF. To Prove: \u00a0 Construction: Draw AL \u22a5\u00a0BC and DM \u22a5\u00a0EF. Proof: Since similar triangles are [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[75,74,71],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Areas Of Two Similar Triangles - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Areas Of Two Similar Triangles\" \/>\n<meta property=\"og:description\" content=\"Areas Of Two Similar Triangles Theorem 1: \u00a0The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides. Given: Two triangles ABC and DEF such that \u2206ABC ~ \u2206DEF. To Prove: \u00a0 Construction: Draw AL \u22a5\u00a0BC and DM \u22a5\u00a0EF. Proof: Since similar triangles are [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/\" \/>\n<meta property=\"og:site_name\" content=\"A Plus Topper\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-12-15T04:30:34+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-12-16T03:56:34+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Veerendra\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"8 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.aplustopper.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/www.aplustopper.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.aplustopper.com\/#website\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"name\":\"A Plus Topper\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/www.aplustopper.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.aplustopper.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png\",\"contentUrl\":\"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage\",\"url\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/\",\"name\":\"Areas Of Two Similar Triangles - A Plus Topper\",\"isPartOf\":{\"@id\":\"https:\/\/www.aplustopper.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage\"},\"datePublished\":\"2022-12-15T04:30:34+00:00\",\"dateModified\":\"2022-12-16T03:56:34+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/www.aplustopper.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Areas Of Two Similar Triangles\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage\"},\"author\":{\"@id\":\"https:\/\/www.aplustopper.com\/#\/schema\/person\/1363657f2432a17193a1f5d6404b8a0e\"},\"headline\":\"Areas Of Two Similar Triangles\",\"datePublished\":\"2022-12-15T04:30:34+00:00\",\"dateModified\":\"2022-12-16T03:56:34+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage\"},\"wordCount\":1628,\"commentCount\":1,\"publisher\":{\"@id\":\"https:\/\/www.aplustopper.com\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png\",\"keywords\":[\"Areas Of Two Similar Triangles\",\"Areas Of Two Similar Triangles Examples\",\"Triangles\"],\"articleSection\":[\"Mathematics\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#respond\"]}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/www.aplustopper.com\/#\/schema\/person\/1363657f2432a17193a1f5d6404b8a0e\",\"name\":\"Veerendra\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/93ba5dedf76e89af65ae2e1d6ac09855?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/93ba5dedf76e89af65ae2e1d6ac09855?s=96&d=mm&r=g\",\"caption\":\"Veerendra\"},\"url\":\"https:\/\/www.aplustopper.com\/author\/veerendra\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Areas Of Two Similar Triangles - A Plus Topper","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/","og_locale":"en_US","og_type":"article","og_title":"Areas Of Two Similar Triangles","og_description":"Areas Of Two Similar Triangles Theorem 1: \u00a0The ratio of the areas of two similar triangles are equal to the ratio of the squares of any two corresponding sides. Given: Two triangles ABC and DEF such that \u2206ABC ~ \u2206DEF. To Prove: \u00a0 Construction: Draw AL \u22a5\u00a0BC and DM \u22a5\u00a0EF. Proof: Since similar triangles are [&hellip;]","og_url":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/","og_site_name":"A Plus Topper","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2022-12-15T04:30:34+00:00","article_modified_time":"2022-12-16T03:56:34+00:00","og_image":[{"url":"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Veerendra","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/www.aplustopper.com\/#organization","name":"Aplus Topper","url":"https:\/\/www.aplustopper.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#logo","inLanguage":"en-US","url":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/www.aplustopper.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/www.aplustopper.com\/#website","url":"https:\/\/www.aplustopper.com\/","name":"A Plus Topper","description":"Improve your Grades","publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/www.aplustopper.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage","inLanguage":"en-US","url":"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png","contentUrl":"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png"},{"@type":"WebPage","@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage","url":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/","name":"Areas Of Two Similar Triangles - A Plus Topper","isPartOf":{"@id":"https:\/\/www.aplustopper.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage"},"datePublished":"2022-12-15T04:30:34+00:00","dateModified":"2022-12-16T03:56:34+00:00","breadcrumb":{"@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/www.aplustopper.com\/"},{"@type":"ListItem","position":2,"name":"Areas Of Two Similar Triangles"}]},{"@type":"Article","@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#article","isPartOf":{"@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage"},"author":{"@id":"https:\/\/www.aplustopper.com\/#\/schema\/person\/1363657f2432a17193a1f5d6404b8a0e"},"headline":"Areas Of Two Similar Triangles","datePublished":"2022-12-15T04:30:34+00:00","dateModified":"2022-12-16T03:56:34+00:00","mainEntityOfPage":{"@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#webpage"},"wordCount":1628,"commentCount":1,"publisher":{"@id":"https:\/\/www.aplustopper.com\/#organization"},"image":{"@id":"https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#primaryimage"},"thumbnailUrl":"https:\/\/c1.staticflickr.com\/9\/8135\/29009415026_c51f1416ae_o.png","keywords":["Areas Of Two Similar Triangles","Areas Of Two Similar Triangles Examples","Triangles"],"articleSection":["Mathematics"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/www.aplustopper.com\/areas-of-two-similar-triangles\/#respond"]}]},{"@type":"Person","@id":"https:\/\/www.aplustopper.com\/#\/schema\/person\/1363657f2432a17193a1f5d6404b8a0e","name":"Veerendra","image":{"@type":"ImageObject","@id":"https:\/\/www.aplustopper.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/93ba5dedf76e89af65ae2e1d6ac09855?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/93ba5dedf76e89af65ae2e1d6ac09855?s=96&d=mm&r=g","caption":"Veerendra"},"url":"https:\/\/www.aplustopper.com\/author\/veerendra\/"}]}},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/654"}],"collection":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/comments?post=654"}],"version-history":[{"count":1,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/654\/revisions"}],"predecessor-version":[{"id":154095,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/posts\/654\/revisions\/154095"}],"wp:attachment":[{"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/media?parent=654"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/categories?post=654"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.aplustopper.com\/wp-json\/wp\/v2\/tags?post=654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}