{"id":2929,"date":"2022-11-21T16:00:21","date_gmt":"2022-11-21T10:30:21","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2929"},"modified":"2022-11-22T16:54:39","modified_gmt":"2022-11-22T11:24:39","slug":"construction-of-perpendicular-bisector-of-a-line-segment","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/construction-of-perpendicular-bisector-of-a-line-segment\/","title":{"rendered":"How Do You Construct A Perpendicular Bisector"},"content":{"rendered":"<h2><strong>Construction Of Perpendicular Bisector Of A Line Segment<\/strong><\/h2>\n<p>A line which is perpendicular to a given line segment (AB) and divides it into two equal halves, i.e., AO = OB is called the perpendicular bisector of AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8064\/29459802383_d79c94f505_o.png\" alt=\"perpendicular-bisector\" width=\"269\" height=\"225\" \/>In figure, XY is the perpendicular bisector of AB since AO = OB and \u2220XOB = 90\u00b0.<\/p>\n<p><strong>Read More:<\/strong><\/p>\n<ul>\n<li><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><a href=\"https:\/\/www.aplustopper.com\/construction-of-similar-triangles\/\"><span style=\"font-weight: 400;\">Construction Of Similar Triangle As Per Given Scale Factor<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-line-segment\/\"><span style=\"font-weight: 400;\">Construction Of A Line Segment<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-bisector-of-an-angle\/\"><span style=\"font-weight: 400;\">Construction Of The Bisector Of A Given Angle<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-angles-using-compass-ruler\/\"><span style=\"font-weight: 400;\">Construction Of An Angle Using Compass And Ruler<\/span><\/a><\/li>\n<\/ul>\n<p><strong>To draw a perpendicular bisector of a line segment<\/strong><br \/>\n<strong>Construction:<\/strong> Draw the perpendicular bisector of a line segment AB = 5.5 cm using a scale and a pair of compasses.<\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Draw a line segment AB of length 5.5 cm.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5250\/30053113246_b0e6337d97_o.png\" alt=\"perpendicular-bisector-construction\" width=\"324\" height=\"41\" \/><\/li>\n<li><strong>Step 2:<\/strong> Taking A as the centre and with any radius more than half of AB, draw an arc on both side of AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8280\/29792590340_075cd47077_o.png\" alt=\"perpendicular-bisector-construction-1\" width=\"325\" height=\"285\" \/><\/li>\n<li><strong>Step 3:<\/strong> Similarly, taking B as the centre and radius as in step 2, draw another arc on both side of AB intersecting the previous arcs at C and D.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/8\/7473\/29459801543_8eaa5caf62_o.png\" alt=\"perpendicular-bisector-construction-2\" width=\"325\" height=\"312\" \/><\/li>\n<li><strong>Step 4:<\/strong> Join C and D crossing AB at O.<br \/>\nHence, CD is the required perpendicular\u00a0bisector of AB.<\/li>\n<\/ul>\n<p><strong>Verification:<\/strong> Measure AO and OB. We find the measurement of AO = OB and also \u2220COB = \u2220COA = 90\u00b0.<\/p>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>Draw a line segment PQ of length 8.4 cm. Draw the perpendicular bisector of this line segment.<br \/>\n<strong>Solution: \u00a0 \u00a0<\/strong>We follow the following steps for constructing the perpendicular bisector of PQ.<br \/>\nSteps of Construction:<br \/>\n<strong>Step I:<\/strong> Draw a line segment PQ = 8.4 cm by using a ruler.<br \/>\n<strong>Step II:<\/strong> With P as centre and radius more than half of PQ, draw two arcs, one on each side of PQ.<br \/>\n<strong>Step III:<\/strong> With Q as centre and the same radius as in step II, draw arcs cutting the arcs drawn in the previous step at L and M respectively.<br \/>\n<strong>Step IV:<\/strong> Draw the line segment with L and M as end-points.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8441\/29584749392_b17f790961_o.png\" alt=\"Perpendicular-Bisector-Of-A-Line-Segment\" width=\"203\" height=\"206\" \/>The line segment LM is the required perpendicular bisector of PQ.<\/p>\n<h3><strong>To draw a perpendicular at a point \u00a0on the line<\/strong><\/h3>\n<p><strong>Construction:<\/strong> Draw a perpendicular at a point on the line segment AB = 5.5 cm using a scale and a<br \/>\npair of compasses.<br \/>\n<strong>Given:<\/strong> A line segment AB of length 5.5 cm and a 1 point P lying on it.<br \/>\n<strong>To construct:<\/strong> A line passing through P being perpendicular to AB<\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Draw a line segment AB of length 5.5 cm and make a point P on it.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8533\/30053113116_778a9a40c7_o.png\" alt=\"perpendicular-bisector-construction-3\" width=\"329\" height=\"42\" \/><\/li>\n<li><strong>Step 2:<\/strong>\u00a0Taking P as the centre and with any convenient radius, draw an arc cutting AB at X and Y.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8120\/29792590200_ef7e7a9b2f_o.png\" alt=\"perpendicular-bisector-construction-4\" width=\"328\" height=\"127\" \/><\/li>\n<li><strong>Step 3:<\/strong> Taking X and Y as centres and with any suitable radius draw arcs cutting each other at Q.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8131\/29459801193_986dab7b97_o.png\" alt=\"perpendicular-bisector-construction-5\" width=\"323\" height=\"198\" \/><\/li>\n<li><strong>Step 4:<\/strong> Join P and Q.<br \/>\nThen PQ is perpendicular to AB passing through the point P.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5553\/30053112866_3dd65a2514_o.png\" alt=\"perpendicular-bisector-construction-6\" width=\"326\" height=\"198\" \/><\/li>\n<\/ul>\n<h3><strong>To draw a perpendicular to a given line from a point lying outside the line<\/strong><\/h3>\n<p><strong>Construction:<\/strong> Draw a perpendicular from a point outside the line segment AB = 5.5 cm.<br \/>\n<strong>Given:<\/strong> A line segment AB of length 5.5 cm and a point Y lying outside the line.<br \/>\n<strong>To construct:<\/strong> A line passing through Y which is perpendicular to AB.<\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Draw a line segment AB of length 5.5 cm and mark point Y outside the line segment AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5737\/29792589980_a8e53ee5b5_o.png\" alt=\"perpendicular-bisector-construction-7\" width=\"323\" height=\"161\" \/><\/li>\n<li><strong>Step 2:<\/strong> Taking Y as the centre and with any suitable radius, draw an arc cutting AB at C and D.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5062\/29792589890_71fcb6474b_o.png\" alt=\"perpendicular-bisector-construction-8\" width=\"324\" height=\"182\" \/><\/li>\n<li><strong>Step 3:<\/strong> Taking C and D as centres and with radius more than half of CD, draw arcs below AB intersecting each other at X.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8406\/29459800653_7027c91d97_o.png\" alt=\"perpendicular-bisector-construction-9\" width=\"325\" height=\"246\" \/><\/li>\n<li><strong>Step 4:<\/strong> Join X and Y.<br \/>\nHence, XY is the required perpendicular to the line segment AB from point Y lying outside the line segment AB.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8352\/29792589710_f1480c3fff_o.png\" alt=\"perpendicular-bisector-construction-10\" width=\"324\" height=\"261\" \/><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Construction Of Perpendicular Bisector Of A Line Segment A line which is perpendicular to a given line segment (AB) and divides it into two equal halves, i.e., AO = OB is called the perpendicular bisector of AB. In figure, XY is the perpendicular bisector of AB since AO = OB and \u2220XOB = 90\u00b0. Read [&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":[1270,2668,292,2669,2670,1271,1272],"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>How Do You Construct A Perpendicular Bisector - 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\/construction-of-perpendicular-bisector-of-a-line-segment\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How Do You Construct A Perpendicular Bisector\" \/>\n<meta property=\"og:description\" content=\"Construction Of Perpendicular Bisector Of A Line Segment A line which is perpendicular to a given line segment (AB) and divides it into two equal halves, i.e., AO = OB is called the perpendicular bisector of AB. In figure, XY is the perpendicular bisector of AB since AO = OB and \u2220XOB = 90\u00b0. 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In figure, XY is the perpendicular bisector of AB since AO = OB and \u2220XOB = 90\u00b0. 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