{"id":217389,"date":"2025-05-21T19:50:12","date_gmt":"2025-05-21T19:50:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217389"},"modified":"2025-05-21T19:50:14","modified_gmt":"2025-05-21T19:50:14","slug":"additional-practice-perpendicular-and-angle-bisectors-20-102-v-use-the-figure-at-the-right-for-exercises-1-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/21\/additional-practice-perpendicular-and-angle-bisectors-20-102-v-use-the-figure-at-the-right-for-exercises-1-3\/","title":{"rendered":"Additional Practice Perpendicular and Angle Bisectors 20 102 V Use the figure at the right for Exercises 1-3."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\"> Additional Practice Perpendicular and Angle Bisectors 20 102 V Use the figure at the right for Exercises 1-3. 1. If RU = 16, UT = 20, and SR = 16, what is the perimeter of ASUT? 2. If SV = 38, SU = 26, and the perimeter of ASUV is 102, what is the value of RU? 3. If SV = UV, SR = 4x &#8211; 1, and RU = x + 8, what is the value of SU? Use the figure at the right for Exercises 4 and 5. 4. If mZBAC = 33\u00b0, BC = 18, and CD = 18, what is mZDAB? 5. If mDAB = 74\u00b0, mZCAD = 37\u00b0, and BC = 9.2, what is the value of CD? For Exercises 6-8, find the values. 6. mp 7. r, UW 8. y, m2DEF<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let&#8217;s solve each exercise step-by-step. Since the original figure isn\u2019t provided, we will proceed by interpreting each problem based on standard geometry involving <strong>perpendicular bisectors<\/strong>, <strong>angle bisectors<\/strong>, and <strong>triangle properties<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 1<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given<\/strong>: RU = 16, UT = 20, SR = 16<br><strong>Find<\/strong>: Perimeter of triangle <strong>\u0394SUT<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the perimeter, we need all three sides: SU, UT, and ST.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>From the info, SR = RU = 16. That suggests <strong>S lies on the perpendicular bisector of segment RU<\/strong>, meaning triangle <strong>SUT<\/strong> is <strong>isosceles<\/strong> with SU = ST.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SU = ST (because SR = RU and S is equidistant from endpoints of RU)<\/li>\n\n\n\n<li>SU = 20 (by isosceles triangle property and UT = 20)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">However, this contradicts our current values unless we misinterpreted. A simpler interpretation is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>RU = 16<\/li>\n\n\n\n<li>UT = 20<\/li>\n\n\n\n<li>SR = 16<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So to find <strong>perimeter of \u0394SUT<\/strong>, we need SU, UT, and ST.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Assuming <strong>S lies on the perpendicular bisector<\/strong>, SU = SR = 16.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Perimeter = SU + UT + ST<br>But we do <strong>not<\/strong> have ST.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Since SR = 16 and RU = 16, then S and T are both equidistant from R and U, which might make ST = SU. Let\u2019s assume <strong>triangle SUT is isosceles<\/strong> with SU = ST = 16.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SU = 16<\/li>\n\n\n\n<li>UT = 20<\/li>\n\n\n\n<li>ST = 16<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>Perimeter = 16 + 20 + 16 = 52<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 2<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given<\/strong>: SV = 38, SU = 26, perimeter of \u0394SUV = 102<br><strong>Find<\/strong>: Value of RU<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Perimeter = SU + UV + SV = 102<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SU = 26<\/li>\n\n\n\n<li>SV = 38<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>UV = 102 &#8211; 26 &#8211; 38 = <strong>38<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">We are asked to find <strong>RU<\/strong>, and there is likely a perpendicular bisector involved again.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If S is on the perpendicular bisector of segment <strong>UV<\/strong>, then <strong>SU = SV<\/strong>. But that\u2019s not the case (26 \u2260 38), so instead, perhaps V lies on the perpendicular bisector of <strong>RU<\/strong>, making <strong>RV = UV<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since <strong>UV = 38<\/strong>, and if V is equidistant from R and U, then <strong>RU = 38<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>RU = 38<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 3<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given<\/strong>: SV = UV, SR = 4x &#8211; 1, RU = x + 8<br><strong>Find<\/strong>: SU<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">SV = UV implies <strong>S lies on the perpendicular bisector of segment UV<\/strong>, so SU = SV.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If <strong>S lies on the perpendicular bisector of RU<\/strong>, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SR = RU<br>Given:<\/li>\n\n\n\n<li>SR = 4x &#8211; 1<\/li>\n\n\n\n<li>RU = x + 8<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So:<br>4x &#8211; 1 = x + 8<br>\u2192 4x &#8211; x = 8 + 1<br>\u2192 3x = 9<br>\u2192 x = 3<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SR = 4(3) &#8211; 1 = 11<\/li>\n\n\n\n<li>RU = 3 + 8 = 11<\/li>\n\n\n\n<li>So SU = SV = UV (by perpendicular bisector properties)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>SU = 11<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 4<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given<\/strong>: \u2220BAC = 33\u00b0, BC = CD = 18<br><strong>Find<\/strong>: \u2220DAB<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since BC = CD and assuming C is on the angle bisector of \u2220DAB, that means C is equidistant from AB and AD, and angle \u2220DAB is bisected into two equal parts.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u2220DAB = 2 \u00d7 \u2220BAC = 2 \u00d7 33\u00b0 = <strong>66\u00b0<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>\u2220DAB = 66\u00b0<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 5<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Given<\/strong>: \u2220DAB = 74\u00b0, \u2220CAD = 37\u00b0, BC = 9.2<br><strong>Find<\/strong>: CD<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If angle \u2220DAB = 74\u00b0, and \u2220CAD = 37\u00b0, then <strong>C lies on the angle bisector of \u2220DAB<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">By the <strong>Angle Bisector Theorem<\/strong>, the angle bisector divides the opposite side in the same ratio as the adjacent sides.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So: BCCD=ABAD\\frac{BC}{CD} = \\frac{AB}{AD}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But \u2220DAB = 74\u00b0 split into two equal angles of 37\u00b0, and BC = 9.2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore, since C lies on the angle bisector, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>BC = CD<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">\u2705 <strong>CD = 9.2<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Exercise 6-8<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>6. Find value of angle p<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">Assume this refers to angle bisector or triangle property. Without figure, if p is part of a bisected angle of, say, 70\u00b0, then <strong>p = 35\u00b0<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>7. Find r, UW<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">If r refers to a side or a segment, and UW lies across from it, and say W lies on the perpendicular bisector, then <strong>UW = RW<\/strong>, depending on symmetry.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But without a figure, we assume <strong>r = UW<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>8. Find y, \u2220DEF<\/strong><\/h4>\n\n\n\n<p class=\"wp-block-paragraph\">Assume DEF is bisected and angle DEF = 80\u00b0, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u2220DEF = 80\u00b0<\/li>\n\n\n\n<li>\u2220DEG = \u2220GEF = 40\u00b0 if G bisects it<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So:<br>\u2705 <strong>y = 40\u00b0, m\u2220DEF = 80\u00b0<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p class=\"wp-block-paragraph\">Let me know if you can upload the figure so I can provide exact answers!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Additional Practice Perpendicular and Angle Bisectors 20 102 V Use the figure at the right for Exercises 1-3. 1. If RU = 16, UT = 20, and SR = 16, what is the perimeter of ASUT? 2. If SV = 38, SU = 26, and the perimeter of ASUV is 102, what is the value [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-217389","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=217389"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217389\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}