{"id":4226,"date":"2023-05-17T09:53:23","date_gmt":"2023-05-17T04:23:23","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=4226"},"modified":"2023-05-17T17:32:32","modified_gmt":"2023-05-17T12:02:32","slug":"ap-board-9th-class-maths-solutions-chapter-4-ex-4-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-4-ex-4-4\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 4 Lines and Angles Ex 4.4 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4<\/h2>\n

Question 1.
\nIn the given triangles, find out \u2220x, \u2220y and \u2220z
\n\"AP
\nSolution:
\nIn fig(i)
\nx\u00b0 = 50\u00b0 + 60\u00b0
\n(\u2235 exterior angle is equal to sum of the opposite interior angles)
\n\u2234 x= 110\u00b0<\/p>\n

In fig (ii)
\nz\u00b0 = 60\u00b0 + 70\u00b0
\n(\u2235 exterior angle is equal to sum of the opposite interior angles)
\n\u2234 z = 130\u00b0<\/p>\n

In fig (iii)
\ny\u00b0 = 35\u00b0 + 45\u00b0 = 80\u00b0
\n(\u2235 exterior angle is equal to sum of the opposite interior angles)<\/p>\n

\"AP<\/p>\n

Question 2.
\nIn the given figure AS \/\/ BT; \u22204 = \u22205, \\(\\overline{\\mathbf{S B}}\\) bisects \u2220AST. Find the measure of \u22201.
\n\"AP
\nSolution:
\nGiven AS \/\/ BT
\n\u22204 = \u22205 and SB bisects \u2220AST.
\n\u2234 By problem
\n\u22202 = \u22203 …………..(1)
\nFor the lines AS \/\/ BT
\n\u22202 = \u22205 ( \u2235alternate interior angles)
\n\u2234 In \u0394BST
\n\u22203 = \u22205 = \u22204
\nHence \u0394BST is equilateral triangle and each of its angle is equal to 60\u00b0.
\n\u2234\u22203 = \u22202 = 60\u00b0 [by eq. (1)]
\nNow \u22201 + \u22202 + \u22203 = 180\u00b0
\n\u22201 + 60\u00b0 + 60\u00b0 = 180\u00b0
\n[ \u2235 angles at a point on a line]
\n\u2234\u22201 = 180\u00b0 – 120\u00b0 = 60\u00b0<\/p>\n

Question 3.
\nIn the given figure AB \/\/ CD; BC \/\/ DE then find the values of x and y.
\n\"AP
\nSolution:
\nGiven that AB \/\/ CD and BC \/\/ DE.
\n\u2234 3x = 105\u00b0 (\u2235 alternate interior angles for AB \/\/ CD)
\nx = \\(\\frac { 105\u00b0 }{ 3 }\\) = 35\u00b0
\nAlso BC \/\/ DE
\n\u2234\u2220D = 105\u00b0
\n(\u2235 alternate interior angles)
\nNow in \u0394CDE
\n24\u00b0 + 105\u00b0 + y = 180\u00b0
\n(\u2235 angle sum property)
\n\u2234 y = 180\u00b0 – 129\u00b0 = 51\u00b0<\/p>\n

\"AP<\/p>\n

Question 4.
\nIn the given figure BE \u22a5 DA and CD \u22a5 DA then prove that m\u22201 = m\u22203.
\n\"AP
\nSolution:
\nGiven that CD \u22a5 DA and BE \u22a5 DA.
\n\u21d2 Two lines CD and BE are perpendicular to the same line DA.
\n\u21d2 CD \/\/ BE (or)
\n\u2220D =\u2220E \u21d2 CD \/\/ BE
\n(\u2235 corresponding angles for CD and BE and DA are transversal)
\nNow m\u22201 = m\u22203
\n(\u2235alternate interior angles for the lines CD \/\/ BE ; DB are transversal)
\nHence proved.<\/p>\n

Question 5.
\nFind the values of x, y for which lines AD and BC become parallel.
\n\"AP
\nSolution:
\nFor the lines AD and BC to be parallel x – y = 30\u00b0 (corresponding angles) ……… (1)
\n2x = 5y ………….(2)
\n(\u2235 alternate interior angles)
\nSolving (1) & (2)
\n\"AP
\ny = \\(\\frac{60}{3}\\) = 20\u00b0
\nSubstituting y = 20\u00b0 in eq. (1)
\nx – 20\u00b0 = 30\u00b0
\n\u21d2 x = 50\u00b0
\n\u2234 x = 50\u00b0 and y = 20\u00b0<\/p>\n

\"AP<\/p>\n

Question 6.
\nFind the values of x and y in the figure.
\n\"AP
\nSolution:
\nFrom the figure y + 140\u00b0 = 180\u00b0
\n(\u2235 linear pair of angles)
\n\u2234 y = 180\u00b0 – 140\u00b0 = 40\u00b0
\nAnd x\u00b0 = 30\u00b0 + y\u00b0
\n(\u2235 exterior angle = sum of the opposite interior angles)
\nx\u00b0 = 30\u00b0 + 40\u00b0 = 70\u00b0<\/p>\n

Question 7.
\nIn the given figure segments shown by arrow heads are parallel. Find the values of x and y.
\n\"AP
\nSolution:
\nFrom the figure
\nx\u00b0 = 30\u00b0 (\u2235 alternate interior angles)
\ny\u00b0 = 45\u00b0 + x\u00b0 (\u2235 exterior angles of a triangle = sum of opp. interior angles)
\ny = 45\u00b0 + 30\u00b0 – 75\u00b0<\/p>\n

Question 8.
\nIn the given figure sides QP and RQ of \u2220PQR are produced to points S and T respectively. If \u2220SPR = 135\u00b0 and \u2220PQT = 110\u00b0, find \u2220PRQ.
\n\"AP
\nSolution:
\nGiven that \u2220SPR = 135\u00b0 and \u2220PQT =110\u00b0
\nFrom the figure
\n\u2220SPR + \u2220RPQ = 180\u00b0
\n\u2220PQT + \u2220PQR = 180\u00b0
\n[\u2235 linear pair of angles]
\n\u21d2 \u2220RPQ = 180\u00b0 – \u2220SPR
\n= 180\u00b0 – 135\u00b0 = 45\u00b0
\n\u21d2 \u2220PQR = 180\u00b0 – \u2220PQT
\n= 180\u00b0-110\u00b0 = 70\u00b0
\nNow in APQR
\n\u2220RPQ + \u2220PQR + \u2220PRQ = 180\u00b0
\n[\u2235 angle sum property]
\n\u2234 45\u00b0 + ’70\u00b0 + \u2220PRQ = 180\u00b0
\n\u2234 \u2220PRQ = 180\u00b0-115\u00b0 = 65\u00b0<\/p>\n

\"AP<\/p>\n

Question 9.
\nIn the given figure \u2220X = 62\u00b0 ; \u2220XYZ = 54\u00b0. In \u0394XYZ. If YO and ZO are the bisectors of \u2220XYZ and \u2220XZY respec-tively find \u2220OZY and \u2220YOZ.
\n\"AP
\nSolution:
\nGiven that \u2220X = 62\u00b0 and \u2220Y = 54\u00b0
\nYO arid ZO are bisectors of \u2220Y and \u2220Z.
\nIn \u0394XYZ
\n\u2220X + \u2220XYZ + \u2220XZY = 180\u00b0 .
\n62\u00b0 + 54\u00b0 + \u2220XZY = 180\u00b0
\n=> \u2220XZY = 180\u00b0- 116\u00b0 = 64\u00b0
\nAlso in \u0394\ud835\udf0fOYZ
\n\u2220OYZ = 1\/2 \u2220XYZ = 1\/2 x 54\u00b0 = 27\u00b0
\n(\u2235 YO is bisector of \u2220XYZ)
\n\u2220OZY = 1\/2 \u2220XZY = 1\/2 x 64\u00b0 = 32
\n(\u2235 OZ is bisector of \u2220XYZ)
\nAnd \u2220OYZ + \u2220OZY + \u2220YOZ = 180\u00b0
\n(\u2235 angle sum property, \u0394OYZ)
\n\u21d2 27 + 32\u00b0 + \u2220YOZ = 180\u00b0
\n\u21d2 \u2220YOZ = 180\u00b0 – 59\u00b0 = 121\u00b0<\/p>\n

Question 10.
\nIn the given figure if AB \/\/ DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0, find \u2220DCE.
\n\"AP
\nSolution:
\nGiven that AB \/\/ DE, \u2220CDE = 53\u00b0;
\n\u2220BAC = 35\u00b0
\nNow \u2220E = 35\u00b0
\n( \u2235 alternate interior angles)
\nNow in \u2206CDE
\n\u2220C + \u2220D + \u2220E = 180\u00b0
\n(\u2235angle sum property, ACDE)
\n\u2234 \u2220DCE + 53\u00b0 + 35\u00b0 = 180\u00b0
\n\u21d2 \u2220DCE = 180\u00b0 – 88\u00b0 = 92\u00b0<\/p>\n

\"AP<\/p>\n

Question 11.
\nIn the given figure if line segments PQ and RS intersect at point T, such that\u2220PRT = 40\u00b0, \u2220RPT = 95\u00b0 and \u2220TSQ = 75\u00b0, find \u2220SQT.
\n\"AP
\nSolution:
\nGiven that \u2220PRT = 40\u00b0; \u2220RPT = 95\u00b0;
\n\u2220TSQ = 75\u00b0
\nIn \u2206PRT \u2220P + \u2220R + \u2220PTR = 180\u00b0
\n(\u2235angle sum property)
\n95\u00b0 + 40\u00b0 + \u2220PTR = 180\u00b0
\n\u21d2 \u2220PTR = 180\u00b0 – 135\u00b0 = 45\u00b0
\nNow \u2220PTR = \u2220STQ
\n(\u2235 vertically opposite angles)
\nIn \u0394STQ \u2220S + \u2220Q + \u2220STQ = 180\u00b0
\n(\u2235 angle sum property)
\n75\u00b0 + \u2220SQT + 45\u00b0 = 180\u00b0
\n\u2234 \u2220SQT = 180\u00b0 – 120\u00b0 = 60\u00b0<\/p>\n

Question 12.
\nIn the given figure, ABC is a triangle in which \u2220B = 50\u00b0 and \u2220C = 70\u00b0. Sides AB and AC are produced. If \u2220 is the measure of angle between the bisec\u00actors of the exterior angles so formed, then find ‘z’.
\n\"AP
\nSolution:
\nGiven that \u2220B = 50\u00b0; \u2220C = 70\u00b0
\nAngle between bisectors of exterior angles B and C is \u2220.
\nFrom the figure
\n50\u00b0 + 2x = 180\u00b0
\n70\u00b0 + 2y = 180\u00b0
\n(\u2235 linear pair of angles)
\n\u2234 2x= 180\u00b0-50\u00b0
\n2x= 130\u00b0
\nx = \\(\\frac{130}{2}\\)
\n= 65\u00b0<\/p>\n

2y= 180\u00b0-70\u00b0
\n2y= 110\u00b0
\nx = \\(\\frac{110\u00b0}{2}\\)
\n= 55\u00b0<\/p>\n

Now in \u0394BOC
\nx + y + \u2220 = 180\u00b0 (\u2235 angle sum property)
\n65\u00b0 + 55\u00b0 + \u2220 = 180\u00b0
\nz = 180\u00b0 -120\u00b0 = 60\u00b0<\/p>\n

\"AP<\/p>\n

Question 13.
\nIn the given figure if PQ \u22a5 PS; PQ \/\/ SR, \u2220SQR = 28\u00b0 and \u2220QRT = 65\u00b0, then find the values of x and y.
\n\"AP
\nSolution:
\nGiven that PQ \u22a5 PS ; PQ \/\/ SR
\n\u2220SQR = 28\u00b0, \u2220QRT = 65\u00b0
\nFrom the figure
\n\u2220QSR = x\u00b0 (\u2235 alt. int. angles for the lines PQ \/\/ SR)
\nAlso 65\u00b0 = x + 28\u00b0 (\u2235 ext. angles = sum of the opp. interior angles)
\n\u2234 x\u00b0 = 65\u00b0 – 28\u00b0 = 37\u00b0
\nAnd x\u00b0 + y\u00b0 = 90\u00b0
\n[ \u2235 PQ \u22a5 PS and PQ \/\/ SR. \u21d2 \u2220P = \u2220S]
\n37\u00b0 + y = 90\u00b0
\n\u2234 y = 90\u00b0 – 37\u00b0 = 53\u00b0<\/p>\n

Question 14.
\nIn the given figure \u0394ABC side AC has been produced to D. \u2220BCD = 125\u00b0 and \u2220A: \u2220B = 2:3, find the measure of \u2220A and \u2220B<\/p>\n

\"AP
\nSolution:
\nGiven that \u2220BCD = 125\u00b0
\n\u2220A : \u2220B = 2 : 3
\nSum of the terms of the ratio
\n\u2220A : \u2220B = 2 + 3 = 5
\nWe know that \u2220A + \u2220B = \u2220BCD
\n(\u2235 exterior angles of triangle is equal to sum of its opp. interior angles)
\n\u2234 \u2220A = \\(\\frac{2}{5}\\) x 125\u00b0 = 50\u00b0
\n\u2220B = \\(\\frac{3}{5}\\) x 125\u00b0 = 75\u00b0<\/p>\n

Question 15.
\nIn the given figure, it is given that, BC \/\/ DE, \u2220BAC = 35\u00b0 and \u2220BCE = 102\u00b0. Find the measure of 0 \u2220BCA i0 \u2220ADE and iii) \u2220CED.
\n\"AP
\nSolution:
\nGiven that BC \/\/ DE ; \u2220BAC = 35\u00b0;
\n\u2220BCE = 102\u00b0<\/p>\n

i) From the figure
\n102\u00b0 + \u2220BCA = 180\u00b0
\n(\u2235 linear pair of angles)
\n\u2234 \u2220BCA = 180\u00b0 – 102\u00b0 = 78\u00b0<\/p>\n

\"AP<\/p>\n

ii) \u2220ADE + \u2220CBD = 180\u00b0
\n(\u2235 interior angles on the same side of the transversal)
\n\u2220ADE + (78\u00b0 + 35\u00b0) = 180\u00b0
\n(\u2235 \u2220CBD = \u2220BAC + \u2220BCA)
\n\u2234 \u2220ADE = 180\u00b0 – 113\u00b0 = 67\u00b0<\/p>\n

iii) From the figure .
\n\u2220CED = \u2220BCA = 78\u00b0
\n(\u2235 corresponding angles)<\/p>\n

Question 16.
\nIn the given figure, it is given that AB = AC; \u2220BAC = 36\u00b0; \u2220ADB = 45\u00b0 and \u2220AEC = 40\u00b0. Find i) \u2220ABC
\ni) \u2220ACB iii) \u2220DAB iv) \u2220EAC.
\n\"AP
\nSolution:
\nGiven that AB = AC; \u2220BAC = 36\u00b0,
\n\u2220ADB = 45\u00b0, \u2220AEC = 40\u00b0
\n(i) & (ii)
\nIn \u2206ABC ; AB = AC
\n\u21d2 \u2220ABC = \u2220 ACB
\nAnd 36\u00b0 + \u2220ABC + \u2220ACB = 180\u00b0
\n(\u2235 angle sum property)
\n\u2234 \u2220ABC = \\(\\frac{180^{\\circ}-36^{\\circ}}{2}=\\frac{144^{\\circ}}{2}=72^{\\circ}\\)
\n\u2220ACB = 72\u00b0<\/p>\n

iii) From the figure
\n\u2220ABD + \u2220ABC = 180\u00b0
\n\u2220ABD = 180\u00b0 – 72\u00b0 = 1086
\nIn \u0394ABD
\n\u2220DAB + \u2220ABD + \u2220D = 180\u00b0
\n\u2220DAB + 108\u00b0 + 45\u00b0 = 180\u00b0
\n\u2220DAB = 180\u00b0 – 153\u00b0 = 27\u00b0<\/p>\n

\"AP<\/p>\n

iv) In \u0394ADE
\n\u2220D + \u2220A + \u2220E = 180\u00b0
\n45\u00b0 + \u2220A + 40\u00b0 = 180\u00b0
\n\u21d2 \u2220A = 180\u00b0 -85\u00b0 = 95\u00b0
\nBut \u2220A = \u2220DAB + 36\u00b0 + \u2220EAC
\n95\u00b0 = 27\u00b0, + 36\u00b0 + \u2220EAC
\n\u2234 \u2220EAC = 95\u00b0 – 63\u00b0 = 32\u00b0<\/p>\n

Question 17.
\nUsing information given in the figure, calculate the values of x and y.
\n\"AP
\nSolution:
\nFrom the figure In \u2206ACB
\n34\u00b0 + 62\u00b0 + \u2220ACB = 180\u00b0
\n(\u2235 angle sum property)
\n\u2234 \u2220ACB = 180\u00b0 – 96\u00b0 = 84\u00b0 .
\nAnd x + \u2220ACB = 180\u00b0
\n(\u2235 linear pair of angles) .
\n\u2234 x + 84\u00b0 = 180\u00b0
\nx = 180\u00b0-84\u00b0 = 96\u00b0
\n(OR)
\nx = 34\u00b0 + 62\u00b0 = 96\u00b0
\n( \u2235 x is exterior angle, \u2206ABC)
\ny = 24\u00b0 + x\u00b0
\n= 24\u00b0 + 96\u00b0 = 120\u00b0
\n(\u2235 y is exterior angle, \u2206DCE)<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.4 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.4 Question 1. In the given triangles, find out \u2220x, \u2220y and \u2220z Solution: In fig(i) x\u00b0 = 50\u00b0 + 60\u00b0 (\u2235 exterior … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4226"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=4226"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4226\/revisions"}],"predecessor-version":[{"id":40506,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/4226\/revisions\/40506"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=4226"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=4226"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=4226"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}