{"id":9748,"date":"2021-04-09T15:14:02","date_gmt":"2021-04-09T09:44:02","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=9748"},"modified":"2021-09-15T11:34:13","modified_gmt":"2021-09-15T06:04:13","slug":"ap-board-7th-class-maths-solutions-chapter-5-ex-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-5-ex-3\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 3"},"content":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions<\/a> Chapter 5 Triangle and Its Properties Ex 3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 3<\/h2>\n

\"AP<\/p>\n

Question 1.
\nFind the value of the unknown ‘x\u2019 in the following triangles.
\n\"AP
\n\"AP
\n\"AP<\/p>\n

Solution:
\ni) In \u0394ABC
\n\"AP
\n\u2220A + \u2220B + \u2220C = 180\u00b0 (angle sum property
\nx\u00b0 + 50\u00b0 + 60\u00b0 = 180\u00b0
\nx\u00b0 + 110\u00b0 = 180\u00b0
\nx\u00b0 = 180\u00b0 – 110\u00b0<\/p>\n

ii) In \u0394PQR,
\n\"AP
\n\u2220P + \u2220Q + \u2220R = 180\u00b0 (angle – sum property)
\n90\u00b0 + 30\u00b0 + x\u00b0 = 180\u00b0
\n120\u00b0 + x\u00b0 – 180\u00b0
\nx\u00b0 = 180\u00b0 – 120\u00b0
\n\u2234 x\u00b0 = 60\u00b0<\/p>\n

iii) In \u0394XYZ,
\n\"AP
\n\u2220X + \u2220Y + \u2220Z = 180\u00b0 (angle – sum property)
\n30\u00b0+110\u00b0 + x\u00b0 =180\u00b0
\n140\u00b0 + x\u00b0 = 180\u00b0
\nx\u00b0 = 180\u00b0 – 140\u00b0
\n\u2234 x\u00b0 = 40\u00b0<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the values of the unknowns \u2018x \u2018and \u2018y \u2018in the following diagrams.
\nSolution:
\ni) In \u0394PQR,
\n\"AP
\nx\u00b0 + 50\u00b0 = 120\u00b0 (exterior angle property)
\nx\u00b0= 120\u00b0- 50\u00b0
\nx\u00b0= 70\u00b0
\nAlso
\n\u2220P + \u2220Q +\u2220R = 180\u00b0 (angle – sum property)
\n70\u00b0 + 50\u00b0 + y\u00b0 = 180\u00b0
\n120\u00b0 + y\u00b0 = 180\u00b0
\ny\u00b0 = 180\u00b0 – 120\u00b0
\ny\u00b0 = 60\u00b0<\/p>\n

(OR)<\/p>\n

y\u00b0 + 120\u00b0 = 1800 (linear pair of angles)
\ny\u00b0 = 180\u00b0- 120\u00b0
\n\u2234 y\u00b0 = 60\u00b0<\/p>\n

ii) In the figure \u0394RST,
\nx\u00b0 = 80\u00b0 (vertically opposite angles)
\nalso \u2220R + \u2220S + \u2220T = 180\u00b0 (angle – sum property)
\n80\u00b0 + 50\u00b0+ y\u00b0= 180\u00b0
\n130\u00b0 + y\u00b0 = 180\u00b0
\ny\u00b0 = 180\u00b0 – 130\u00b0
\n\u2234 y\u00b0 = 50\u00b0
\n\"AP<\/p>\n

iii) m \u0394MAN,
\nx\u00b0 = \u2220M + \u2220A (exterior angle property)
\nx\u00b0 = 50\u00b0 + 60\u00b0
\nx\u00b0 = 110\u00b0
\nAlso x\u00b0 + y\u00b0 = 180\u00b0
\n110\u00b0+ y\u00b0= 180\u00b0
\ny\u00b0 = 180\u00b0 – 110\u00b0
\ny\u00b0 = 70\u00b0
\n\"AP<\/p>\n

(OR) in \u0394MAN,
\n\u2220M + \u2220A + \u2220N = 180\u00b0 (angle – sum property )
\n50\u00b0 + 60\u00b0 + y\u00b0 = 180\u00b0
\n110\u00b0 + y\u00b0 = 180\u00b0
\ny\u00b0 = 180\u00b0- 110\u00b0
\n\u2234 y\u00b0 = 70\u00b0<\/p>\n

iv) In the figure \u0394ABC,
\nx\u00b0 = 60\u00b0 (vertically opposite angles)
\n\u2220A + \u2220B + \u2220ACB = 180\u00b0 (angle – sum property)
\ny\u00b0 + 30\u00b0 + 60\u00b0 = 180\u00b0
\ny\u00b0 + 90\u00b0 = 180\u00b0
\ny\u00b0 = 180\u00b0 – 90\u00b0
\n\u2234 y\u00b0 = 90\u00b0
\n\"AP<\/p>\n

v) In the figure \u0394EFG,
\ny\u00b0 = 90\u00b0 (vertically opposite angles)
\nAlso in \u0394EFG;
\n\u2220F + \u2220E + \u2220G = 180\u00b0 (angle – sum property)
\n\u2234 x\u00b0 + x\u00b0 + y\u00b0 = 180
\n2x\u00b0 + 90\u00b0 = 180\u00b0
\n2x\u00b0 = 180\u00b0- 90\u00b0
\n2x\u00b0 = 90\u00b0
\nx\u00b0 = \\(\\frac{90^{\\circ}}{2}\\)
\n\u2234 x\u00b0 = 45\u00b0
\n\"AP<\/p>\n

vi) In the figure \u0394LET,
\n\u2220L = \u2220T = \u2220E = x\u00b0 (vertically opposite angles)
\nAlso in \u0394LET
\n\u2220L + \u2220E + \u2220T = 180\u00b0 (angle – sum property)
\nx\u00b0 + x\u00b0 + x\u00b0 = 180\u00b0
\n3x\u00b0 = 180\u00b0
\nx\u00b0 = \\(\\frac{180^{\\circ}}{3}\\)
\nx\u00b0 = 60\u00b0
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 3.
\nFind the measure of the third angle of triangles whose two angles are given below:
\n(i) 38\u00b0 , 102\u00b0
\n(ii) 116\u00b0, 30\u00b0
\n(iii) 40\u00b0, 80\u00b0
\nSolution:
\n(i) 38\u00b0 , 102\u00b0
\nLet the third angle be x\u00b0 then
\n38\u00b0 + 102\u00b0 + x\u00b0 = 180\u00b0 (angle – sum property)
\n140\u00b0 + x\u00b0 = 180\u00b0
\nx\u00b0 = 180\u00b0 – 140\u00b0 = 40\u00b0<\/p>\n

(ii) 116\u00b0, 30\u00b0
\nLet the third angle be x\u00b0
\nthen 116\u00b0 + 300 + x\u00b0 = 180\u00b0 (angle – sum property)
\n146\u00b0 + x = 180\u00b0
\nx= 180\u00b0- 146\u00b0 = 34\u00b0<\/p>\n

(iii) 40\u00b0, 80\u00b0
\nLet the third angle be x\u00b0 then
\n40C + 80\u00b0 + x\u00b0 = 180\u00b0 (angle-sum property)
\n120\u00b0+x\u00b0 = 180\u00b0
\nx\u00b0 = 180 – 120\u00b0 – 60\u00b0.<\/p>\n

Question 4.
\nIn a right-angled triangle, one acute angle is 30\u00b0. Find the other acute angle.
\nSolution:
\nGiven triangle is right angled triangle.
\nLet the third angle be x\u00b0
\nthen 90\u00b0 +30\u00b0 + x\u00b0 = 180\u00b0 (angle – sum property)
\n120\u00b0 + x\u00b0 – 180\u00b0
\nx\u00b0 = 180\u00b0 – 120\u00b0 = 60\u00b0<\/p>\n

\"AP<\/p>\n

Question 5.
\nState true or false for each of the following statements.
\n(i) A triangle can have two right angles.
\n(ii) A triangle can have two acute angles.
\n(iii) A triangle can have two obtuse angles.
\n(iv) Each angle of a triangle can be less than 60\u00b0.
\nSolution:
\n(i) A triangle can have two right angles. – False
\n(ii) A triangle can have two acute angles. – True
\n(iii) A triangle can have two obtuse angles. – False
\n(iv) Each angle of a triangle can be less than 60\u00b0. – False<\/p>\n

Question 6.
\nThe angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.
\nSolution:
\nGiven that ratio of the angle = 1: 2: 3
\nSum of the terms of the ratio – 1 + 2 + 3 = 6
\nSum of the angles of a triangle = 1800
\n\u2234 1st angle = \\(\\frac { 1 }{ 6 }\\) x 180\u00b0 = 30\u00b0 (angle-sum property)
\n\u2234 2nd angle = \\(\\frac { 2 }{ 6 }\\) x 180\u00b0 = 60\u00b0 (angle – sum property)
\n\u2234 3rd angle = \\(\\frac { 3 }{ 6 }\\) x 180\u00b0 = 90\u00b0 (angIe – sum property)<\/p>\n

Question 7.
\nIn the figure, \\(\\overline{\\mathrm{DE}} \\| \\overline{\\mathrm{BC}}\\) , \u2220A = 300 and \u2220B = 50\u00b0. Find the values of x, y and z.
\nSolution:
\n\"AP
\ny\u00b0 = 50\u00b0 (correspondIng angles)
\nx\u00b0 = z\u00b0 (corresponding angles)
\nAlso in MBC;
\n\u2220A + \u2220B +\u2220C = 180\u00b0 (angIe – sum property)
\n30\u00b0 + 50\u00b0 + z\u00b0 = 180\u00b0
\n80\u00b0 + z\u00b0 = 180\u00b0
\nz\u00b0 = 180\u00b0 – 80\u00b0
\n\u2234 z\u00b0 = 100\u00b0
\n\u2234 x\u00b0 = 100\u00b0; y\u00b0 = 50\u00b0; z\u00b0 = 100\u00b0<\/p>\n

\"AP<\/p>\n

Question 8.
\nIn the figure, \u2220ABD = 3 \u2220DAB and \u2220BDC = 96\u00b0. Find \u2220ABD.
\nSolution:
\n\"AP
\nIn the figure
\n\u2220ABD + \u2220DAB = 96\u00b0 (exterior angle property)
\n3\u2220DAB + \u2220DAB = 96\u00b0 (given)
\n4\u2220DAB = 96\u00b0
\n\u2220DAB = \\(\\frac{96^{\\circ}}{4}\\) = 24\u00b0
\n\u2220ABD = 3 x 24\u00b0 = 72\u00b0<\/p>\n

Question 9.
\nIn \u0394PQR \u2220P= 2 \u2220Q and 2 \u2220R =3 \u2220Q , calculate the angles of \u0394PQR.
\nSolution:
\nIn \u0394PQR
\n\u2220P + \u2220Q + \u2220R = 180\u00b0 also
\n\u2220P:\u2220Q:\u2220R = 2\u2220Q:\u2220Q: \\(\\frac { 3 }{ 2 }\\)\u2220Q = 4 : 2 : 3
\nSum of the term of the ratio = 4 + 2 + 3 = 9
\n\u2220P = \\(\\frac { 4 }{ 9 }\\) x 180\u00b0 = 80\u00b0
\n\u2220Q = \\(\\frac { 2 }{ 9 }\\) x 180\u00b0 = 40\u00b0
\n\u2220R = \\(\\frac { 3 }{ 9 }\\) x 180\u00b0 = 60\u00b0<\/p>\n

Question 10.
\nIf the angles of a triangle are in the ratio 1:4: 5, find the angles.
\nSolution:
\nGiven that ratio of the angles = 1 : 4 : 5
\nSum of the terms of the ratio = 1 + 4 + 5 = 10
\nSum of the angles 180\u00b0
\n\u2234 1st angIe = \\(\\frac { 1 }{ 10 }\\) x 180\u00b0 = 18\u00b0
\n2nd angle \\(\\frac { 4 }{ 10 }\\) x 180\u00b0 = 72\u00b0
\n3rd angle = \\(\\frac { 5 }{ 10 }\\) x 180\u00b0= 90\u00b0<\/p>\n

\"AP<\/p>\n

Question 11.
\nThe acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle.
\nSolution:
\nGiven that ratio of acute angles = 2 : 3
\nSum of the terms of the ratio = 2 + 3 = 5
\nSum of the acute angles = 90\u00b0
\n\u2234 1st acute angle x 900 = 36\u00b0
\n2nd acute angle = x 900 = 540
\n\u2234 Angles of the triangle = 36\u00b0, 54\u00b0 and 90\u00b0<\/p>\n

Question 12.
\nIn the figure, \u2206PQR is right angled at Q, \\(\\overline{\\mathrm{ML}} \\| \\overrightarrow{\\mathrm{RQ}}\\) and \u2220LMR = 130\u00b0. Find \u2220LPM, \u2220PML and \u2220PRQ.
\n\"AP
\nSolution:
\n\u2220PRQ + \u2220LMR = 180\u00b0 (int, angles on the same side of the transversal)
\n\u2220PRQ + 130\u00b0 = 18 0\u00b0
\n\u2234 \u2220PRQ = 180\u00b0 – 130\u00b0= 50\u00b0
\nIn \u2206PRQ,
\n\u2220P + \u2220R +\u2220Q = 180\u00b0 (angle – sum property)
\n\u2220LPM + 50\u00b0 + 90\u00b0 = 180\u00b0
\n\u2220LPM + 140\u00b0 = 180\u00b0
\n\u2234 \u2220LPM = 180\u00b0 – 140\u00b0 = 40\u00b0
\nAlso \u2220PML + \u2220LMR = 180\u00b0 (Linear pair of angles)
\n\u2234 \u2220PML + 130\u00b0 = 180\u00b0
\n\u2220PML = 180\u00b0 – 130\u00b0 = 50\u00b0
\n(or) \u2220PML = \u2220PRQ = 50\u00b0 (corresponding angles)<\/p>\n

\"AP<\/p>\n

Question 13.
\nIn Figure ABCDE, find \u22201 +\u22202 + \u22203 + \u22204 + \u22205.
\n\"AP
\nSolution:
\nIn \u2206ABC, \u2220BAC + \u2220ACB + \u2220ABC = 180\u00b0 ………….(1)
\nIn \u2206ACD, \u2220CAD + \u2220ADC + \u2220ACD = 180\u00b0 ……………. (2)
\nIn \u2206ADE, \u2220DAE + \u2220ADE + \u2220AED = 180\u00b0 …………(3)
\nAdding (1), (2) and (3) we get
\n(\u2220BAC + \u2220ACB + \u2220ABC) + \u2220CAD+ (\u2220ADC + \u2220ACD + \u2220DAE + \u2220ADE + \u2220AED = 180\u00b0+ 180\u00b0 180\u00b0
\n(\u2220BAC + \u2220CAD + \u2220DAE) + \u2220ABC + (\u2220ACB + \u2220ACD) + \u2220ADC + \u2220ADE) + \u2220AED = 180\u00b0 + 180\u00b0 + 180\u00b0
\nHene
\n\u22201 + \u22202 + \u22203 + \u22204 + \u22205 = 540\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 3 Textbook Questions and Answers. AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 3 Question 1. Find the value of the unknown ‘x\u2019 in the following triangles. Solution: i) In \u0394ABC \u2220A + …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9748"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=9748"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/9748\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=9748"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=9748"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=9748"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}