{"id":8401,"date":"2024-01-04T10:19:35","date_gmt":"2024-01-04T04:49:35","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=8401"},"modified":"2024-01-05T17:36:08","modified_gmt":"2024-01-05T12:06:08","slug":"samacheer-kalvi-8th-maths-solutions-term-3-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-8th-maths-solutions-term-3-chapter-3-ex-3-2\/","title":{"rendered":"Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2"},"content":{"rendered":"

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2<\/h2>\n

Miscellaneous and Practice Problems<\/span><\/p>\n

Question 1.
\nIdentify the centroid of \u0394PQR.
\n\"Samacheer
\nSolution:
\nIn \u0394PQR, PT = TR \u21d2 QT is a median from vertex Q.
\nQS = SR \u21d2 PS is a median from vertex P.
\nQT and PS meet at W and therefore W is the centroid of \u0394PQR.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nName the orthocentre of \u0394PQR.
\n\"Samacheer
\nSolution:
\n\u2220P = 90\u00b0
\nThis is a right triangle
\n\u2234 orthocentre = P [\u2234 In right triangle orthocentre is the vertex containing 90\u00b0]<\/p>\n

Question 3.
\nIn the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.
\n\"Samacheer
\nSolution:
\nGiven A is the midpoint of YZ.
\n\u2234 ZA = AY
\nG is the centroid of \u0394XYZ centroid divides each median in a ratio 2 : 1 \u21d2 XG : GA = 2 : 1
\n\\(\\frac{XG}{GA}\\) = \\(\\frac{2}{1}\\)
\n\\(\\frac{XG}{3}\\) = \\(\\frac{2}{1}\\)
\nXG = 2 \u00d7 3
\nXG = 6 cm
\nXA = XG + GA
\n= 6 + 3 \u21d2 XA = 9 cm<\/p>\n

Challenging Problems<\/span><\/p>\n

Question 4.
\nFind the length of an altitude on the hypotenuse of a right angled triangle of legs of length 15 feet and 20 feet.
\nSolution:
\n\"Samacheer
\nSince \u2220B = 90\u00b0
\nUsing pythagoras theorem
\nAC\u00b2 = AB\u00b2 + BC\u00b2 = 20\u00b2 + 15\u00b2
\n= 400 + 225
\nAC\u00b2 = 625<\/sup>
\nAC\u00b2 = 255
\n<\/sup>AC = 25
\nArea of a triangle \u0394ABC = \\(\\frac{1}{2}\\) \u00d7 15 \u00d7 20 = 150 feet\u00b2
\nAgain Area of \u0394ABC = \\(\\frac{1}{2}\\) \u00d7 AC \u00d7 BD
\n150 = \\(\\frac{1}{2}\\) \u00d7 25 \u00d7 BD
\nBD = \\(\\frac{2 \u00d7 150}{25}\\) = \\(\\frac{300}{25}\\)
\nBD = 12 feet
\n\u2234 Length of the altitude on the hypotenuse of the right angled triangle is 12 feet.<\/p>\n

Question 5.
\nIf I is the incentre of \u0394XYZ, \u2220IYZ = 30\u00b0 and \u2220IZY = 40\u00b0, find \u2220YXZ.
\n\"Samacheer
\nSolution:
\nSince I is the incentre of \u0394XYZ
\n\u2220IYZ = 30\u00b0 \u21d2 \u2220IYX = 30\u00b0
\n\u2220IZY = 40\u00b0 \u21d2 \u2220IZX = 40\u00b0
\n\u2234 \u2220XYZ = \u2220XYI + \u2220IYZ = 30\u00b0 + 30\u00b0
\n\u2220XYZ = 60\u00b0
\nlll ly \u2220XYZ = \u2220XZI + \u2220IZY = 40\u00b0 + 40\u00b0
\n\u2220XYZ = 80\u00b0
\nBy angle sum property of a triangle
\n\u2220XZY + \u2220XYZ + \u2220YXZ = 180\u00b0
\n80\u00b0 + 60\u00b0 + \u2220YXZ = 180\u00b0
\n140\u00b0 + \u2220YXZ = 180\u00b0
\n\u2220YXZ = 180\u00b0 – 140\u00b0
\n\u2220YXZ = 40\u00b0<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nIn \u0394DEF, DN, EO, FM are medians and point P is the centroid. Find the following.
\n(i) If DE = 44, then DM = ?
\n(ii) If PD = 12, then PN = ?
\n(iii) If DO = 8, then FD = ?
\n(iv) If OE = 36, then EP = ?
\n\"Samacheer
\nSolution:
\nGiven DN, EO, FM are medians.
\n\u2234 FN = EN
\nDO = FO
\nEM = DM
\n(i) If DE = 44, then
\nDM = \\(\\frac{44}{2}\\) = 22
\nDM = 22<\/p>\n

(ii) If PD = 12, PN = ?
\n\\(\\frac{PD}{PN}\\) = \\(\\frac{2}{1}\\)
\n\\(\\frac{12 }{PN}\\) = \\(\\frac{2}{1}\\) \u21d2 PN = \\(\\frac{12}{2}\\) = 6
\nPN = 6<\/p>\n

(iii) If DO = 8, then
\nFD = DO + OF = 8 + 8
\nFD = 16<\/p>\n

(iv) If OE = 36,
\nthen \\(\\frac{EP}{PO}\\) = \\(\\frac{2}{1}\\)
\n\\(\\frac{EP}{2}\\) = PO
\nOE = OP + PE
\n36 = \\(\\frac{PE}{2}\\) + PE
\n36 = \\(\\frac{PE}{2}\\) + \\(\\frac{2PE}{2}\\)
\n36 = \\(\\frac{3PE}{2}\\)
\nPE = \\(\\frac{36 \u00d7 2}{3}\\)
\nPE = 24<\/p>\n

\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Geometry Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2 Miscellaneous and …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[4],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8401"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=8401"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8401\/revisions"}],"predecessor-version":[{"id":60290,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/8401\/revisions\/60290"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=8401"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=8401"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=8401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}