{"id":8460,"date":"2024-09-18T10:15:58","date_gmt":"2024-09-18T04:45:58","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=8460"},"modified":"2024-09-19T10:00:26","modified_gmt":"2024-09-19T04:30:26","slug":"samacheer-kalvi-8th-maths-guide-chapter-5-ex-5-2","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-8th-maths-guide-chapter-5-ex-5-2\/","title":{"rendered":"Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.2"},"content":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide<\/a> Pdf Chapter 5 Geometry Ex 5.2 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.2<\/h2>\n

Question 1.
\nFill in the blanks:
\n(i) If in a \u2206 PQR, PR2<\/sup> = PQ2<\/sup> + QR2<\/sup>, then the right angle of \u2206 PQR is at the vertex _______ .
\nAnswer:
\nQ
\nHint:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

(ii) If \u2018l\u2018 and ‘m\u2019 are the legs and \u2018n\u2019 is the hypotenuse of a right angled triangle then, l2<\/sup> = _______ .
\nAnswer:
\nn2<\/sup> – m2<\/sup>
\nHint:
\n\"Samacheer<\/p>\n

(iii) If the sides of a triangle are in the ratio 5:12:13 then, it is _______ .
\nAnswer:
\na right angled triangle
\nHint:
\n132<\/sup> = 169
\n52<\/sup> = 25
\n122<\/sup> = 144
\n169 = 25 + 144
\n\u2234 132<\/sup> = 52<\/sup> + 122<\/sup><\/p>\n

(iv) The medians of a triangle cross each other at _______ .
\nAnswer:
\nCentroid<\/p>\n

(v) The centroid of a triangle divides each medians in the ratio _______ .
\nAnswer:
\n2 : 1<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nSay True or False.
\n(i) 8, 15, 17 is a Pythagorean triplet.
\nAnswer:
\nTrue
\nHint:
\n172<\/sup> = 289
\n152<\/sup> = 225
\n82<\/sup> = 64
\n64 + 225 = 289 \u21d2 172<\/sup> = 152<\/sup> + 82<\/sup><\/p>\n

(ii) In a right angled triangle, the hypotenuse is the greatest side.
\nAnswer:
\nTrue
\nHint:
\n\"Samacheer<\/p>\n

(iii) In any triangle the centroid and the incentre are located inside the triangle.
\nAnswer:
\nTrue<\/p>\n

\"Samacheer<\/p>\n

(iv) The centroid, orthocentre, and incentre of a triangle are collinear.
\nAnswer:
\nTrue<\/p>\n

(v) The incentre is equidistant from all the vertices of a triangle.
\nAnswer:
\nFalse<\/p>\n

Question 3.
\nCheck whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
\n(i) 8, 15, 17
\nAnswer:
\nTake a = 8 b = 15 and c = 17
\nNow a2<\/sup> + b2<\/sup> = 82<\/sup> + 152<\/sup> = 64 + 225 = 289
\n172 = 289 = c2<\/sup>
\n\u2234 a2<\/sup> + b2<\/sup> = c2<\/sup>
\nBy the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
\n\u2234 Ans: yes<\/p>\n

\"Samacheer<\/p>\n

(ii) 12, 13, 15
\nAnswer:
\n(ii) 12, 13. 15
\nTake a = 12,b = 13 and c = 15
\nNow a2<\/sup> + b2<\/sup> = 122<\/sup> + 132<\/sup> = 144 + 169 = 313
\n152<\/sup> = 225 \u2260 313
\nBy the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
\n\u2234 Ans: No.<\/p>\n

(iii) 30, 40, 50
\nAnswer:
\nTake a = 30, b = 40 and c = 50
\nNow a2<\/sup> + b2<\/sup> = 302<\/sup> + 402<\/sup> = 900 + 1600 = 2500
\nC2<\/sup> = 502<\/sup> = 2500
\n\u2234 a2<\/sup> + b2<\/sup> = c2<\/sup>
\nBy the converse of Pythagoras theorem, the triangle with given measures is a right
\nangled triangle.
\n\u2234 Ans: yes<\/p>\n

(iv) 9, 40, 41
\nAnswer:
\nTake a = 9, b = 40 and c = 41
\nNow a2<\/sup> + b2<\/sup> = 92<\/sup> + 402<\/sup> = 81 + 1600 = 1681
\nc2<\/sup> = 412<\/sup> = 1681
\n\u2234 a2<\/sup> + b2<\/sup> = c2<\/sup>
\nBy the converse of Pythagoras theorem, the triangle with given measures is a right
\nangled triangle.
\n\u2234 Ans: yes<\/p>\n

\"Samacheer<\/p>\n

(v) 24, 45, 51
\nAnswer:
\nTake a = 24,b = 45 and c = 51
\nNow a2<\/sup> + b2<\/sup> = 242<\/sup> + 452<\/sup> = 576 + 2025 = 2601
\nc2<\/sup> = 512<\/sup> = 2601
\n\u2234 a2<\/sup> + b2<\/sup> = c2<\/sup>
\nBy the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
\n\u2234 Ans: yes<\/p>\n

Question 4.
\nFind the unknown side in the following triangles.
\n(i)
\n\"Samacheer
\nAnswer:
\nFrom \u2206 ABC, by Pythagoras theorem
\nBC2<\/sup> = AB2<\/sup> + AC2<\/sup>
\nTake AB2<\/sup> + AC2<\/sup> = 92<\/sup> + 402<\/sup> = 81 + 1600 = 1681
\nBC2<\/sup> = AB2<\/sup> + AC2<\/sup> = 1681 = 412<\/sup>
\nBC2<\/sup> = 412<\/sup> \u21d2 BC = 41
\n\u2234 x = 41<\/p>\n

\"Samacheer<\/p>\n

(ii)
\n\"Samacheer
\nAnswer:
\nFrom \u2206 PQR, by Pythagoras theorem.
\nPR2<\/sup> = PQ2<\/sup> + QR2<\/sup>
\n342<\/sup> = y2<\/sup> + 302<\/sup>
\n\u21d2 y2<\/sup> = 342<\/sup> – 302<\/sup>
\n= 1156 – 900
\n= 256 = 162<\/sup>
\ny2<\/sup> = 162<\/sup> \u21d2 y = 16<\/p>\n

(iii)
\n\"Samacheer
\nAnswer:
\nFrom \u2206 XYZ, by Pythagoras theorem,
\n= YZ2<\/sup> = XY2<\/sup> + XZ2<\/sup>
\n\u21d2 XY2<\/sup> = YZ2<\/sup> – XZ2<\/sup>
\nZ2<\/sup> = 392<\/sup> – 362<\/sup>
\n= 1521 – 1296
\n= 225 = 152<\/sup>
\nz2<\/sup> = 152<\/sup>
\n\u21d2 z = 15<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nAn isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.
\nAnswer:
\n\"Samacheer
\nIn an isosceles triangle the altitude dives its base into two equal parts.
\nNow in the figure, \u2206ABC is an isosceles triangle with AD as its height
\nIn the figure, AD is the altitude and \u2206ABD is a right triangle.
\nBy Pythagoras theorem,
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n\u21d2 AD2<\/sup> = AB2<\/sup> – BD2<\/sup>
\n= 132<\/sup> – 122<\/sup> = 169 – 144 = 25
\nAD2<\/sup> = 25 = 52<\/sup>
\nHeight: AD = 5cm<\/p>\n

Question 6.
\nFind the distance between the helicopter and the ship.
\n\"Samacheer
\nAnswer:
\nFrom the figure AS is the distance between the helicopter and the ship.
\n\u2206 APS is a right angled triangle, by Pythagoras theorem,
\nAS2<\/sup> = AP2<\/sup> + PS2<\/sup>
\n= 802<\/sup> + 1502<\/sup> = 6400 + 22500 = 28900 = 1702<\/sup>
\n\u2234 The distance between the helicopter and the ship is 170 m<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nIn triangle ABC, line I, is a perpendicular bisector of BC.
\nIf BC = 12cm, SM = 8cm, find CS.
\n\"Samacheer
\nAnswer:
\nGiven l1<\/sub>, is the perpendicular bisector of BC.
\n\u2234 \u2220SMC = 90\u00b0and BM = MC
\nBC = 12cm
\n\u21d2 BM + MC = 12cm
\nMC + MC = 12cm
\n2MC = 12
\nMC = \\(\\frac{12}{2}\\)
\nMC = 6cm
\nGiven SM = 8 cm
\nBy Pythagoras theorem SC2<\/sup> = SM2<\/sup> + MC2<\/sup>
\nSC2<\/sup> = 82<\/sup> + 62<\/sup>
\nSC2<\/sup> = 64 + 36
\nCS2<\/sup> = 100
\nCS2<\/sup> = 102<\/sup>
\nCS = 10 cm<\/p>\n

Question 8.
\nIdentify the centroid of \u2206PQR.
\n\"Samacheer
\nAnswer:
\nIn \u2206PQR, 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 \u2206PQR.<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nName the orthocentre of \u2206PQR.
\n\"Samacheer
\nAnswer:
\nThis is a right triangle
\n\u2234 orthocentre = P [\u2235 In right triangle orthocentre is the vertex containing 90\u00b0]<\/p>\n

Question 10.
\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
\nAnswer:
\nGiven A is the midpoint of YZ.
\n\u2234 ZA = AY
\nG is the centroid of XYZ centroid divides each median in a ratio 2 : 1 \u21d2 XG : GA = 2:1
\n\\(\\frac{\\mathrm{XG}}{\\mathrm{GA}}=\\frac{2}{1}\\)
\n\\(\\frac{\\mathrm{XG}}{3}=\\frac{2}{1}\\)
\nXG = 2 \u00d7 3
\nXG = 6 cm
\nXA = XG + GA = 6 + 3 \u21d2 XA = 9cm<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nIf I is the incentre of \u2206XYZ, \u2220IYZ = 30\u00b0 and \u2220IZY = 40\u00b0, find \u2220YXZ.
\n\"Samacheer
\nAnswer:
\nSince I is the incentre of AXYZ
\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
\n\"Samacheer
\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

Objective Type Questions<\/p>\n

Question 12.
\nIf \u2206GUT is isosceles and right angled, then \u2220TUG is ______ .
\n\"Samacheer
\n(A) 30\u00b0
\n(B) 40\u00b0
\n(C) 45\u00b0
\n(D) 55\u00b0
\nAnswer:
\n(C) 45\u00b0
\nHint:
\n\u2220U \u2220T = 45\u00b0 (\u2235 GUT is an isosceles given)
\n\u2234 \u2220TUG = 45\u00b0<\/p>\n

Question 13.
\nThe hypotenuse of a right angled triangle of sides 12cm and 16cm is ______ .
\n(A) 28 cm
\n(B) 20 cm
\n(C) 24 cm
\n(D) 21 cm
\nAnswer:
\n(B) 20 cm
\nHint:
\nSide take a = 12 cm
\nb = 16cm
\nThe hypotenuse c2<\/sup> = a2<\/sup> + b2<\/sup>
\n= 122<\/sup> + 162<\/sup>
\n= 144 + 256
\nc2<\/sup> = 400 \u21d2 c = 20 cm<\/p>\n

\"Samacheer<\/p>\n

Question 14.
\nThe area of a rectangle of length 21cm and diagonal 29 cm is ______ .
\n(A) 609 cm2<\/sup>
\n(B) 580 cm2<\/sup>
\n(C) 420 cm2<\/sup>
\n(D) 210 cm2<\/sup>
\nAnswer:
\n(C) 420 cm2<\/sup>
\n\"Samacheer
\nlength = 21 cm
\ndiagonal = 29 cm
\nBy the converse of Pythagoras theorem,
\nAB2<\/sup> + BC2<\/sup> = AC2<\/sup>
\n212<\/sup> + x2<\/sup> = 292<\/sup>
\nx2<\/sup> = 841 – 441 400 = 202<\/sup>
\nx = 20 cm
\nNow area of the rectangle = length \u00d7 breadth.
\nie AB \u00d7 BC = 21 cm \u00d7 20 cm = 420 cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 15.
\nThe sides of a right angled triangle are in the ratio 5:12:13 and its perimeter is 120 units then, the sides are .
\n(A) 25, 36, 59
\n(B) 10, 24, 26
\n(C) 36, 39, 45
\n(D) 20, 48, 52
\nAnswer:
\n(D) 20,48,52
\nHint:
\nThe sides of a right angled triangle are in the ratio 5 : 12 : 13
\nTake the three sides as 5a, 12a, 13a
\nIts perimeter is 5a + 12a + 13a = 30a
\nIt is given that 30a = 120 units
\na = 4 units
\n\u2234 the sides 5a = 5 \u00d7 4 = 20 units
\n12a = 12 \u00d7 4 = 48 units
\n13a = 13 \u00d7 4 = 52 units<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.2 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.2 Question 1. Fill in the blanks: (i) If in a \u2206 PQR, PR2 = PQ2 + QR2, then the right angle …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[10],"tags":[],"class_list":["post-8460","post","type-post","status-publish","format-standard","hentry","category-class-8"],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8460"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=8460"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8460\/revisions"}],"predecessor-version":[{"id":40866,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/8460\/revisions\/40866"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=8460"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=8460"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=8460"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}