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4, , , , \\, , Z, XA, ps, , , , , ele) 1 aS angles, , , , Recall, 1. i. Striangles, AABC, APO. ABP!, , ii, AAP, ACQR, and APOR, 2, i, AABC, ABD. AABP, AA’, ii, ZAPB, ZACB, zaps), iii, AB, BC. AD, AP, BRC), iv, APB, AADE., , i. ABPC, AA, iv, No such triangles exist, , , , ji, ACDP, ABCD, SAPD, ABPC, AADC iii, ACDP, ACD, AADP, , vy, APCD, AADP, BBCP, ABP, , , , Exercise 8.1, 1, Acute-angled triangle ~ ¢, CObtuse-angled triangle —a. d, Right-angled triangle ~b, 2. Scalene triangle a d, Isosceles triangle ¢. €, Equilateral triangle ~ b, , Solution Set, , , , 3, a. Sides ~ PQ, OX, PX and Angles ~ 2P. £0. 2X, b. Sides ~ RS, ST, RT and Angles - 28, 2S. 2T, Scalene triangle - a, 4, Isosceles triangle - ¢, Equilateral triangle - b, 5. Acute-angled triangle ~a, d, Obtuse-angled triangle ~ €, Right-angled triangle ~b, DE and EF, a x4 43 +67 = 180°, =p x= 180-110 = 70°, b. 70-440 + third angle = 180°, = third angle = 180 ~ 110 = 70°, ‘Also third angle +x = 180°, =x= 180-70 = 110°, 4+ $0+30 = 180°, = x= 180~ 80 = 100°, | xt De + 3 = 180°, = Gx = 180 = r= 30”, Here x +50 = 180°, , = x= 180— 50 = 130°, In an isosceles triangle, angles oppos ides are equal., , =9 50-40-40 = 180? @, , = 2x = 180-50 = 130° wo, , ‘We know that in any sve ‘two sides is greater than the third side., a, Here6+4>4, 444 44+6>6, So, it is possible ‘riangle with these sides., , er than 7, So, itis not Passdble to have triangle with these sides., , , , , , , , , , ¢. Here 2+8 is not greater than 11, So, itis not possible to have triangle with these sides., , d. Here 7.1 +5.6 = 12.7, 564 64> 7:1, 6447.1 > 56, So, it is possible to have triangle with theses sides., , Let the vertical angle of an isosceles triangle be x°., , We have each of the two equal angles is 2x*., , Now, sum of the angles of a triangle = 180°, , ord e+ 2e= 180°, , Soi 282 WO 236, , te So, angles of the triangle are x= 36°, 2x= 2(36) = 72°., , es, , , , , , ‘Solution Set

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10. Ina right-angled triangle, we have one acute angle = 47°, , , , Lot other acute angle be =, Now, sum of the angles of a triangle = 180°, 390447 +x = 180, , , , 11, Here 64° + 36° + 80° = 180°, So, they can form the angles of a triangle., , 12. No, as 5 +6 is not greater than 11, , 13, Here ZPTQ = 32°, ZRST = 94° and ZORS = 43°, , , , , , , , In AQRS, Z50R + OR: (exterior angle property), = 250K +43" = 98°, = Z50R=94" 43° = 51°, Also in APOT, 2), IPQ + £PTQ= LSQR Ceucriorangle propery), = £1PQ=51"~32"= 19° <, , 14, We knw that in a twiange sum of two sides > third side, a. InAXIZ,XY +25 XZ wey, In APAY, PX + PY> XY 0) >, 6. In APXZ, PX + PZ> XZ 2) oe, aes o, , Adding equ.(1), (2) and (3), we get \Y, , PX4+ PY s PY + PL + PL-4 PX > XY ¢, , “9 UPR + PY + PD)> V+ YZ, , 15, Wehave to of mesures of nels ¢hfeanl e234, ie, , aan &, , second angle = —*> QP oor, , second ange = —2 ee, , ‘Third angle cee, 16. We have in AATETAP = 8 cm, AE = 7 cm and PE = 6.4.cm, , Now, the median is the line segment joining the vertex to midpoint of opposite side ofa triangle, Also we have in AAPE, AT, PI, EM are medians. So, M, T and I are midpoints of sides AP, PE and, EA respectively, , So, first angl, , , , , , Solution Set, , , , , , 18, We know that in a triangle sum of two sides > third side, So, 4 +9> third side, => third side < 13 os), Also, in a triangle difference of two sides < thied side, So, 9—4 < thd side, = 5 < third side 2), From inequality (1) and (2), we get, 5 < third side < 13, So, among the given options the value between 5 and 13 are 9 em, 8 em and 12 em and so either of, , , , them can be length of third side of the triangle., Mental Maths, aF bF cF aT eT a T, iw ke Le Oy Lt, , Exercise 8.2 oe, , 1. Ina right-angled triangle, (hypotenuse, a. Here t= 68+ 8, , = GFR = (F64EE = /100 =10em S, Here 12848 = 1 S, = Vi 12? = I@= 1 = V5, c, Here x +8? = 10°, , Vi0?=8" = J100=64 = J3@Pp em, Fa 6-1, e. Here x= Vi IGT + JIGS = 95 + 95 =5+5= 100m, , , , sum of squares of two s, , , , =