The Triangle and its Properties

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The Triangle and its Properties Learn and Remember 1. Triangles. triangle is a closed, three sided figure obtained by three non-collinear points in a plane and symbol used to denote a triangle is. 2.

1 The Triangle and its Properties Learn and Remember 1. Triangles. triangle is a closed, three sided figure obtained by three non-collinear points in a plane and symbol used to denote a triangle is. 2. Three angles and three sides are the si elements ofa triangle. 3. Median of a triangle is a line-segment joining a verte to the mid-point of its opposite side. triangle has three medians. 4. ltitude of a triangle is a perpendicular line-segment from a verte to its opposite side. Types of triangles: 5. ccding to interi angles Types of triangles Property/Definition Diagram cute-angled Each of the angle of B a triangle is less D than 90 i.e., a < 90, b < 90 andc < 90. C Right-angled One of the angle is equal to 90, then it is called as rightangled triangle. Rest two angles are complementary to each other. B C Obtuse-angled One of the angle is obtuse (i.e., greater than 90 ) then it is called as obtuseangled triangle. c

104 MTHEMTiCS-V 6. ccding to the length of sides Scalene triangle triangle in which none ac of the three sides are equal, is called a scalene triangle (all the three angles are also B c different) i.

2 104 MTHEMTiCS-V 6. ccding to the length of sides Scalene triangle triangle in which none ac of the three sides are equal, is called a scalene triangle (all the three angles are also B c different) i.e., a 1= b 1= c. sosceles triangle in which.dc atleasttwo triangle sides are equal is called an isosceles triangle. n this triangle, the angle oppositeto the congruent B=BC sides are also equal. i.e., LB=LC,!fB =C,then LB = LC. Equilateral Triangle triangle in which all the three sides are equal, is called as an equilateral triangle. n this triangle, each angle is congruent and equal to 60. z.e., l \: z s: D = BC = C, then LB = LC = 60..Dc B= BC =C L = LB = Le = When a side of a triangle is produced then an eteri angle is fmed. 8. property of eteri angles: The measure of any eteri angle of a triang12 is equal to the sum of the measure of its interi opposite angles. 9. The angle sum property of a triangle: The total measure of the three angles of a triangle is Sum of any two sides is always greater than the third side. U. The difference ofany two sides is always less than the third side. 12. Greater angle has a greater side opposite to it and smaller angle has a smaller side opposite to it i.e., if two sides of a triangle are not congruent then the angle opposite to the greater side is greater. THE TRNGLE ND TS PROPERTES Let a, band c be the three sides of MBC and c is the largest side then B» if c 2 <a 2 + b 2, the triangle is acute-angled triangle. a b» if c 2 = a 2 + b 2, the triangle is right-angled triangle. < C c» if c 2 > a 2 + b 2, the triangle is obtuse-angled triangle. 14. triangle has atleast two acute angles. E F 15. Pythagas property: n a right-angled triangle, the square of the hypotenuse equals to the sum of the square of its sides. i.e., (Hypotenuser' B C D = (Base)2 + (ltitudej'' 16. Hypotenuse is the longest side in a right-angled triangle. TEXTBOOK QUESTONS SOLVED Eercise 6.1 (Page No. 116) 1. n LU'QR, D is the mid-point of QR. PM is _ PDis _ sqm=mr? p at i \ ',0 Sol. Given, QD = DR. So, PM is altitude. M 0 PD is median. No, QM 1= MR. Q2. Draw rough sketches f the following: (a) n MBe, BE is a median. (b) n LU'QR, PQ and PR are altitudes of the triangle. (c) n 6XYZ, YL is an altitude in the eteri of the triangle.

$(a) Draw BE is a median in MBC, B E=EC \ E "C (b) Draw PQ and PR are the altitudes of the PQR, Llw R (c) YL is an altitude in the eteri of the t,xyz, y Q3. L1 Sol. Q2.$

3 106 MTHEMTiCS-V THE TRNGLE ND TS PROPERTES 107 Sol. (a) Draw BE is a median in MBC, B E=EC \ E "C (b) Draw PQ and PR are the altitudes of the PQR, Llw R (c) YL is an altitude in the eteri of the t,xyz, y Q3. L1 Sol. Q2. (iv) (v) (vi) We know that sum of interi opposite angles = eteri angle From Fig. (i), = From Fig. (ii), = = 120 = 110 From Fig. (iii), = From Fig. (iv), = = 70 = 120 From Fig. (v), = From Fig. (vi), = = 100 = 90. Find the value of the unknown interi angle in the following figures: Sol. X Z L Verify by drawing a diagram if the median altitude of a isosceles triangle can be same. sosceles triangle means any two sides are same. Take MBC and draw the median when B = C. L is the median and altitude of the given triangle. B&C L and (i) (ii) (iii) Eercise 6.2 (Page No ) Ql. Find the value of the unknown eteri angle in the following diagrams: (i) (ii) (iii) Sol. U0 M n s we know that sum of interi opposite angles = Eteri angle From Fig. (i), + 50 = 115 = = 65 From Fig. (ii), 70 + = 100 = =30

108 MTHEMTiCS-V From Fig. (iii), + 90 = 125 From Fig. (iv), 60 + = 120 = 125 _ 90 X = 120-60 From Fig. (u), 30 + = 80 X = 80-30 = 35 =60 From Fig. (vi), + 35 = 75 = 75-35 = 50 = 40. Eercise 6.

4 108 MTHEMTiCS-V From Fig. (iii), + 90 = 125 From Fig. (iv), 60 + = 120 = 125 _ 90 X = From Fig. (u), 30 + = 80 X = = 35 =60 From Fig. (vi), + 35 = 75 = = 50 = 40. Eercise 6.3 (Page No ) Ql. Find the value of the unknown in the following diagrams: Si 1 'e (i) R 30" 170 (iv) (u) (vi) Sol. From Fig. (i), in C, LBC + LCB + LBC = 180 (By angle Rumproperty of a triangle.) J = = 180 = = 70. From Fig. (ii), in PQR, LRPQ + LPQR + LRPQ = = = 180 = = 60. z (ii) (iii) nl:j THE TRNGLE ND TS PROPERTES 109 Q2. From Fig. (iii), in SKYZ, LXYZ + LXYZ + LYZX = = = 180 = = 40. From Fig. (iv), in given isosceles triangle, = = = = X= -- = From Fig. (u), in given equilateral triangle, From Fig.(vi), + + = = 180 = 10 = 600. in given right-angled triangle, = = = = = - 3 = 30. Find the values of the unknowns and y in the following diagrams: (i) (ii) (iii)

110 MTHEMTiCS-V (iv) (V) (vi) Sol. From Fig. (i), 50 + = 120 (Eteri angle property of a triangle.) :::::} = 120-50 = 70.

5 110 MTHEMTiCS-V (iv) (V) (vi) Sol. From Fig. (i), 50 + = 120 (Eteri angle property of a triangle.) :::::} = = 70...(i) Now, y = 180 (By angle sum property of a triangle) y = 180 [From equation (i)] :::::} 120 +y = 180 :::::} y = y = 60 Thus, = 70 andy = 60. From Fig. (ii), y = 80...(i) (Vertically opposite angles) Now, 50 +y + = 180 :::::} = 180 [From equation (i)] :::::} = 180 :::::} = = 50 Thus, = 50 andy = 80. From Fig. (iii), = (Eteri angle property of a triangle.) = 110 Now, y = 180 (By angle sum property of atriangle.) :::::} y = 180 :::::} y = y = 70 Thus, = 110 andy = 70. From Fig. (iv), = 60 (Vertically opposite angles)...(i) Now, 30 + y + = 180 (By angle sum property ofa triangle.) :::::}30 + y + 60 = 180 [From equation (i)] THE TRNGLE ND TS PROPERTES 111 :::::} y + 90 = 180 :::::} Y = y = 90 Thus, = 60 andy = 90. From Fig. (u), y = 90 Now, y + + = 180 :::::} = 180 :::::} 2 = :::::} 2 = = - =45 2 Thus, = 45 andy = 90. From Fig. (vi), =y Now, + + y = 180 :::::} 2 + = 180 :::::} 3 = 180 = 180 = y = 60 Thus, = 60 andy = 60. Eercise 6.4 (Page No ) (Vertically opposite angles)...(i) [From equation (i)] (Vertically opposite angles)...(i) [From equation (i)] [From equation Q1. s it possible to have a triangle with the following sides? m (iii) 6 cm, 3 cm, 2 cm. Sol. Suppose such triangles are possible whose the sum of the lengths of any two sides would be greater than the length of third side. Let us check this: (i) 2 cm, 3 cm, 5 cm s > 5? s > 3? s > 2? But this triangle No, it is equal. is not possible. (ii) 3 cm, 6 cm, 7 cm s > 7? s > 3? s > 6? (i)] Thus, this triangle is possible.

$s (i) OP + OQ > PQ? (ii) OQ + OR > QR? -a (iii) OR + OP > RP? P Q Since sum of two sides is greater than third side. Sol. Join the OR, OQ and OP. (i) s OP + OQ > PQ? \ Yes, POQ fm the triangle. l.<?$

6 112 MTHEMTiCS-V (iii) 6 cm, 3 cm, 2 cm s > 2? s > 3? s > 6? No. t is less than 6. So, this triangle is not possible. Q2. Take any point 0 in the interi R of a triangle PQR. s (i) OP + OQ > PQ? (ii) OQ + OR > QR? -a (iii) OR + OP > RP? P Q Since sum of two sides is greater than third side. Sol. Join the OR, OQ and OP. (i) s OP + OQ > PQ? \ Yes, POQ fm the triangle. l.<? (ii) s OQ + OR> QR? p,----'-- Q (iii) Yes, ROQ fm the triangle. s OR + OP > RP? Yes, ROP fm the triangle. Q3. M is a median of a trianglebc. s B +BC + C> 2M? (Consider the sides of triangles MBM and MMC.) BC 11 : M 11 -c Sol. We know, the sum of length of any two sides ina triangle would be greater than the length of third side. n MBM, B + BM >M (i) n MMC, C + MC > M (ii) dding equations (i) and (ii), Thus, it is true. B+BM+C+MC>M+M B +C + (BM + MC) > 2M B +C +BC>2M THE TRNGLE ND TS PROPERTES Q4. BCD is a quadrilateral. s B + BC + CD + D > C + BD? OC B Sol. We know that sum of length of any two sides in a triangle would be greater than the length of third side. n MBC, B + BC >C and MDC, D + DC >C n DCB, DC + CB > DB and MDB, D + B > DB dding equations (i), tii), (iii) and (iv), B+OO+@+OC+OC++@+B>+)+roB+Dm (B +B) + (BC + BC) + (D + D) + (DC + DC) > 2(C+DB) 2(B + BC +D + DC) > 2(C + DB) B + BC +D + DC > C + DB B + BC + CD + D>C + BD. Q5. BCD is quadrilateral. s B + BC + CD + D < 2(C + BD)? Sol. We draw a quadrilateralbcd andc, DB intersect at point O. 01<.C < VB (i)...(ii)...(iii)...(iv) We know that sum of length of any two sides in a triangle would be greater than the length of third side. n MOB, n OC, n COD, n MOD, dding equations (z),(ii), (iii) B<O+OB BC < OB + OC CD < OC + OD D<OD+O and (iv), B+BC+CD+D<O+OB+OB+OC+OC+OD+OD+O B + BC + CD + D < B + 20C + 20D...(i)...(ii)...(iii)...(iv)

7 114 B + BC + CD + D< 2[(O + OC) + (DO + OB)] Thus, B + BC + CD + D < 2(C + BD). Proved. MTHEMTiCS-V Q6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall? Sol. We know that the sum of two sides of a triangle is always greater than the third side. Given, two sides of triangle are 12 cm and 15 cm. Therefe, the third side is less than = 27 cm. nd the third side cannot be less than the difference of the two sides. Thus, the third side has to be me than = 3 cm. The length ofthe third side could be any length greater than 3 cm but less than 27 cm. Eercise 6.5 (Page No. 130) Q1. PQR is a triangle, right-angled at P. f PQ = 10 cm and PR = 24 cm, find QR. Sol. Given, PQ = 10 cm, PR = 24 cm Let QR be cm. n right-angled QPR, (hypotenuser' = (base)2 + (altitude)2 (By Pythagas property) Q / 10 cm P (QR)2= (PQ)2+ (PR)2 2 = (10)2+ (24)2 2 = = 676 =...)676= 26 R (Given) Thus, the length of QR is 26 cm. Q2. BC is a triangle, right-angled at C. f B = 25 cm and C = 7 cm, find BC. Sol. Given, B = 25 cm and C = 7 cm 24 cm Let BC be cm. n right-angled MCB, (hypotenuself = (baser' + (altitude)2 (By Pythagas property) THE TRNGLE ND TS PROPERTES Now, Thus, the length of BC is 24 cm. 115 (B)2= (C)2+ (BC)2 (25)2= LJ B (7) = '/, = 2 2 = 576 C 7cm =...)576= 24 cm Q3. 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall. 12m Sol. LetC be the ladder andbe the window. 15 m m/'1'2 Given, C = 15 m, B = 12 m, CB = a m L--.d n right-angled C, CaB (hypotenussf = (base)2+ (altitudaf (By Pythagas property) Now, (C)2= (CB)2+ (B)2 (15)2= (a)2 + (12)2 a 2 = (15)2_ (12)2 a 2 = a 2 = 81 a = 181 = 9 Thus, the distance ofthe foot of the ladder from the wall is 9 m. Q4. Which of the following can be the sides of a right triangle? (i) 2.5 cm, 6.5 cm, 6 cm. (iii) 1.5 cm, 2 cm, 2.5 cm. (ii) 2 cm, 2 cm, 5 cm. n the case of right-angled triangles, identify the right angles. Sol. s we consider, the larger side in the given sides of triangles be the hypotenuse.

116 MTHEMTiCS-V nd also using Pythagas property, (hypotenuser' = (base)2 + (altituder' (i) 2.5 cm, 6.5 cm, 6 cm n MBC, (C)2 = (B)2 + (CB)2 6cm Take L.H.S. = (6.5)2 = 42.25 and RH.S. = (6)2 + (2.

8 116 MTHEMTiCS-V nd also using Pythagas property, (hypotenuser' = (base)2 + (altituder' (i) 2.5 cm, 6.5 cm, 6 cm n MBC, (C)2 = (B)2 + (CB)2 6cm Take L.H.S. = (6.5)2 = and RH.S. = (6)2 + (2.5)2 C ( " B 2.5 cm = = L.H.S. = RH.S. Then given sides are the sides of the right-angled triangle. Right angle lies on the opposite side of 6.5 cm in MBC right-angled at B. (ii) 2 cm, 2 cm, 5 cm (5)2=22+22 Take L.H.S. = (5)2 = 25 and RH.S. = = = 8 L.H.S. *- LJ RH.S. These are not the sides of the right-angled triangle. (iii) 1.5 cm, 2 cm, 2.5 cm P n LlPQR, (PR)2 = (PQ)2 + (RQ)2 Take L.H.S. = (2.5)2 = v '1-. cm and RH.S. = (1.5) + (2) = = 6.25 R Q L.H.S. = RH.S. 1.5 cm Then given sides are the sides of a right-angled triangle. Right angle lies on the opposite side of 2.5 cm in PQR right-angled at Q. Q5. tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the iginal height of the tree. Sol. Let'CB represent the tree befe it broken at the point C and let the top ' touch the ground at after it broke. Then, MBC is a right-angled triangle at B such that : ' B = 12 and BC = 5 : Using Pythagas property in MBC, we : have. (C)2 = (B)2 + (BC? LJC = (12)2 + (5)2 5 m = C2 = m B THE TRNGLE ND TS PROPERTES C = -/169 = 13 Hence, the total height of the tree = C + CB = = 18 m. Q6. ngles Q and R of a M>QR are 25 and 65. Write which of the following is true: (i) PQ2 + QR2 = Rp2 (ii) PQ2 + Rp 2 = QR2 (iii) Rp2 + QR2 = PQ2. Q /")25 0 W (\ R Sol. First we find the third angle of the given triangle Now, in PQR, LQPR = 180 PQR (By angle sum property of triangle.) 90 + LQPR = 180 LQPR = LQPR = 90 Now the given triangle is right angled at P So, by the property of Pythagas, (hypotenuser' = (baser' + (altitudej'' (QR)2 = (PR)2 + (QP)2. Thus, (ii) option is true. Q7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm. s, R Sol. Draw a rough figure of the rectangle,// PQRS.>:" G rven, diagona = 41 cm, S de = 40 cm.,/ ", 41 cm Let another side be, p,/ Q 40 cm Now, in right-angled PQR, (PR)2 = (RQ)2 + (PQ)2 (By Pythagas property) (41)2 = X2 + (40)2 2 = (41)2 _ (40)2 2 = = 81 =.J8i = 9 Therefe, the another side of the rectangle is 9 cm. Now, we find the perimeter of the rectangle through given sides. p 117

9 118 MTHEMTiCS-V We know that, Perimeter of the rectangle = 2(length + breadth) = 2(9 + 40) = 2(49) = 98 Thus, the perimeter of the rectangle is 98 cm. Q8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter. Sol. s we know that the diagonals of ",.c rhombus bisect at right angle to each other. From the rough diagram of rhombus. V v Given, C and DB is 30 cm and 16 cm respectively. F the calculation we have, OD= DB 16 - = - =8cm 2 2 C 30 OC = - = = 15 cm n right-angled f1doc, (DCP = (0D)2 + (OC)2 (By Pythagas property) (DC)2 = (8)2 + (15)2 DC 2 = DC 2 =289 DC = 289 DC= 17 Therefe, the side of the rhombus is 17 cm. We know, the perimeter of rhombus = 4 side = 4 17 = 68. Thus, the perimeter of the rhombus is 68 cm. DD

Q1. The sum of the lengths of any two sides of a triangle is always (greater/lesser) than the length of the third side. askiitians

Class: VII Subject: Math s Topic: Properties of triangle No. of Questions: 20 Q1. The sum of the lengths of any two sides of a triangle is always (greater/lesser) than the length of the third side. Greater