This pdf includes:- Triangles and parallel line MCQ Examples Mental Maths Corner
1. MATHEMATICS IN EVERYDAY LIFE–6 Chapter 11 : Triangles and Parallel Lines ANSWER KEYS EXERCISE 11.1 4. (i) 12 cm, 13 cm, 15 cm Perimeter of triangle = sum of the lengths of 1. P three sides = 12 cm + 13 cm + 15 cm = 40 cm (ii) 6 cm, 6 cm, 5 cm Perimeter of triangle = sum of the lengths of three sides Q R = 6 cm + 6 cm + 5 cm (i) The side opposite to Q is PR. = 17 cm (ii) The vertex opposite to the side PQ is R. (iii) 8 cm, 12 cm, 14 cm (iii) The angle opposite to the side QR is P. Perimeter of triangle = sum of the lengths of (iv) The side opposite to the vertex P is QR. three sides 2. No, we cannot draw a triangle with three collinear = 8 cm + 12 cm + 14 cm points. In this case, sum of two sides is equal to the = 34 cm third side. (iv) 6 cm, 5 cm, 10 cm 3. (i) 7, 8, 16 Perimeter of triangle = sum of the lengths of For a triangle, sum of length of any two sides is three sides always greater than the third side, i.e., = 6 cm + 5 cm + 10 cm 7 + 8 < 16 = 21 cm 8 + 16 > 7 5. (i) P 7 + 16 > 8 Since, 7 + 8 < 16 does not satisfy the condition. So, the numbers do not denote the lengths of side of a triangle. (ii) 3, 2, 4 3+2>4 70° x 2+4>3 Q R 3+4>2 70° + x = 180° (Linear pair) Hence, the numbers denote the lengths of side of x = 180° – 70° a triangle. x = 110° (iii) 5, 7, 12 A 5 + 7 = 12 (ii) 5 + 12 > 7 7 + 12 > 5 Since, 5 + 7 = 12 does not satisfy the condition. So, the numbers do not denote the lengths of side of a triangle. x 130° B C Mathematics In Everyday Life-6 1
2. x + 130° = 180° (Linear pair) (ii) In right angled triangle ACB, C = 90°. x = 180° – 130° A x = 50° x 6. P 55° B C Q R A + B + C = 180° (angle sum property) S (i) Three triangles : PSR, PQS, PQR x + 55° + 90° = 180° (ii) Name of seven angles : QPS, SPR, QPR, x + 145° = 180° PQS, PRS, PSQ, PSR x= 180° – 145° (iii) Name of six line segments : PQ, PR, QR, PS, QS, SR x = 35 (iv) PQS and PQR have Q as common. (iii) In PQR, EXERCISE 11.2 P 1. Let the two angles of a triangle be 2x and 3x. ° 25 Third angle = 60° Therefore, x 2x + 3x + 60° = 180° (Angle sum property) 30° Q R 5x + 60° = 180° 5x = 180° – 60° PQR + QRP + RPQ = 180° (Angle sum property) 5x = 120° 30° + x + 25° = 180° 120 x= x + 55° = 180° 5 x = 24° x = 180° – 55° Henc e, other two angles of the triangle are x = 125 2 × 24° = 48° and 3 × 24° = 72°. 2. (i) P A 130° x (iv) 3x 2x B C 2x 120° extA + BAC = 180° (Linear pair) Q R 130° + BAC = 180° extR + PRQ = 180° (Linear pair) BAC = 180° – 130° 120° + PRQ = 180° BAC = 50° PRQ = 180° – 120° = 60° Now, in ABC, Now In PQR, BAC + ACB + CBA = 180° Sum of the interior angles of a triangle is 180°. (Angle sum property) QPR + PRQ + RQP = 180° x + 60° + 2x = 180° 50° + 2x + 3x = 180° 3x + 60° = 180° 50° + 5x = 180° 3x = 180° – 60° 5x = 180° – 50° 3x = 120° 5x = 130° 120 130 x= = 40° x= 3 5 x = 40 x = 26 2 Answer Keys
3. (v) (iii) 48°, 62°, 70° P 110° Sum of the three angles = 48° + 62° + 70° = 180° Hence, a triangle with these given angles is 60° x possible. Q R (iv) 35°, 43°, 130° Sum of the three angles = 35° + 43° + 130° ext P + QPR = 180° (Linear pair) = 208° 180° 110° + QPR = 180° Hence, it is not possible to have a triangle with QPR = 180° – 110° these given angles. QPR = 70° (v) 40°, 50°, 80° In PQR, QPR + PRQ + RQP = 180° Sum of the three angles = 40° + 50° + 80° (Angle sum property) = 170° 180° 70° + x + 60° = 180° Hence, it is not possible to have a triangle with these given angles. x = 180° – 130° (vi) 35°, 95°, 90° x = 50 Sum of the three angles = 35° + 95° + 90° (vi) A = 220° 180° x Hence, it is not possible to have a triangle with these given angles. 150° x B C 4. A ext B + ABC = 180° (Linear pair) 150° + ABC = 180° 3x ABC = 180° – 150° ABC = 30° In ABC, ABC + ACB + CBA = 180° 2x 120° (Angle sum property) B C 30° + x + x = 180° Let the opposite interior angles of a ABC be 2x and 30° + 2x = 180° 3x. 2x = 180° – 30° ext C + ACB = 180° (Linear pair) 2x = 150° 120° + ACB = 180° 150 x = = 75° ACB = 180° – 120° 2 ACB = 60° Hence, x = 75 In ABC, 3. (i) 88°, 42°, 120° BAC + ACB + CBA = 180° Sum of the angles of a triangle is 180°. 3x + 60° + 2x = 180° Sum of the three angles = 88° + 42° + 120° 5x + 60° = 180° = 250° 180° Hence, it is not possible to have a triangle with 5x = 180° – 60° these given angles. 120 (ii) 50°, 50°, 50° x= = 24° 5 Sum of the angles of a triangle is 180°. Hence, Sum of the three angles = 50° + 50° + 50° BAC = 3 24° = 72° = 150° 180° CBA = 2 24° = 48° Hence, it is not possible to have a triangle with these given angles. ACB = 60°. Mathematics In Everyday Life-6 3
4. 5. Let equal angles of a ABC be x. Therefore, EXERCISE 11.3 sum of interior angles of a triangle = 180° A Q A + B + C = 180° 30° + x + x = 180° 30° 30° + 2x = 180° P 2x = 180° – 30° 150 1. C D x= = 75° 2 x x No, the line segments PQ and CD do not intersect. In Hence, each of equal angles = 75°. B C this case both line segments will intersect, when 6. (i) No, the reason a triangle can’t have 2 obtuse produced towards the left. So, they are not parallel angles is because it would add upto more than to each other, since the distance between them is not 180°, which is not possible in a triangle as sum same throughout. of the interior angles of a triangle is 180°. Q S (ii) No, the reason a triangle can’t have 2 right 2. m angles is because the sum of the angles would add upto more than 180°. (iii) No, because all the angles of an equilateral triangle are same and equal to 60° (acute). (iv) No, an equilateral triangle has all the angles 90° 90° equal to 70° which lead to the sum of angles of n triangle is 70° + 70° + 70° = 210° 180°. P R 7. (i) In ABC, AB = AC = 4 cm. So, it is an isosceles Since, m || n, QP n, SR n and PQ = 2.5 cm. triangle. The distance between two parallel lines m and n (ii) In PQR, PQ = 12 cm, QR = 5 cm, RP = 13 cm, remains same throughout. since, all sides are of different lengths. So, it is a So, RS = PQ = 2.5 cm scalene triangle. Hence, RS = 2.5 cm (iii) In MNP, MN = NP = PM = 7.5 cm, all sides are of equal lengths. So, it is an equilateral triangle. 3. (i) Two lines perpendicular to the same line are 8. Let the angles of a triangle be x, 3x and 5x. parallel. (True) Sum of the three angles of a triangle = 180° (ii) Distance between two parallel lines is not same everywhere. (False) x + 3x + 5x = 180° (iii) No parallel line segments intersect. (True) 9x = 180° (iv) If l m, n m and l n, then l || n. (True) 180 x= = 20° (v) Railway tracks are parallel to each other. (True) 9 4. (i) Hence, angles of the triangle are 20°, 60° and 100°. A 9. (i) P D E Q R D F Altitude = AD; Median = PF ( F is mid point of QR) B C (ii) A DE || BC (ii) S R C E B D Altitude = AE; Median = AD P Q ( D is mid point of BC) SR || PQ and SP || RQ 4 Answer Keys
5. (iii) S R MENTAL MATHS CORNER Fill in the blanks: 1. All the sides of a scalene triangle are of different lengths. 2. The angles opposite to equal sides of an isosceles P Q triangle are equal. SR || PQ 3. The region inside the triangle is called its interior. 4. Parallel lines always lie in the same plane. MULTIPLE CHOICE QUESTIONS 5. A triangle can have one right angle. 1. A perpendicular drawn from the vertex to the 6. An equiangular triangle is also called an equilateral opposite side of a triangle is known as an altitude. triangle. Hence, option (a) is correct. 7. A triangle has 3 altitudes and 3 medians. 2. The sum of the angles of a triangle is 180°. 8. A line intersecting two or more given lines at Hence, option (d) is correct. different points is called a transversal. 3. Sum of the angles of a triangle is 180°. Third angle = 180° – (55° + 35°) REVIEW EXERCISE = 180° – 90° = 90° 1. Sides of a triangle are 7 cm, 5.5 cm and 6.5 cm. Third angle = 90° Perimeter of triangle = sum of the lengths of sides Hence, option (b) is correct. Perimeter of triangle = 7 cm + 5.5 cm + 6.5 cm 4. In PQR, P = 40°, Q = 50° = 19 cm P + Q + R = 180° Hence, the perimeter of triangle is 19 cm. (Angle sum property) 2. (i) 8.4, 16.1, 4.5 40° + 50° + R = 180° 8.4 + 16.1 > 4.5 R = 180° – 90° = 90° 8.4 + 4.5 < 16.1 Since, R is 90°. So, PQR is a right angled triangle. Hence, option (c) is correct. 16.1 + 4.5 > 8.4 5. Total number of parts of a triangle is 6. (3 sides, 3 angles) Since, 8.4 + 4.5 < 16.1 Hence, option (a) is correct. So, the given numbers do not denote the sides of 6. A triangle has three altitudes. a triangle. Hence, option (c) is correct. (ii) 7, 16, 8 7. The sum of the lengths of the sides of a triangle is 7 + 16 > 8 known as its perimeter. 7 + 8 < 16 Hence, option (b) is correct. 8 + 16 > 7 8. Line segments joining the vertices to the mid-points Since 7 + 8 < 16, so the given numbers do not of opposite sides of a triangle are known as medians. denote the sides of a triangle. Hence, option (c) is correct. (iii) 7, 5, 8 9. A triangle whose two sides are equal is known as isosceles. 7+5>8 Hence, option (c) is correct. 7+8>5 10. The angle formed 8+5>7 N between North and Sum of the lengths of any two sides is Sourth-East directions is greater than the third side. an obtuse angle. Hence, the given numbers denote the sides of a Henc e, option (a) is triangle. W E correct. S-E S Mathematics In Everyday Life-6 5
6. 3. S R In PQR, PQ = QR = 2.5 cm, PR = 4.2 cm. Since, two sides PQ and QR are of equal length. So, it is an isosceles triangle. 5. (i) A P Q 58° In PQS, SPQ + PQS + QSP = 180° ...(i) In QRS, SQR + QRS + RSQ = 180° ...(ii) Adding equations (i) and (ii), we get 58° 64° B C SPQ + PQS + QSP + SRQ In ABC, A = B = 58°and each angle is less + SQR + RSQ = 360° than 90°. Also, BC = AC. or, LMN is an acute angled or an isosceles SPQ + PQS + SQR + SRQ triangle. + PSQ + QSR = 360° (ii) P 4. (i) A 60° 6 cm 7 cm 60° 60° B C Q R 7 cm In PQR, P = Q = R = 60° or QR = PR = PQ . In ABC, AB = 6 cm, BC = AC = 7 cm. Since, each angles less than 90°. Since, two sides BC and AC are of equal length. PQR is an acute angled or an equilateral So, it is an isosceles triangle. triangle. (ii) P (iii) X m 11 52° c cm 11 Q R 11 cm In PQR, PQ = PR = QR = 11 cm. Since, all the sides are of equal length. 60° 68° Y Z So, it is an equilateral triangle. In XYZ, X = 52°, Y = 60°, Z = 68° all are (iii) P acute angles. Also, XY YZ ZX So, XYZ is an acute angled or scalene triangle. 6. (i) 45°, 80°, 60° 4.2 cm The sum of the angles of a triangle is 180°. 2.5 cm Sum of the given angles = 45° + 80° + 60° = 185° 180°. R So, it is not possible to have a triangle with these Q 2.5 cm angles. 6 Answer Keys
7. (ii) 30°, 50°, 90° 9. l || m, QP l and SR l. The sum of the given angles = 30° + 50° + 90° Q S = 170° 180° m So, it is not possible to have a triangle with these angles. 3.5 cm (iii) 52°, 52°, 76° The sum of the given angles = 52° + 52° + 76° = 180° l P R Hence, it is possible to have a triangle with these The distance between two parallel lines always given angles. remains same throughout. 7. In XYZ, XY = YZ = ZX = 5 cm SR = QP = 3.5 cm It is an equilateral triangle. Hence, SR = 3.5 cm. X = Y = Z = x(let) 10. C In XYZ, X + Y + Z = 180° x + x + x = 180° E 3x = 180° 180 x= = 60° A 3 B F X = Y = Z = 60°. 8. P D 60° Hence, AD, BE and CF are three altitudes of ABC. HOTS QUESTIONS 1. If the three sides of a triangle are equal, it is necessary that if any two of the sides are chosen, 50° Q R they are also of the same length. Thus equilateral S triangles are isosceles. In PQR, This does not imply, that all isosceles triangles have to be equilateral. PQR + QRP + RPQ = 180° (Angle sum property) 2. (a) Line segment is a part of the line that has two PQS + 50° + 60° = 180° ( PQR = PQS) end points. PQS + 110° = 180° (b) Line is a straight path that goes on indefinitely PQS = 180° – 110° in two opposite directions. (c) Ray is a straight path that goes on indefinitely in PQS = 70° one direction. In RPS, (d) AB means, a ray AB. RPS + PSR + SRP = 180° (e) Parallel lines are lines that are same distance RPS + 90° + 50° = 180° ( SRP = QRP) apart and never meet. RPS + 140° = 180° Hence, (a) (iv), (b) (iii), (c) (i), (d) (ii), (e) (v) RPS = 180° – 140° = 40° Hence, PQS= 70° and RPS= 40°. Mathematics In Everyday Life-6 7
8. First of all, count all the triangles whose vertices are vertices of pentagon. There are exactly 10 of these, one for every different sets of three vertices. ACD, BDE, CEA, DAB, EBC, BCD, ABC, CDE, DEA, A EAB. Now, count triangles using exactly two adjacent vertices of the pentagon and one inside the pentagon. Each pair of adjacent vertices has three such triangles. So, there are 15 more. F G E B JCD, ICD, HCD, IBC, HBC, GBC, HAB, GAB, FAB, GEA, FEA, JEA, FDE, JDE, IDE. J H Now, count triangle which are formed using two non-adjacent vertices I of the pentagon and on inside vertex. There is one triangle for each such pair, and there are 5 pairs. So we get 5 more triangles. D C FBD, GCE, HDA, IEB, JAC Now, count the triangles using one pentagon vertex and two inside vertices. There is only one for each vertex. So, 5 more. DJI, EFJ, AGF, BHG, CIH There are no triangle using only inside vertices. Total number of triangles = 10 + 15 + 5 + 5 = 35. 8 Answer Keys
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