{"id":28716,"date":"2022-01-30T21:14:13","date_gmt":"2022-01-30T15:44:13","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=28716"},"modified":"2023-12-22T13:15:27","modified_gmt":"2023-12-22T07:45:27","slug":"the-triangles-and-its-properties-class-7-extra-questions-maths-chapter-6","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/maths\/the-triangles-and-its-properties-class-7-extra-questions-maths-chapter-6\/","title":{"rendered":"Class 6 Maths Chapter 7 Extra Questions – The Triangles and its Properties"},"content":{"rendered":"

<\/span>The Triangles and its Properties Class 7 Extra Questions Very Short Answer Type<\/span><\/h2>\n

Question 1. In \u2206ABC, write the following:<\/strong>
\n(a) Angle opposite to side BC.<\/strong>
\n(b) The side opposite to \u2220ABC.<\/strong>
\n(c) Vertex opposite to side AC.<\/strong>
\n\"The<\/p>\n

Solution:
\n(a) In \u2206ABC, Angle opposite to BC is \u2220BAC
\n(b) Side opposite to \u2220ABC is AC
\n(c) Vertex opposite to side AC is B<\/p>\n

\n

Also Check: CBSE Syllabus for Class 1 to 12<\/a><\/strong><\/p>\n<\/header>\n

\n
<\/div>\n<\/div>\n

Question 2. Classify the following triangle on the bases of sides<\/strong>
\n\"The<\/p>\n

Solution:
\n(i) PQ = 5 cm, PR = 6 cm and QR = 7 cm
\nPQ \u2260 PR \u2260 QR
\nThus, \u2206PQR is a scalene triangle.
\n(ii) AB = 4 cm, AC = 4 cm
\nAB = AC
\nThus, \u2206ABC is an isosceles triangle.
\n(iii) MN = 3 cm, ML = 3 cm and NL = 3 cm
\nMN = ML = NL
\nThus, \u2206MNL is an equilateral triangle.<\/p>\n

Question 3.
\nIn the given figure, name the median and the altitude. Here E is the midpoint of BC.
\nSolution
\nIn \u2206ABC, we have
\n\"The<\/p>\n

AD is the altitude.
\nAE is the median.<\/p>\n

Question 4. In the given diagrams, find the value of x in each case.<\/strong>
\n\"The<\/p>\n

Solution:
\n(i) x + 45\u00b0 + 30\u00b0 = 180\u00b0 (Angle sum property of a triangle)
\n\u21d2 x + 75\u00b0 \u2013 180\u00b0
\n\u21d2 x = 180\u00b0 \u2013 75\u00b0
\nx = 105\u00b0
\n(ii) Here, the given triangle is right angled triangle.
\nx + 30\u00b0 = 90\u00b0
\n\u21d2 x = 90\u00b0 \u2013 30\u00b0 = 60\u00b0
\n(iii) x = 60\u00b0 + 65\u00b0 (Exterior angle of a triangle is equal to the sum of interior opposite angles)
\n\u21d2 x = 125\u00b0<\/p>\n

Question 5.Which of the following cannot be the sides of a triangle?<\/strong>
\n(i) 4.5 cm, 3.5 cm, 6.4 cm<\/strong>
\n(ii) 2.5 cm, 3.5 cm, 6.0 cm<\/strong>
\n(iii) 2.5 cm, 4.2 cm, 8 cm<\/strong>
\nSolution:
\n(i) Given sides are, 4.5 cm, 3.5 cm, 6.4 cm
\nSum of any two sides = 4.5 cm + 3.5 cm = 8 cm
\nSince 8 cm > 6.4 cm (Triangle inequality)
\nThe given sides form a triangle.<\/p>\n

(ii) Given sides are 2.5 cm, 3.5 cm, 6.0 cm
\nSum of any two sides = 2.5 cm + 3.5 cm = 6.0 cm
\nSince 6.0 cm = 6.0 cm
\nThe given sides do not form a triangle.<\/p>\n

(iii) 2.5 cm, 4.2 cm, 8 cm
\nSum of any two sides = 2.5 cm + 4.2 cm = 6.7 cm
\nSince 6.7 cm < 8 cm
\nThe given sides do not form a triangle.<\/p>\n

Question 6. In the given figure, find x.<\/strong>
\n\"The
\nSolution:
\nIn \u2206ABC, we have
\n5x \u2013 60\u00b0 + 2x + 40\u00b0 + 3x \u2013 80\u00b0 = 180\u00b0 (Angle sum property of a triangle)
\n\u21d2 5x + 2x + 3x \u2013 60\u00b0 + 40\u00b0 \u2013 80\u00b0 = 180\u00b0
\n\u21d2 10x \u2013 100\u00b0 = 180\u00b0
\n\u21d2 10x = 180\u00b0 + 100\u00b0
\n\u21d2 10x = 280\u00b0
\n\u21d2 x = 28\u00b0
\nThus, x = 28\u00b0<\/p>\n

Question 7. One of the equal angles of an isosceles triangle is 50\u00b0. Find all the angles of this triangle.<\/strong>
\nSolution:
\nLet the third angle be x\u00b0.
\nx + 50\u00b0 + 50\u00b0 = 180\u00b0
\n\u21d2 x\u00b0 + 100\u00b0 = 180\u00b0
\n\u21d2 x\u00b0 = 180\u00b0 \u2013 100\u00b0 = 80\u00b0
\nThus \u2220x = 80\u00b0<\/p>\n

Question 8.
\nIn \u0394ABC, AC = BC and \u2220C = 110\u00b0. Find \u2220A and \u2220B.
\n\"The
\nSolution:
\nIn given \u0394ABC, \u2220C = 110\u00b0
\nLet \u2220A = \u2220B = x\u00b0 (Angle opposite to equal sides of a triangle are equal)
\nx + x + 110\u00b0 = 180\u00b0
\n\u21d2 2x + 110\u00b0 = 180\u00b0
\n\u21d2 2x = 180\u00b0 \u2013 110\u00b0
\n\u21d2 2x = 70\u00b0
\n\u21d2 x = 35\u00b0
\nThus, \u2220A = \u2220B = 35\u00b0<\/p>\n

<\/span>The Triangles and its Properties Class 7 Extra Questions Short Answer Type<\/span><\/h3>\n

Question 9. Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?<\/strong>
\nSolution:
\nLet the length of the third side be x cm.
\nCondition I: Sum of two sides > the third side
\ni.e. 4 + 7 > x \u21d2 11 > x \u21d2 x < 11
\nCondition II: The difference of two sides less than the third side.
\ni.e. 7 \u2013 4 < x \u21d2 3 < x \u21d2 x > 3
\nHence the possible value of x are 3 < x < 11
\ni.e. x < 3 < 11<\/p>\n

Question 10. Find whether the following triplets are Pythagorean or not?<\/strong>
\n(a) (5, 8, 17)<\/strong>
\n(b) (8, 15, 17)<\/strong>
\nSolution: (a) Given triplet: (5, 8, 17)
\n172<\/sup> = 289
\n82<\/sup> = 64
\n52<\/sup> = 25
\n82<\/sup> + 52<\/sup> = 64 + 25 = 89
\nSince 89 \u2260 289
\n52<\/sup> + 82<\/sup> \u2260 172<\/sup>
\nHence (5, 8, 17) is not Pythagorean triplet.<\/p>\n

(b) Given triplet: (8, 15, 17)
\n172<\/sup> = 289
\n152<\/sup> = 225
\n82<\/sup> = 64
\n152<\/sup> + 82<\/sup> = 225 + 64 = 289
\n172<\/sup> = 152<\/sup> + 82<\/sup>
\nHence (8, 15, 17) is a Pythagorean triplet.<\/p>\n

Question 11. In the given right-angled triangle ABC, \u2220B = 90\u00b0. Find the value of x.<\/strong>
\n\"The
\nSolution: In \u0394ABC, \u2220B = 90\u00b0
\nAB2<\/sup> + BC2<\/sup> = AC2<\/sup> (By Pythagoras property)
\n(5)2<\/sup> + (x \u2013 3)2<\/sup> = (x + 2)2<\/sup>
\n\u21d2 25 + x2<\/sup> + 9 \u2013 6x = x2<\/sup> + 4 + 4x
\n\u21d2 -6x \u2013 4x = 4 \u2013 9 \u2013 25
\n\u21d2 -10x = -30
\n\u21d2 x = 3
\nHence, the required value of x = 3<\/p>\n

Question 12. AD is the median of a \u0394ABC, prove that AB + BC + CA > 2AD (HOTS)<\/strong>
\n\"The
\nSolution: In \u0394ABD,
\nAB + BD > AD \u2026(i)
\n(Sum of two sides of a triangle is greater than the third side)
\nSimilarly, In \u0394ADC, we have
\nAC + DC > AD \u2026(ii)
\nAdding (i) and (ii), we have
\nAB + BD + AC + DC > 2AD
\n\u21d2 AB + (BD + DC) + AC > 2AD
\n\u21d2 AB + BC + AC > 2AD
\nHence, proved.<\/p>\n

Question 13. The length of the diagonals of a rhombus is 42 cm and 40 cm. Find the perimeter of the rhombus.<\/strong>
\n\"The
\nSolution: AC and BD are the diagonals of a rhombus ABCD.
\nSince the diagonals of a rhombus bisect at the right angle.
\nAC = 40 cm
\nAO = \\(\\frac { 40 }{ 2 }\\) = 20 cm
\nBD = 42 cm
\nOB = \\(\\frac { 42 }{ 2 }\\) = 21 cm
\nIn right angled triangle AOB, we have
\nAO2<\/sup> + OB2<\/sup> = AB2<\/sup>
\n\u21d2 202<\/sup> + 212<\/sup> = AB2<\/sup>
\n\u21d2 400 + 441 = AB2<\/sup>
\n\u21d2 841 = AB2<\/sup>
\n\u21d2 AB = \u221a841 = 29 cm.
\nPerimeter of the rhombus = 4 \u00d7 side = 4 \u00d7 29 = 116 cm
\nHence, the required perimeter = 116 cm<\/p>\n

Question 14. The sides of a triangle are in the ratio 3 : 4 : 5. State whether the triangle is right-angled or not.<\/strong>
\nSolution:
\nLet the sides of the given triangle are 3x, 4x and 5x units.
\nFor right angled triangle, we have
\nSquare of the longer side = Sum of the square of the other two sides
\n(5x)2<\/sup> = (3x)2<\/sup> + (4x)2<\/sup>
\n\u21d2 25x2<\/sup> = 9x2<\/sup> + 16x2<\/sup>
\n\u21d2 25x2<\/sup> = 25x2<\/sup>
\nHence, the given triangle is a right-angled.<\/p>\n

Question 15. A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.<\/strong>
\nSolution:
\nHere, OA = 320 km
\nAB = 240 km
\nOB = ?
\n\"The
\nClearly, \u2206OBA is right angled triangle
\nOB2<\/sup> = OA2<\/sup> + AB2<\/sup> (By Pythagoras property)
\n\u21d2 OB2<\/sup> = 3202<\/sup> + 2402<\/sup>
\n\u21d2 OB2<\/sup> = 102400 + 57600
\n\u21d2 OB2<\/sup> = 160000
\n\u21d2 OB = \u221a160000 = 400 km.
\nHence the required shortest distance = 400 km.<\/p>\n

<\/span>The Triangles and its Properties Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type<\/span><\/h3>\n

Question 16. In the following figure, find the unknown angles a and b, if l || m.<\/strong>
\n\"The
\nSolution:
\nHere, l || m
\n\u2220c = 110\u00b0 (Corresponding angles)
\n\u2220c + \u2220a = 180\u00b0 (Linear pair)
\n\u21d2 110\u00b0 + \u2220a = 180\u00b0
\n\u21d2 \u2220a = 180\u00b0 \u2013 110\u00b0 = 70\u00b0
\nNow \u2220b = 40\u00b0 + \u2220a (Exterior angle of a triangle)
\n\u21d2 \u2220b = 40\u00b0 + 70\u00b0 = 110\u00b0
\nHence, the values of unknown angles are a = 70\u00b0 and b = 110\u00b0<\/p>\n

Question 17. In figure (i) and (ii), Find the values of a, b and c.<\/strong>
\n\"The
\nSolution:
\n(i) In \u2206ADC, we have
\n\u2220c + 60\u00b0 + 70\u00b0 = 180\u00b0 (Angle sum property)
\n\u21d2 \u2220c + 130\u00b0 = 180\u00b0
\n\u21d2 \u2220c = 180\u00b0 \u2013 130\u00b0 = 50\u00b0
\n\u2220c + \u2220b = 180\u00b0 (Linear pair)
\n\u21d2 50\u00b0 + \u2220b = 180\u00b0
\n\u21d2 \u2220 b = 180\u00b0 \u2013 50\u00b0 = 130\u00b0
\nIn \u2206ABD, we have
\n\u2220a + \u2220b + 30\u00b0 = 180\u00b0 (Angle sum property)
\n\u21d2 \u2220a + \u2220130\u00b0 + 30\u00b0 = 180\u00b0
\n\u21d2 \u2220a + 160\u00b0 = 180\u00b0
\n\u21d2 \u2220a = 180\u00b0 \u2013 160\u00b0 = 20\u00b0
\nHence, the required values are a = 20\u00b0, b = 130\u00b0 and c = 50\u00b0<\/p>\n

(ii) In \u2206PQS, we have
\n\u2220a + 60\u00b0 + 55\u00b0 = 180\u00b0(Angle sum property)
\n\u21d2 \u2220a + 115\u00b0 = 180\u00b0
\n\u21d2 \u2220a = 180\u00b0 \u2013 115\u00b0
\n\u21d2 \u2220a = 65\u00b0
\n\u2220a + \u2220b = 180\u00b0 (Linear pair)
\n\u21d2 65\u00b0 + \u2220b = 180\u00b0
\n\u21d2 \u2220b = 180\u00b0 \u2013 65\u00b0 = 115\u00b0
\nIn \u2206PSR, we have
\n\u2220b + \u2220c + 40\u00b0 = 180\u00b0 (Angle sum property)
\n\u21d2 115\u00b0 + \u2220c + 40\u00b0 = 180\u00b0
\n\u21d2 \u2220c + 155\u00b0 = 180\u00b0
\n\u21d2 \u2220c = 180\u00b0 \u2013 155\u00b0 = 25\u00b0
\nHence, the required angles are a = 65\u00b0, b = 115\u00b0 and c = 25\u00b0<\/p>\n

Question 18. I have three sides. One of my angle measure 15\u00b0. Another has a measure of 60\u00b0. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I? [NCERT Exemplar]<\/strong>
\nSolution:
\nSince I have three sides.
\nIt is a triangle i.e. three-sided polygon.
\nTwo angles are 15\u00b0 and 60\u00b0.
\nThird angle = 180\u00b0 \u2013 (15\u00b0 + 60\u00b0)
\n= 180\u00b0 \u2013 75\u00b0 (Angle sum property)
\n= 105\u00b0
\nwhich is greater than 90\u00b0.
\nHence, it is an obtuse triangle.<\/p>\n

<\/span>Class 6 Maths Chapter 7 Extra Questions FAQs<\/span><\/h2>\n","protected":false},"excerpt":{"rendered":"

The Triangles and its Properties Class 7 Extra Questions Very Short Answer Type Question 1. 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