{"id":27663,"date":"2021-07-01T14:25:45","date_gmt":"2021-07-01T08:55:45","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27663"},"modified":"2021-07-01T14:25:45","modified_gmt":"2021-07-01T08:55:45","slug":"ncert-solutions-for-class-7-maths-chapter-6-intext-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-7-maths-chapter-6-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 6 The Triangles and Its Properties InText Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions<\/h2>\n

NCERT In-text Question Page No. 113 & 114<\/span>
\nQuestion 1.
\nWrite the six elements (i.e. the 3 sides and the 3 angles) of \u0394ABC.
\nAnswer:
\nSix elements of \u0394ABC are: \u2220A, \u2220B, \u2220C, \\(\\overline{\\mathrm{AB}}\\), \\(\\overline{\\mathrm{BC}}\\), and \\(\\overline{\\mathrm{CA}}\\).<\/p>\n

Question 2.
\nWrite the:
\n(i) Side opposite to the vertex Q of \u0394PQR
\n(ii) Angle opposite to the side LM of \u0394LMN
\n(iii) Vertex opposite to the side RT of \u0394RST
\nAnswer:
\n(i) The side opposite to the vertex Q is \\(\\overline{\\mathrm{PR}}\\).
\n\"NCERT<\/p>\n

(ii) The angle opposite to the side LM is \u2220N.
\n\"NCERT<\/p>\n

(iii) Vertex opposite to the side RT is \u2018S\u2019.
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nLook at figures and classify each of the triangles according to its.
\n(a) Sides
\n(b) Angles
\n\"NCERT
\n\"NCERT<\/p>\n

Answer:
\n(i) (a) Since \\(\\overline{\\mathrm{AC}}=\\overline{\\mathrm{BC}}\\) = 8 cm
\n\u2234 \u0394ABC is an isosceles triangle,<\/p>\n

(b) Since all angles of \u2234ABC are less than 90\u00b0.
\n\u2234 It is an acute triangle.<\/p>\n

(ii) (a) Since PQ \u2260 QR \u2260 RP
\n\u2234 \u0394PQR is a scalene triangle.<\/p>\n

(b) Since \u2220R = 90\u00b0
\n\u2234 \u0394PRQ is a right triangle.<\/p>\n

(iii) (a) In \u0394LMN, LN = MN = 7 cm
\n\u2234 \u0394LMN is an isosceles triangle,<\/p>\n

(b) In \u0394LMN, \u2220N > 90\u00b0
\n\u2234 \u0394LMN is an obtuse triangle.<\/p>\n

(iv) (a) In \u0394RST, RS = ST = TR = 5.2 cm
\n\u2234 It is an equilateral triangle.<\/p>\n

(b) All the angles of \u0394RST are acute.
\n\u2234 It is an acute triangle.<\/p>\n

(v) (a) In \u0394ABC, \\(\\overline{\\mathrm{AB}}=\\overline{\\mathrm{BC}}\\) = 3 cm
\n\u2234 It is an isosceles triangle.<\/p>\n

(b) In \u0394ABC, \u2220B > 90\u00b0
\n\u2234 It is an obtuse triangle.<\/p>\n

(vi) (a) In \u0394PQR, latex]\\overline{\\mathrm{PQ}}=\\overline{\\mathrm{QR}}[\/latex] = 6 cm
\n\u2234 It is an isosceles triangle.<\/p>\n

(b) In \u0394PQR, \u2220Q = 90\u00b0
\n\u2234 It is a right triangle.<\/p>\n

\"NCERT<\/p>\n

NCERT In-text Question Page No. 118<\/span>
\nQuestion 1.
\nAn exterior angle of a triangle is of measure 70\u00b0 and one of its interior opposite angles is for measure 25\u00b0. Find the measure of the other interior opposite angle.
\nAnswer:
\nExterior angle = 70\u00b0
\nInterior opposite angles are 25\u00b0 and x.
\n\u2234 x + 25\u00b0 = 70\u00b0
\n[Using the exterior angle property of a triangle]
\nor x = 70\u00b0- 25\u00b0= 45\u00b0
\n\u2234 The required interior opposite angle = 45\u00b0.<\/p>\n

Question 2.
\nThe two interior opposite angles of an exterior angle of triangle are 60\u00b0 and 80\u00b0. Find the measure of the exterior angle.
\nAnswer:
\nInterior angles are 60\u00b0 and 80\u00b0.
\n\u2235 [Exterior angle] = 60\u00b0 + 80\u00b0 = 140\u00b0<\/p>\n

Question 3.
\nIs something wrong in this diagram?
\n\"NCERT
\nAnswer:
\nWe know that an exterior angle of a triangle is equal to the sum of interior opposite angles. Here interior angles are 50\u00b0 each and exterior angle is 50\u00b0.
\n\u2234 This triangle cannot be formed.
\n[\u2235 50\u00b0 \u2260 50\u00b0 + 50\u00b0]<\/p>\n

\"NCERT<\/p>\n

NCERT In-text Question Page No. 122<\/span>
\nQuestion 1.
\nTwo angles of a triangle are 30\u00b0 and 80\u00b0. Find the third angle.
\nAnswer:
\nLet the third angle be x.
\n\u2234 Using the angle sum property of a triangle,
\n30\u00b0 + 80\u00b0 + x = 180\u00b0
\nor x + 110\u00b0 = 180\u00b0
\nor x = 180\u00b0- 110\u00b0 = 70\u00b0
\n\u2234 The measure of the required third angle is 70\u00b0.<\/p>\n

Question 2.
\nOne of the angles of a triangle is 80\u00b0 and the other two angles are equal. Find the measure of each of the equal angles.
\nAnswer:
\nLet each of the equal angles be x
\nx + x + 80\u00b0 = 180\u00b0
\n(angle sum property of a triangle)
\n2x + 80\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 80\u00b0
\n2x = 100\u00b0
\nx = \\(\\frac{100^{\\circ}}{2}\\) = 50\u00b0
\n\u2234 The required measure of each of the equal angle is 50\u00b0.<\/p>\n

Question 3.
\nThe three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
\nAnswer:
\nLet the three angles of a triangle be x, 2x and x
\nx + 2x + x = 180\u00b0
\n(using the angle sum property)
\n2x + 2x = 180\u00b0
\n4x = 180\u00b0
\nx \\(\\frac{180^{\\circ}}{4}\\) = 45\u00b0
\nThus, the three angles are 45\u00b0, 90\u00b0 and 45\u00b0. It is an isosceles triangle (two angles are equal, opposite sides also equal)
\nAs one angle is 90\u00b0.
\n\u2234 It is a right-angled triangle.<\/p>\n

\"NCERT<\/p>\n

NCERT In-text Question Page No. 123 & 124<\/span>
\nQuestion 1.
\nFind angle x in each figure:
\n\"NCERT
\n\"NCERT
\n\"NCERT
\n\"NCERT
\n\"NCERT
\nAnswer:
\n(i) Since the two sides in the triangle are equal, the base angle opposite to equal sides are equal.
\n\u2234 x = 40\u00b0<\/p>\n

(ii) Since the two sides of the triangle are equal, the base angles opposite to equal sides are equal so, the other angle = 45\u00b0
\nSum of three angles of a triangle = 180\u00b0
\n45\u00b0 + 45\u00b0 + x = 180\u00b0
\nx + 90\u00b0 = 180\u00b0
\nx = 180\u00b0 – 90\u00b0
\nx = 90\u00b0<\/p>\n

(iii) Since the two sides of the triangle are equal, the base angles opposite to equal sides are equal.
\nx = 50\u00b0<\/p>\n

(iv) Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles is 180\u00b0.
\n100\u00b0 + x + x = 180\u00b0
\n2x = 180\u00b0 – 100\u00b0
\n2x = 80\u00b0
\nx = \\(\\frac{80^{\\circ}}{2}\\) = 40\u00b0<\/p>\n

(v) Base angles opposite to the equal sides of an isosceles triangles are equal and the sum of the measure of three angles of triangle is 180\u00b0.
\nx + x + 90\u00b0 = 180\u00b0
\n2x + 90\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 90\u00b0
\n2x = 90\u00b0
\nx = \\(\\frac{90^{\\circ}}{2}\\) = 45\u00b0<\/p>\n

(vi) Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of triangle is 180\u00b0.
\nx + x + 40\u00b0 = 180\u00b0
\n2x + 40\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 40\u00b0
\n2x = 140\u00b0
\nx = \\(\\frac{140^{\\circ}}{2}\\)
\n= 70\u00b0
\nx = 70\u00b0<\/p>\n

(vii) In the figure, two sides of the triangle are equal.
\n\u2234 The base angles opposite to equal sides are equal one of the base angle = x
\nother base angle = x
\nNow, x\u00b0 and 120\u00b0 form a linear pair = 180\u00b0
\nx + 120\u00b0 = 180\u00b0
\nx = 180\u00b0 -120\u00b0
\n= 60\u00b0
\nThus, the value of x = 60\u00b0<\/p>\n

(viii) In the figure, two sides of the triangle are equal.
\nsince, one of the base angles = x
\n\u2234 The other base angle = x
\nSince, exterior angle is equal to sum of the interior opposite angles
\n\u2234 x + x = 110\u00b0
\n2x = 110\u00b0
\nx = \\(\\frac{110^{\\circ}}{2}\\) = 55\u00b0<\/p>\n

(ix) Two sides of the triangle are equal.
\n\u2234 The base angles opposite to the equal sides are equal.
\nSince, one of the base angle = x
\n\u2234 The other base angle = x
\nAlso, the vertically opposite angles 30\u00b0 and x are equal.
\n\u2234 x = 30\u00b0<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind angles x and y in each figure.
\n\"NCERT
\nAnswer:
\n(i) Two sides of the triangle are equal.
\n\u2234 The base angles opposite to the equal sides are equal.
\nThe other base angle is y.
\nNow y and 120\u00b0 form a linear pair.
\n\u2234 y + 120\u00b0 = 180\u00b0
\ny = 180\u00b0 – 120\u00b0 = 60\u00b0
\n\u2234 x + y + y = 180\u00b0
\n(sum of the three angles = 180\u00b0)
\nx + 60\u00b0 + 60\u00b0= 180\u00b0
\nx+ 120\u00b0 = 180\u00b0
\nx = 180\u00b0 – 120\u00b0
\n= 60\u00b0
\nThus x = 60\u00b0 and y = 60\u00b0<\/p>\n

(ii) Two sides of a triangles are equal.
\n\u2234 The base angles opposite to equal sides are equal. Since one of the base angle is \u2018x\u2019
\n\u2234 The other base angle = x
\nThe given triangle is a right-angled triangle.
\nx + x + 90\u00b0 = 180\u00b0
\n2x + 90\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 90\u00b0
\n2x = 90\u00b0
\nx = \\(\\frac{90^{\\circ}}{2}\\) = 45\u00b0
\nNow x and y form a linear pair
\nx + y = 180\u00b0
\n45\u00b0+ y = 180\u00b0
\ny = 180\u00b0 – 45\u00b0 = 135\u00b0
\nThus, x = 45\u00b0 and y = 135\u00b0<\/p>\n

(iii) In the given figure, two sides of a triangle are equal.
\n\u2234 The base angles are x and x
\nThe third angle = 92\u00b0
\n(vertically opposite angles are equal)
\nx + x + 92\u00b0 = 180\u00b0
\n(Sum of the three angles of a triangle is 180\u00b0)
\n2x + 92\u00b0 = 180\u00b0
\n2x = 180\u00b0 – 92\u00b0
\n2x = 88\u00b0
\nx = \\(\\frac{88^{\\circ}}{2}\\) = 44\u00b0
\nNow, x and y form a linear pair
\nx + y = 180\u00b0
\n44 + y = 180\u00b0
\ny = 180\u00b0 – 44\u00b0 = 136\u00b0
\nThus, x = 44\u00b0 and y = 136\u00b0<\/p>\n

\"NCERT<\/p>\n

NCERT In-text Question Page No. 129 & 130<\/span>
\nQuestion 1.
\nFind the unknown length x in the following figures.
\n\"NCERT
\n\"NCERT
\n\"NCERT
\n(i) In the given right-angled triangle, the longest side (hypotenuse) is x
\nx2<\/sup> = 32<\/sup> + 42<\/sup>
\n(using pythagoras property)
\n= 9 + 16
\n= 25
\nor x2<\/sup> = 52<\/sup>
\nx = \\(\\sqrt{25}\\)
\n= 5
\nx = 5<\/p>\n

(ii) The given figure is a right-angled triangle.
\nx2<\/sup> = 62<\/sup> + 82<\/sup>
\n(using pythagoras property)
\n= 36 + 64
\nx2<\/sup> = 100
\nx2<\/sup> = 102<\/sup>
\n\u2234 x = 10<\/p>\n

(iii) The given figure is a right-angled triangle.
\nx2<\/sup> = 82<\/sup> + 152<\/sup>
\n(using pythagoras property)
\nx2<\/sup> = 64 + 225
\nx2<\/sup> = 289
\nx2<\/sup> = 172<\/sup>
\nx = 17<\/p>\n

(iv) The given figure is a right-angled triangle
\nx2<\/sup> = 72<\/sup> + 242<\/sup>
\n(using pythagoras property)
\nx2<\/sup> = 49 + 576
\nx2<\/sup> = 625
\nx2<\/sup> = 252<\/sup>
\nx = 25<\/p>\n

(v) The given figure can be labelled as \u0394ABC and the altitude is AD. Consider the right-angled triangle ABD
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup>
\n(using pythagoras property)
\n\"NCERT
\n372<\/sup> = 122<\/sup> + BD2<\/sup>
\n1369 = 144 + BD2<\/sup>
\n1369 – 144 = BD2<\/sup>
\n1225 = BD2<\/sup>
\n352<\/sup> = BD2<\/sup>
\n\u2235 BD = 35
\nIn the right-angled triangle ADC
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup>
\n(using pythagoras property)
\n372<\/sup> = 122<\/sup> + DC2<\/sup>
\n1369 = 144 +DC2<\/sup>
\n1369 – 144 = DC2<\/sup>
\n1225 = DC2<\/sup>
\n352<\/sup> = DC2<\/sup>
\n\u2234 DC = 35
\n\u2235 BC = BD + DC
\n= 35 + 35
\nx = 70<\/p>\n

(vi) In the given right-angled triangle
\nx2<\/sup> = 122<\/sup> + 52<\/sup>
\n(using the pythagoras property)
\n= 144 + 25
\nx2<\/sup> = 169
\nx2<\/sup> = 132<\/sup>
\n\u2234 x = 13<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions NCERT In-text Question Page No. 113 & 114 Question 1. Write the six elements …<\/p>\n

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NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties InText Questions NCERT In-text Question Page No. 113 & 114 Question 1. 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