Section outline

    • Solutions to Ex. 6.1
      Q1. In ∆ PQR, D is the mid-point of QR. 

      Ans:

      PM is the altitude.

      PD is the median.

      No, QM is not equal to MR.

      Q2. Draw rough sketches for the following:

      Ans:

      (i)

      (ii)

      (iii)

       

       

       

    • Solutions to Ex. 6.2
      Q1. Find the value of the unknown exterior angle x in the following diagrams:

      Ans:

      (i) Exterior angle = Sum of interior opposite angles

      x = 50° + 70° = 120° 

      The exterior angle is 120°.

      (ii) Exterior angle = Sum of interior opposite angles

      x = 65° + 45° = 110° 

      The exterior angle is 110°.

      (iii) Exterior angle = Sum of interior opposite angles

      x = 30° + 40°  = 70°

      The exterior angle is 70°.

      (iv) Exterior angle = Sum of interior opposite angles

      x = 60° + 60° = 120° 

      The exterior angle is 120°.

      (v) Exterior angle = Sum of interior opposite angles

      x = 50° + 50° = 100° 

      The exterior angle is 100°.

      (vi) Exterior angle = Sum of interior opposite angles

      x = 30° + 60° = 90° 

      The exterior angle is 90°.

      Q2. Find the value of the unknown interior angle x in the following figures:

      Ans:

      (i) Sum of interior opposite angles = Exterior angle

      x + 50° = 115° 

      x = 115° - 50° = 65° 

      The interior angle is 65° 

      (ii) Sum of interior opposite angles = Exterior angle

      x + 70° = 100° 

      x = 100° - 70° = 30° 

      The interior angle is 30° 

      (iii) Sum of interior opposite angles = Exterior angle

      x + 90° = 125° 

      x = 125° - 90° = 35° 

      The interior angle is 35° 

      (iv) Sum of interior opposite angles = Exterior angle

      x + 60° = 120° 

      x = 120° - 60° = 60° 

      The interior angle is 60° 

      (v) Sum of interior opposite angles = Exterior angle

      x + 30° = 85° 

      x = 80° - 30° = 65° 

      The interior angle is 50° 

      (vi) Sum of interior opposite angles = Exterior angle

      x + 35° = 75° 

      x = 75° - 35° = 40° 

      The interior angle is 40° 

       

    • Solutions to Ex. 6.3
      Q1. Find the value of the unknown x in the following diagrams:

      Ans:

      (i) By angle sum property of a triangle,

      x + 50° + 60° = 180° 

      x = 180° - 110°

      x = 70°  

      (ii) By angle sum property of a triangle,

      x + 30° + 90° = 180° 

      x = 180° - 120°

      x = 60°  

      (iii) By angle sum property of a triangle,

      x + 30° + 110° = 180° 

      x = 180° - 140°

      x = 40°  

      (iv) By angle sum property of a triangle,

      x + x + 50° = 180° 

      2x = 180° - 50° = 130°  

      x = 130° / 2

      x = 65°

      (v) By angle sum property of a triangle,

      x + x + x= 180° 

      3x = 180° 

      x = 180° / 3

      x = 60°  

      (vi) By angle sum property of a triangle,

      x + 2x + 90° = 180° 

      3x = 180° - 90° = 90°  

      x = 90° / 3

      x = 30° 

      Q2. Find the values of the unknowns x and y in the following diagrams:

      Ans:

      (i) By exterior angle property of a triangle,

      x + 50° = 120°

      x = 120° - 50°

      x = 70°

      By angle sum property of a triangle,

      70° + y + 50°  = 180°     

      y + 120° = 180°

      y = 180° - 120°

      y = 60°

      (ii) y = 80° (vertically opposite angle)

      By angle sum property of a triangle,

      x + 80° + 50° = 180° 

      x + 130° = 180° 

      x = 50°

      (iii) By exterior angle property of a triangle,

      x = 50° + 60°

      x = 110°

      By angle sum property of a triangle,

      y + 50° + 60° = 180° 

      y = 180° - 110° 

      y = 70°

      (iv) x = 60° (vertically opposite angle)

      By angle sum property of a triangle,

      y + 60° + 30° = 180° 

      y = 180° - 90° 

      y = 90° 

      (v) y = 90° (vertically opposite angle)

      By angle sum property of a triangle,

      x + x + 90° = 180° 

      2x = 180° - 90° = 90°

      x = 90° / 2

      x = 45°

      (vi) As x = y (vertically opposite angle)

      By angle sum property of a triangle,

      x + x + x = 180°

      3x = 180°

      x = 180°/3

      x = 60° = y

       

    • Solutions to Ex. 6.4
      Q1. Is it possible to have a triangle with the following sides?

      Ans:

      (i)  The sum of two sides is,

      2 + 3 = 5cm

      Third side = 5 cm

      As, the sum of the length two sides is not greater than the length of the third side.

      This triangle is not possible.

      (ii) Sum of any two sides,

      3 + 6 >7

      3 + 7 > 6 &

      7 + 6 > 3

      As, the sum of the length of any two sides is greater than the length of the third side.

      This triangle is possible.

      (iii) Sum of any two sides,

      3 + 2  = 5cm which is less than 6cm

      As, the sum of the length two sides is not greater than the length of the third side.

      This triangle is not possible.

      Q2. Take any point O in the interior of a triangle PQR. Is

      Ans:

      (i) OP, OQ & PQ are sides of ∆OPQ.

      The sum of the length of two sides of a triangles is always greater than the length of the third side.

      Yes, OP + OQ > PQ

      (ii) OQ, OR & QR are sides of ∆OQR.

      The sum of the length of two sides of a triangles is always greater than the length of the third side.

      Yes, OQ + OR > QR

      (iii) OP, OR & RP are sides of ∆OPR.

      The sum of the length of two sides of a triangles is always greater than the length of the third side.

      Yes, OR + OP > RP

      Q3. AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM?

      Ans:

      As the sum of the length of two sides of a triangle is always greater than the length of the third side,

      So, in ∆ABM,

      AB + BM > AM ...... (1)

      So, in ∆ACM,

      AC + CM > AM ....... (2)

      Adding equations (1) & (2),

      AB + BM +AC +CM > 2AM

      As, BM + CM = BC

      AB + BC + CA > 2AM is proved

      Q4. ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?

      Ans:

      As the sum of the length of two sides of a triangle is always greater than the length of the third side,

      In ΔABC,

      AB + BC > AC ...... (1)

      In ΔCDA,

      CD + DA > AC ....... (2)

      In  ΔBCD,

      BC + CD > BD ....... (3)

      In ΔBAD,

      AB + DA > BD ...... (4)

      Adding equations 1, 2, 3 & 4,

      2 (AB + BC + CD + DA) > 2(AC + BD)

      Dividing both sides by 2

      AB + BC + CD + DA > AC + BD

      Q5. ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)

      Ans:

      As the sum of the length of two sides of a triangle is always greater than the length of the third side,

      In ΔAOB,

      AO + OB > AB ...... (1)

      In ΔBOC,

      BO + OC > BC ....... (2)

      In  ΔCDO,

      CO + OD > CD ....... (3)

      In ΔADO,

      AO + OD > DA ...... (4)

      Adding equations 1, 2, 3 & 4,

      AO + OB + BO + OC + CO + OD + AO + OD > AB + BC + CD + DA

      2(AO + OC) + 2(BO +OD) > AB + BC + CD + DA

      2AC + 2BD > AB + BC + CD + DA

      2 (AC + BD) > AB + BC + CD + DA

      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?

      Ans:

      Given: Length of two sides = 12cm, 15cm

      As the sum of length of two sides of a triangle is always greater than the length of the third side,

      12 + 15 = 27

      Hence, the third side has to be less than 27cm.

      The third side has to be greater than difference of the length of the two sides,

      15 - 12 = 3

      The third side has to be greater than 3cm and less than 27cm.

       

    • Solutions to Ex. 6.5
      Q1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR

      Ans: ∆PQR is a right angled triangle,

      By Pythagoras property,

      PQ2 + PR2 = QR2

      102 + 242 = QR2

      QR2 = 100 + 576 = 676

      QR = 26cm

      Q2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

      Ans: 

      ∆ABC is a right angled triangle,

      By Pythagoras property,

      AC2 + BC2 = AB2

      72 + BC2 = 252

      BC2 = 625 - 49 = 576

      = \( \sqrt[2]{576} \)

      BC = 24cm

      Q3. A 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.

      Ans: 
      NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 3

      By Pythagoras property,

      a2 + 122 = 152

      a2 + 144 = 225

      a2 = 225 -144 = 81

      a = 9m

      The distance of the foot of the ladder from the wall is 9m.

      Q4. Which of the following can be the sides of a right triangle? In the case of right-angled triangles, identify the right angles.

      Ans:

      (i) 2.5 cm, 6.5 cm, 6 cm.

      6.52 = 42.25

      2.52 + 62 = 6.25 + 36 = 42.25

      Thus, 2.52 + 62 = 6.52

      As, Pythagoras property holds true,

      This is a right angled triangle.

      The right angle is the angle opposite to the 6.5cm side (hypotenuse).

      (ii) 2 cm, 2 cm, 5 cm

      52 = 25

      22 + 22 = 4 + 4 = 8

      Thus, 22 + 22 \( \neq \) 52

      This cannot be a right angled triangle.

      (iii) 1.5 cm, 2cm, 2.5 cm

      2.52 = 6.25

      1.52 + 22 = 2.25 + 4 = 6.25

      Thus, 1.52 + 22 = 2.52

      As, Pythagoras property holds true,

      This is a right angled triangle.

      The right angle is the angle opposite to the 2.5cm side (hypotenuse).

      Q5. A 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 original height of the tree.

      Ans: 

      Given: Height at which tree is cut (AC) = 5cm

         Distance from base (AB) = 12m

         

      By Pythagoras property,

      AC2 + AD2 = CD2

      52 + 122 = CD2

      CD2 = 25 + 144 = 169

      CD = 13m = BC

      Height of the tree = AC + BC 

      = 5 + 13 = 18m

      Original height of the tree is 18m.

      Q6. Angles Q and R of a ∆PQR are 25º and 65º. Write which of the following is true:

      Ans:

      By angle sum property of a triangle,

      ∠P + ∠Q + ∠R = 180º

      ∠P + 25º + 65º = 180º

      ∠P = 180º - 90º = 90º

      Thus, ∆PQR is a right angled triangle,

      By Pythagoras property,

      (i) PQ2 + QR2 = RP2 is not true.

      (ii) PQ2 + RP2 = QR2 is true.

      (iii) RP2 + QR2  = PQ2 is not true.

      Q7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

      Ans: 

      Given: Length (AB) = 40cm

         Diagonal (AC) = 41cm

      NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 6

      As ∠B = 90º,

      By Pythagoras property,

      AB2 + BC2 = AC2

      402 + BC2 = 412

      BC2 = 1681 - 1600 = 81

      BC = 9cm

      The breadth of the rectangle is 9cm.

      Perimeter = 2 (l + b)

      = 2 (40 + 9) = 2 x 49

      Perimeter = 98cm

      Q8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

      Ans:

      Given: Diagonal AC = 16cm

        Diagonal BD = 30cm

      NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 7

      The diagonals of rhombus bisect & are perpendicular to each other at O.

      AO = 16/2 = 8cm

      OD = 30/2 = 15cm

      ∆AOD is a right angled triangle,

      By Pythagoras property,

      AO2 + OD2 = AD2

      82 + 152 = AD2

      AD2 = 64 + 225 = 289

      AD = 17

      Side of the rhombus = 17cm

      Perimeter = 4 x 17 = 68cm

      The perimeter of rhombus is 68cm.