Chapter 8 Construction of Triangles Ex 8.6

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 8 Construction of Triangles Ex 8.6

Question 1.
Construct ∆ABC in which BC = 7 cm, ∠C = 50° and AC + AB = 8 cm.
Solution.
We are given base BC = 7 cm, the sum of other two sides AB + AC = 8 cm and ∠ACB = 50°.
We are required to construct of ∆ABC.
Steps of construction:
  1. Draw a line segment BC = 7 cm a base line.
    RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles 1
  2. Construct ∠BCY = 50°.
  3. From CY cut off a line segment of length 8 cm and mark the point D.
  4. Join CD.
  5. Draw the perpendicular bisector of BD, intersecting CD at A.
  6. Join BA.
    Thus, ∆ABC is the required triangle.
Question 2.
Construct a triangle PQR in which PQ = 6 cm, ∠Q = 60° and QR + PR = 8 cm.
Solution.
Steps of construction:
  1. Draw a line segment PQ = 6 cm as base line.
    RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles 2
  2. At Q, construct an angle of 60° and produce it.
  3. From QY, cut off a line segment of length 8 cm and mark the point M.
  4. Join PM.
  5. Draw perpendicular bisector of PM, intersecting QM at R.
  6. Join P to R.
    Thus ∆PQR is the required triangle.
Question 3.
Construct a triangle PQR in which QR = 5 cm, ∠R = 40° and PR – PQ = 1 cm.
Solution.
It is given that QR = 5 cm, the difference between two sides PR – PQ = 1 cm and base angle ∠R = 40°.
Steps of construction:
  1. Draw a line segment QR = 5 cm.
    RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles 3
  2. At R, construct at angle of 40° and produce it.
  3. By taking R as centre, draw an arc of radius 1 cm cutting RT at S.
  4. Join QS.
  5. Also, draw ⊥ bisector of QS which meets RT at R
  6. Join P to Q.
    Hence, ∆PQR is the required triangle.
Question 4.
Construct a triangle ABC having its perimeter 12 cm and base angles 50° and 70°.
Solution.
Perimeter of ∆ABC is given
i. e., AB + BC + CA = 12 cm, ∠B = 50° and ∠C = 70°.
We are required to construct the ∆ABC.
Steps of construction:
  1. Draw a ray PX and cut off a line segment PQ = 12 cm from it.
  2. At P, construct ∠YPB = 25° with the help of protractor i.e. [\frac { 1 }{ 2 } x 50° ]
  3. At Q, construct ∠ZQP = 35°
    RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles 4
  4. Draw perpendicular bisectors of AP intersecting PQ at B.
  5. Draw perpendicular bisectors of AQ intersecting PQ at C.
  6. Join AB and AC.
    Then, ∆ABC is the required triangle.
Question 5.
Constructs ∆ABC in which BC = 8 cm, and medians AD and CF are 6 cm and 7.5 cm respectively.
Solution.
Steps of construction:
  1. Draw BC = 8 cm and get mid-point of BC as D.
    RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles 5
  2. AD = 6 cm, let G be centroid then GD = \frac { 1 }{ 3 } x 6 =2 cm. Taking D as centre, draw an arc of radius 2 cm.
  3. CG = \frac { 2 }{ 3 } x CF = \frac { 2 }{ 3 } x 7.5 = 5 cm. From centre C, cut an arc of radius 5 cm to intersect previous arc at G.
  4. Draw CGF = 7.5 cm.
  5. Draw BF and extend.
  6. From centre D, draw an arc of radius AD = 6 cm which intersects BF (produced) at A.
  7. Join AC.
    Hence, ∆ABC is the required triangle.

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