RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

Construct ∆ ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.

Answer 1:

Steps of construction:

  1. Draw a line segment BC of length 4 cm.
  2. Draw ∠CBX such that ∠CBX=50°.
  3. Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 70°°.
  4. Let CY and BX intersect at A.
  5. ABC is the required triangle.

RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

 

Question 2:

Draw ∆ ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.

Answer 2:

∠ABC+∠BCA+∠CAB=180°
∠BCA=180°−∠ABC−∠CAB
∠BCA=180°−100°=80°


Steps of construction:

  1. Draw a line segment BC of length 8 cm.
  2. Draw ∠CBX such that ∠CBX = 50°.
  3. Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°°.
  4. Let CY and BX intersect at A.

RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

 

Question 3:

Draw ∆ PQR in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.

Answer 3:

Steps of construction:

  1. Draw a line segment QR = 4.5 cm.
  2. Draw ∠RQX = 80° and ∠QRY =55°.
  3. Let QX and RY intersect at P so that PQR is the required triangle.
  4. With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
  5. With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
  6. Join MN; MN is the required perpendicular bisector of QR.

RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

 

Question 4:

Construct ∆ ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°.

Answer 4:

Steps of construction:

  1. Draw a line segment AB = 6.4 cm.
  2. Draw ∠BAX = 45°∠BAX = 45°.
  3. Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°°.

Let AX and BY intersect at C; ABC is the required triangle.

RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev

 

Question 5:

Draw ∆ ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.

Answer 5:

We can see that ∠A+∠B+∠C = 180°. Therefore ∠C = 180°° − 60°° − 90°° = 30°.

Steps of construction:

  1. Draw a line segment AC = 6 cm.
  2. Draw ∠ACX=30°.
  3. Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°°.
  4. Join CX and AY. Let these intersect at B.
  5. ABC is the required triangle where angle ∠ABC = 60°

RD Sharma Solutions - 17.4, Constructions, Class 7, Math Class 7 Notes | EduRev 

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