RD Sharma solution class 7 chapter 17 Construction Exercise 17.4

Exercise 17.4

Pages-17.6

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.
               

Question 2:

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

Answer 2:

ABC+BCA+CAB=180°BCA=180°-ABC-CABBCA=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.
    
      

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.

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°.
  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.

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°.

 

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