Constructions (Exercise 17.4) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download

 Page 1


  
 
 
 
  
 
 
 
        
 
1. Construct ?ABC in which BC = 4 cm, ?B = 50
o
 and ?C = 70
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 4 cm. 
2. Draw ?CBX such that ?CBX=50
o
. 
3. Draw ?BCY with Y on the same side of BC as X such that ?BCY=70
o
. 
4. Let CY and BX intersects at A. 
5. ABC is the required triangle. 
 
2. Draw ?ABC in which BC = 8 cm, ?B = 50
o
 and ?A = 50
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 8 cm. 
2. Draw ?CBX such that ?CBX = 50
o
. 
Page 2


  
 
 
 
  
 
 
 
        
 
1. Construct ?ABC in which BC = 4 cm, ?B = 50
o
 and ?C = 70
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 4 cm. 
2. Draw ?CBX such that ?CBX=50
o
. 
3. Draw ?BCY with Y on the same side of BC as X such that ?BCY=70
o
. 
4. Let CY and BX intersects at A. 
5. ABC is the required triangle. 
 
2. Draw ?ABC in which BC = 8 cm, ?B = 50
o
 and ?A = 50
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 8 cm. 
2. Draw ?CBX such that ?CBX = 50
o
. 
  
 
 
 
  
 
 
 
3. Draw ?BCY with Y on the same side of BC as X such that ?BCY = 80
o
. 
4. Let CY and BX intersects at A. 
 
3. Draw ?ABC in which ?Q = 80
o
, ?R = 55
o
 and QR = 4.5 cm. Draw the perpendicular 
bisector of side QR. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment QR = 4.5 cm. 
2. Draw ?RQX = 80° and ?QRY = 55
o
. 
3. Let QX and RY intersects at P so that PQR is the required triangle. 
4. With Q as center and radius more than 4.5 cm, draw arcs on either sides of QR. 
5. With R as center and radius more than 4.5 cm, draw arcs intersecting the previous 
arcs at M and N. 
6. Join MN 
7. MN is the required perpendicular bisector of QR. 
 
4. Construct ?ABC in which AB = 6.4 cm, ?A = 45
o
 and ?B = 60
o 
 
Solution: 
Page 3


  
 
 
 
  
 
 
 
        
 
1. Construct ?ABC in which BC = 4 cm, ?B = 50
o
 and ?C = 70
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 4 cm. 
2. Draw ?CBX such that ?CBX=50
o
. 
3. Draw ?BCY with Y on the same side of BC as X such that ?BCY=70
o
. 
4. Let CY and BX intersects at A. 
5. ABC is the required triangle. 
 
2. Draw ?ABC in which BC = 8 cm, ?B = 50
o
 and ?A = 50
o
. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment BC of length 8 cm. 
2. Draw ?CBX such that ?CBX = 50
o
. 
  
 
 
 
  
 
 
 
3. Draw ?BCY with Y on the same side of BC as X such that ?BCY = 80
o
. 
4. Let CY and BX intersects at A. 
 
3. Draw ?ABC in which ?Q = 80
o
, ?R = 55
o
 and QR = 4.5 cm. Draw the perpendicular 
bisector of side QR. 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment QR = 4.5 cm. 
2. Draw ?RQX = 80° and ?QRY = 55
o
. 
3. Let QX and RY intersects at P so that PQR is the required triangle. 
4. With Q as center and radius more than 4.5 cm, draw arcs on either sides of QR. 
5. With R as center and radius more than 4.5 cm, draw arcs intersecting the previous 
arcs at M and N. 
6. Join MN 
7. MN is the required perpendicular bisector of QR. 
 
4. Construct ?ABC in which AB = 6.4 cm, ?A = 45
o
 and ?B = 60
o 
 
Solution: 
  
 
 
 
  
 
 
 
 
Steps of construction: 
1. Draw a line segment AB = 6.4 cm. 
2. Draw ?BAX = 45
o
. 
3. Draw ?ABY with Y on the same side of AB as X such that ?ABY = 60
o
. 
4. Let AX and BY intersects at C. 
5. ABC is the required triangle. 
 
5. Draw ?ABC in which AC = 6 cm, ?A = 90
o
 and ?B = 60
o 
 
Solution: 
 
Steps of construction: 
1. Draw a line segment AC = 6 cm. 
2. Draw ?ACX = 30
o
. 
3. Draw ?CAY with Y on the same side of AC as X such that ?CAY = 90
o
. 
4. Join CX and AY. Let these intersects at B. 
5. ABC is the required triangle where angle ?ABC = 60
o
.