05 - Question bank - Construction - Class 10 - Maths Class 10 Notes | EduRev

Crash Course for Class 10 Maths by Let's tute

Class 10 : 05 - Question bank - Construction - Class 10 - Maths Class 10 Notes | EduRev

 Page 1


   CONSTRUCTION 
 
 
 
1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. 
Solution: 
   
 
  
 
 
 
 
 
 
STEPS OF CONSTRUCTION: 
 1) Draw  AB= 6.4cm 
 2) Draw any ray AX , making an acute angle with AB. 
 3) Mark 5(=m+n, m =3, n =2) points A
1
, A2, A3, A4 and A
5
 on the ray AX such that A A
1
= A
1
A
2
, =A
2
A
3
= A
3
A
4
 
= A
4
A
5 
 
4) Join A
5
 to B 
 5)  Through the point A
3
draw a line parallel to A
5
B intersecting the line AB at the point C. 
   Then AC:CB=3:2 
 Proof: 
 Since 
5
AB is parallel to 
3
AC we can say that  
 
3
35
AA AC
CB A A
?     (Basic Proportionality Theorem)  
 By Construction 
3
35
AA 3
A A 2
? 
 
AC 3
CB 2
?? 
 Hence C divides AB in the Ratio 3:2. 
 
 
2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. 
Solution: 
Steps of Construction: 
 1) Draw AB=7.8cm. 
 2) Draw any ray AX making an acute angle with AB. 
 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 
 4) Mark the points
1
A , 
2
A, 
3
A, 
4
A, and 
5
A on AX and 
1
B, 
2
B,
3
B, ...... 
8
B on BY such that 
  
1
AA = 
12
AA = 
23
AA = 
34
AA = 
45
AA = 
1
BB = 
12
BB = 
23
BB =......= 
78
BB 
  
Page 2


   CONSTRUCTION 
 
 
 
1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. 
Solution: 
   
 
  
 
 
 
 
 
 
STEPS OF CONSTRUCTION: 
 1) Draw  AB= 6.4cm 
 2) Draw any ray AX , making an acute angle with AB. 
 3) Mark 5(=m+n, m =3, n =2) points A
1
, A2, A3, A4 and A
5
 on the ray AX such that A A
1
= A
1
A
2
, =A
2
A
3
= A
3
A
4
 
= A
4
A
5 
 
4) Join A
5
 to B 
 5)  Through the point A
3
draw a line parallel to A
5
B intersecting the line AB at the point C. 
   Then AC:CB=3:2 
 Proof: 
 Since 
5
AB is parallel to 
3
AC we can say that  
 
3
35
AA AC
CB A A
?     (Basic Proportionality Theorem)  
 By Construction 
3
35
AA 3
A A 2
? 
 
AC 3
CB 2
?? 
 Hence C divides AB in the Ratio 3:2. 
 
 
2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. 
Solution: 
Steps of Construction: 
 1) Draw AB=7.8cm. 
 2) Draw any ray AX making an acute angle with AB. 
 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 
 4) Mark the points
1
A , 
2
A, 
3
A, 
4
A, and 
5
A on AX and 
1
B, 
2
B,
3
B, ...... 
8
B on BY such that 
  
1
AA = 
12
AA = 
23
AA = 
34
AA = 
45
AA = 
1
BB = 
12
BB = 
23
BB =......= 
78
BB 
  
CONSTRUCTION 
 
 
 
 
 
5) Join 
5
A to 
8
B . Name the point it intersects AB as C. 
  Then AC: CB=5:8. 
 
 
 
 
 
 
 
 
 
 
 
Proof: 
 ? 
5
AA C ~? 
8
BB C 
 ?
5
8
AA AC 5
BB CB 8
??
 
 
 
3) Construct a triangle similar to a given triangle ABC with its sides equal to 
  
3
4
of the corresponding sides of ?ABC. 
Solution: 
   
 
 
 
 
 
 
 
 
 
 
Page 3


   CONSTRUCTION 
 
 
 
1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. 
Solution: 
   
 
  
 
 
 
 
 
 
STEPS OF CONSTRUCTION: 
 1) Draw  AB= 6.4cm 
 2) Draw any ray AX , making an acute angle with AB. 
 3) Mark 5(=m+n, m =3, n =2) points A
1
, A2, A3, A4 and A
5
 on the ray AX such that A A
1
= A
1
A
2
, =A
2
A
3
= A
3
A
4
 
= A
4
A
5 
 
4) Join A
5
 to B 
 5)  Through the point A
3
draw a line parallel to A
5
B intersecting the line AB at the point C. 
   Then AC:CB=3:2 
 Proof: 
 Since 
5
AB is parallel to 
3
AC we can say that  
 
3
35
AA AC
CB A A
?     (Basic Proportionality Theorem)  
 By Construction 
3
35
AA 3
A A 2
? 
 
AC 3
CB 2
?? 
 Hence C divides AB in the Ratio 3:2. 
 
 
2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. 
Solution: 
Steps of Construction: 
 1) Draw AB=7.8cm. 
 2) Draw any ray AX making an acute angle with AB. 
 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 
 4) Mark the points
1
A , 
2
A, 
3
A, 
4
A, and 
5
A on AX and 
1
B, 
2
B,
3
B, ...... 
8
B on BY such that 
  
1
AA = 
12
AA = 
23
AA = 
34
AA = 
45
AA = 
1
BB = 
12
BB = 
23
BB =......= 
78
BB 
  
CONSTRUCTION 
 
 
 
 
 
5) Join 
5
A to 
8
B . Name the point it intersects AB as C. 
  Then AC: CB=5:8. 
 
 
 
 
 
 
 
 
 
 
 
Proof: 
 ? 
5
AA C ~? 
8
BB C 
 ?
5
8
AA AC 5
BB CB 8
??
 
 
 
3) Construct a triangle similar to a given triangle ABC with its sides equal to 
  
3
4
of the corresponding sides of ?ABC. 
Solution: 
   
 
 
 
 
 
 
 
 
 
 
 
CONSTRUCTION 
 
 
 
 
 
 
Steps of Construction 
1)   Given ?ABC 
2)  Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 
3) Mark four points 
1
B,
2
B,
3
B and 
4
B on BX such that   
1
BB = 
12
BB = 
23
BB = 
34
BB 
4) Join   
4
B to C and draw a line parallel to 
4
BC through 
3
B ( smaller of 3 & 4 in 3/4)intersecting  
BC at C’ 
5)  Draw a line C’A’ through C’ parallel to AC intersecting AB at A’.  
 A’BC’ is the required triangle.  
 
4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle  
 whose sides are 
7
5
 of the corresponding sides of the first triangle. 
Solution:    
  
     
  
  
 
 
 
 
 
Steps Of Construction 
 1) Draw a line segment BC of length 9 cm. 
 2)  With B as centre and radius 7 cm draw an arc.  
 3) With C as centre and radius 8 cm draw another arc on the same side 
   and name the point of intersection as A.  
 4) Join AB and AC to get first triangle ABC.  
 5)  Now draw a ray BX such that it makes an acute angle with BC. 
Page 4


   CONSTRUCTION 
 
 
 
1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. 
Solution: 
   
 
  
 
 
 
 
 
 
STEPS OF CONSTRUCTION: 
 1) Draw  AB= 6.4cm 
 2) Draw any ray AX , making an acute angle with AB. 
 3) Mark 5(=m+n, m =3, n =2) points A
1
, A2, A3, A4 and A
5
 on the ray AX such that A A
1
= A
1
A
2
, =A
2
A
3
= A
3
A
4
 
= A
4
A
5 
 
4) Join A
5
 to B 
 5)  Through the point A
3
draw a line parallel to A
5
B intersecting the line AB at the point C. 
   Then AC:CB=3:2 
 Proof: 
 Since 
5
AB is parallel to 
3
AC we can say that  
 
3
35
AA AC
CB A A
?     (Basic Proportionality Theorem)  
 By Construction 
3
35
AA 3
A A 2
? 
 
AC 3
CB 2
?? 
 Hence C divides AB in the Ratio 3:2. 
 
 
2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. 
Solution: 
Steps of Construction: 
 1) Draw AB=7.8cm. 
 2) Draw any ray AX making an acute angle with AB. 
 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 
 4) Mark the points
1
A , 
2
A, 
3
A, 
4
A, and 
5
A on AX and 
1
B, 
2
B,
3
B, ...... 
8
B on BY such that 
  
1
AA = 
12
AA = 
23
AA = 
34
AA = 
45
AA = 
1
BB = 
12
BB = 
23
BB =......= 
78
BB 
  
CONSTRUCTION 
 
 
 
 
 
5) Join 
5
A to 
8
B . Name the point it intersects AB as C. 
  Then AC: CB=5:8. 
 
 
 
 
 
 
 
 
 
 
 
Proof: 
 ? 
5
AA C ~? 
8
BB C 
 ?
5
8
AA AC 5
BB CB 8
??
 
 
 
3) Construct a triangle similar to a given triangle ABC with its sides equal to 
  
3
4
of the corresponding sides of ?ABC. 
Solution: 
   
 
 
 
 
 
 
 
 
 
 
 
CONSTRUCTION 
 
 
 
 
 
 
Steps of Construction 
1)   Given ?ABC 
2)  Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 
3) Mark four points 
1
B,
2
B,
3
B and 
4
B on BX such that   
1
BB = 
12
BB = 
23
BB = 
34
BB 
4) Join   
4
B to C and draw a line parallel to 
4
BC through 
3
B ( smaller of 3 & 4 in 3/4)intersecting  
BC at C’ 
5)  Draw a line C’A’ through C’ parallel to AC intersecting AB at A’.  
 A’BC’ is the required triangle.  
 
4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle  
 whose sides are 
7
5
 of the corresponding sides of the first triangle. 
Solution:    
  
     
  
  
 
 
 
 
 
Steps Of Construction 
 1) Draw a line segment BC of length 9 cm. 
 2)  With B as centre and radius 7 cm draw an arc.  
 3) With C as centre and radius 8 cm draw another arc on the same side 
   and name the point of intersection as A.  
 4) Join AB and AC to get first triangle ABC.  
 5)  Now draw a ray BX such that it makes an acute angle with BC. 
CONSTRUCTION 
 
 
 
 
 
 
 6)  Mark 7 points on BX  
1
B,
2
B, ......
7
B such that  
1
BB = 
12
BB = 
23
BB =......= 
67
BB 
 7)  Join 
5
B to C and draw a line B
7
C’ parallel to 
5
B C through 
7
B intersecting BC extended  
    at C’. 
 8) Draw a line C’A’ parallel to AC through C’ intersecting BA at A’.  
  ?A’BC’ is the required triangle.  
  ?ABC ~?A’BC’ 
  
AB AC BC
A'B A'C' BC'
? ? ? 
  
5
7
BB BC
BC' BB
? 
5
7
? 
  
BC' 7
BC 5
? 
     
A'B A'C' BC' 7
AB AC BC 5
? ? ? ? 
 
 
5)  Construct an isosceles triangle whose base is 5 cm and altitude is 4 cm.   
      Then another triangle whose sides are 
1
1
2
 times the corresponding sides of the isosceles triangle. 
Solution: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Page 5


   CONSTRUCTION 
 
 
 
1) Draw a line segment of length 6.4cm and divide in the ratio 3:2. 
Solution: 
   
 
  
 
 
 
 
 
 
STEPS OF CONSTRUCTION: 
 1) Draw  AB= 6.4cm 
 2) Draw any ray AX , making an acute angle with AB. 
 3) Mark 5(=m+n, m =3, n =2) points A
1
, A2, A3, A4 and A
5
 on the ray AX such that A A
1
= A
1
A
2
, =A
2
A
3
= A
3
A
4
 
= A
4
A
5 
 
4) Join A
5
 to B 
 5)  Through the point A
3
draw a line parallel to A
5
B intersecting the line AB at the point C. 
   Then AC:CB=3:2 
 Proof: 
 Since 
5
AB is parallel to 
3
AC we can say that  
 
3
35
AA AC
CB A A
?     (Basic Proportionality Theorem)  
 By Construction 
3
35
AA 3
A A 2
? 
 
AC 3
CB 2
?? 
 Hence C divides AB in the Ratio 3:2. 
 
 
2) Draw a line segment of length 7.8cm and divide in the ratio 5:8. 
Solution: 
Steps of Construction: 
 1) Draw AB=7.8cm. 
 2) Draw any ray AX making an acute angle with AB. 
 3) Draw a ray BY parallel to AX making ABY ? equal to BAX ? . 
 4) Mark the points
1
A , 
2
A, 
3
A, 
4
A, and 
5
A on AX and 
1
B, 
2
B,
3
B, ...... 
8
B on BY such that 
  
1
AA = 
12
AA = 
23
AA = 
34
AA = 
45
AA = 
1
BB = 
12
BB = 
23
BB =......= 
78
BB 
  
CONSTRUCTION 
 
 
 
 
 
5) Join 
5
A to 
8
B . Name the point it intersects AB as C. 
  Then AC: CB=5:8. 
 
 
 
 
 
 
 
 
 
 
 
Proof: 
 ? 
5
AA C ~? 
8
BB C 
 ?
5
8
AA AC 5
BB CB 8
??
 
 
 
3) Construct a triangle similar to a given triangle ABC with its sides equal to 
  
3
4
of the corresponding sides of ?ABC. 
Solution: 
   
 
 
 
 
 
 
 
 
 
 
 
CONSTRUCTION 
 
 
 
 
 
 
Steps of Construction 
1)   Given ?ABC 
2)  Draw and ray BX making an acute angle with BC on the side opposite to the vertex A. 
3) Mark four points 
1
B,
2
B,
3
B and 
4
B on BX such that   
1
BB = 
12
BB = 
23
BB = 
34
BB 
4) Join   
4
B to C and draw a line parallel to 
4
BC through 
3
B ( smaller of 3 & 4 in 3/4)intersecting  
BC at C’ 
5)  Draw a line C’A’ through C’ parallel to AC intersecting AB at A’.  
 A’BC’ is the required triangle.  
 
4) Construct a triangle with sides 7 cm, 8 cm and 9 cm. And then another triangle  
 whose sides are 
7
5
 of the corresponding sides of the first triangle. 
Solution:    
  
     
  
  
 
 
 
 
 
Steps Of Construction 
 1) Draw a line segment BC of length 9 cm. 
 2)  With B as centre and radius 7 cm draw an arc.  
 3) With C as centre and radius 8 cm draw another arc on the same side 
   and name the point of intersection as A.  
 4) Join AB and AC to get first triangle ABC.  
 5)  Now draw a ray BX such that it makes an acute angle with BC. 
CONSTRUCTION 
 
 
 
 
 
 
 6)  Mark 7 points on BX  
1
B,
2
B, ......
7
B such that  
1
BB = 
12
BB = 
23
BB =......= 
67
BB 
 7)  Join 
5
B to C and draw a line B
7
C’ parallel to 
5
B C through 
7
B intersecting BC extended  
    at C’. 
 8) Draw a line C’A’ parallel to AC through C’ intersecting BA at A’.  
  ?A’BC’ is the required triangle.  
  ?ABC ~?A’BC’ 
  
AB AC BC
A'B A'C' BC'
? ? ? 
  
5
7
BB BC
BC' BB
? 
5
7
? 
  
BC' 7
BC 5
? 
     
A'B A'C' BC' 7
AB AC BC 5
? ? ? ? 
 
 
5)  Construct an isosceles triangle whose base is 5 cm and altitude is 4 cm.   
      Then another triangle whose sides are 
1
1
2
 times the corresponding sides of the isosceles triangle. 
Solution: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
CONSTRUCTION 
 
 
 
 
 
Steps Of Construction 
 1) Draw BC =5cm. Draw perpendicular bisector of BC and name the  
  mid-point of BC as O 
2) With O as centre and radius 4 cm draw an arc intersecting the perpendicular  bisector of BC. 
Name the point of intersection as A.  
 3)  Join AB and AC. This is the triangle ABC.  
 4)  Now draw a ray BX making an acute angle with BC. 
 5) Locate 3 equal points B
1
, B
2
 and B
3
 such that BB
1
=B
1
B
2 
= B
2
B
3
   
 6) Join B
2
C and through B
3
 draw a line B
3
C’ parallel to B
2
C intersecting BC extended at C’.  
 7) Through C’ draw a line A’C’ parallel to AC intersecting BA at A’. 
  Thus ?A’BC’ is the required triangle  
 
6)  Draw a triangle ABC with sides BC=9 cm and AB=6 cm and ABC ? = 55 ? . Then  
 Construct a triangle whose sides are 
3
4
 of the corresponding sides of the triangle ABC.  
Solution: 
  
 
 
 
 
      
  
  
 
 
 
Steps Of Construction 
 1) Draw BC= 9cm. 
 2) At the point B draw a line BA =4cm making an angle of 55 ?. Join A to C to get ?ABC 
 3) Draw a ray BX making acute angle with the line segment BC and divide it into  
  4 equal parts BB
1
, BB
2
, BB
3
 & BB
4
.   
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of Class 10

Dynamic Test

Content Category

Related Searches

Previous Year Questions with Solutions

,

05 - Question bank - Construction - Class 10 - Maths Class 10 Notes | EduRev

,

Semester Notes

,

Objective type Questions

,

video lectures

,

05 - Question bank - Construction - Class 10 - Maths Class 10 Notes | EduRev

,

practice quizzes

,

MCQs

,

Extra Questions

,

Sample Paper

,

shortcuts and tricks

,

study material

,

Summary

,

pdf

,

Important questions

,

past year papers

,

mock tests for examination

,

Viva Questions

,

ppt

,

Exam

,

Free

,

05 - Question bank - Construction - Class 10 - Maths Class 10 Notes | EduRev

;