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

## 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
.
```
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