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# RS Aggarwal Solutions: Exercise 11A -Constructions Class 10 Notes | EduRev

## Class 10 : RS Aggarwal Solutions: Exercise 11A -Constructions Class 10 Notes | EduRev

``` Page 1

1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB
such that
3
5
AP
AB
.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 points (greater of 3 and 5)
1 2 3 4
, , , A A A A and
5
A such that
1 1 2 2 3 3 4 4 5
A A A A A A A A A A
Step 4: Join
5
. A B
Step 5: From
3
, A draw
3
A P parallel to
5
A B (draw an angle equal to
5
AA B ), meeting A B in
P.
Here, P is the point on AB such that
3
2
A P
P B
or
3
.
5
A P
A B
2. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7.6 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 8 13points
1 2 3 4 5 6 7 8 9 10 11 12
, , , , , , , , , , , A A A A A A A A A A A A and
13
A such that
1 1 2 2 3 3 4 4 5 6 7 8 9 9 10 10 11 11 12 12 13
. A A A A A A A A A A A A A A A A A A A A A A
Step 4: Join
13
. A B
Step 5: From
5
, A draw
5
A P parallel to
13
A B (draw an angle equal to
13
A A B ), meeting A B
in P.
Page 2

1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB
such that
3
5
AP
AB
.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 points (greater of 3 and 5)
1 2 3 4
, , , A A A A and
5
A such that
1 1 2 2 3 3 4 4 5
A A A A A A A A A A
Step 4: Join
5
. A B
Step 5: From
3
, A draw
3
A P parallel to
5
A B (draw an angle equal to
5
AA B ), meeting A B in
P.
Here, P is the point on AB such that
3
2
A P
P B
or
3
.
5
A P
A B
2. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7.6 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 8 13points
1 2 3 4 5 6 7 8 9 10 11 12
, , , , , , , , , , , A A A A A A A A A A A A and
13
A such that
1 1 2 2 3 3 4 4 5 6 7 8 9 9 10 10 11 11 12 12 13
. A A A A A A A A A A A A A A A A A A A A A A
Step 4: Join
13
. A B
Step 5: From
5
, A draw
5
A P parallel to
13
A B (draw an angle equal to
13
A A B ), meeting A B
in P.

Here, P is the point on A B which divides it in the ratio 5 :8.
Length of 2.9 AP c m (Approx)
Length of 4.7 B P c m (Approx)
3. Construct a P Q R , in which PQ = 6 cm, QR = 7 cm and PR =- 8 cm. Then, construct
another triangle whose sides  are
4
5
times the corresponding sides of P Q R
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . Q R c m
Step 2: With Q as center and radius 6 cm, draw an arc.
Step 3: With R as center and radius 8cm, draw an arc cutting the previous arc at P
Step 4: Join P Q and P R . Thus, P Q R is the required triangle.
Step 5: Below QR, draw an acute angle . R Q X
Step 6: Along OX, mark five points
1 2 3 4
, , , R R R R and
5
R such that
1 1 2 2 3 3 4 4 5
. Q R R R R R R R R R
Step 7: Join
5
. R R
Step 8: From
4
, R draw
4 5
'|| R R R R meeting Q R at R
Step 9: From R P R P R meeting P Q .
Page 3

1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB
such that
3
5
AP
AB
.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 points (greater of 3 and 5)
1 2 3 4
, , , A A A A and
5
A such that
1 1 2 2 3 3 4 4 5
A A A A A A A A A A
Step 4: Join
5
. A B
Step 5: From
3
, A draw
3
A P parallel to
5
A B (draw an angle equal to
5
AA B ), meeting A B in
P.
Here, P is the point on AB such that
3
2
A P
P B
or
3
.
5
A P
A B
2. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7.6 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 8 13points
1 2 3 4 5 6 7 8 9 10 11 12
, , , , , , , , , , , A A A A A A A A A A A A and
13
A such that
1 1 2 2 3 3 4 4 5 6 7 8 9 9 10 10 11 11 12 12 13
. A A A A A A A A A A A A A A A A A A A A A A
Step 4: Join
13
. A B
Step 5: From
5
, A draw
5
A P parallel to
13
A B (draw an angle equal to
13
A A B ), meeting A B
in P.

Here, P is the point on A B which divides it in the ratio 5 :8.
Length of 2.9 AP c m (Approx)
Length of 4.7 B P c m (Approx)
3. Construct a P Q R , in which PQ = 6 cm, QR = 7 cm and PR =- 8 cm. Then, construct
another triangle whose sides  are
4
5
times the corresponding sides of P Q R
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . Q R c m
Step 2: With Q as center and radius 6 cm, draw an arc.
Step 3: With R as center and radius 8cm, draw an arc cutting the previous arc at P
Step 4: Join P Q and P R . Thus, P Q R is the required triangle.
Step 5: Below QR, draw an acute angle . R Q X
Step 6: Along OX, mark five points
1 2 3 4
, , , R R R R and
5
R such that
1 1 2 2 3 3 4 4 5
. Q R R R R R R R R R
Step 7: Join
5
. R R
Step 8: From
4
, R draw
4 5
'|| R R R R meeting Q R at R
Step 9: From R P R P R meeting P Q .

Here, ' ' P Q R is the required triangle, each of whose sides are
4
5
times the corresponding
sides of . P Q R
4. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides
are
7
5
of the corresponding sides of the first triangle.
Sol:
Steps of Construction :
Step 1: Draw a line segment 4 . BC c m
Step 2: With B as center, draw an angle of 90 .
Step 3: With B as center and radius equal to 3 cm, cut an arc at the right angle and name it
A.
Step 4: Join A B and A C.
Thus, ABC is obtained.
Step 5: Extend B C to D, such that
7
75 4 5.6 .
5
B D B C c m c m
Step 6: Draw || , D E C A cutting AB produced to E.
Thus, E B D is the required triangle, each of whose sides is
7
5
the corresponding sides of
. A B C
5. Construct a A B C with BC = 7 cm, 60 B and AB = 6 cm. Construct another triangle
whose sides are
3
4
times the corresponding sides of A B C
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . BC c m
Step 2: At B, draw 60 . XBC
Step 3: With B as center and radius 6 cm, draw an arc cutting the ray B X at A.
Step 4: Join A C. Thus, A B C is the required triangle.
Step 5: Below B C, draw an acute angle . Y B C
Step 6: Along B Y, mark four points
1 2 3
, , B B B and
4
B such that
1 1 2 2 3 3 4
. B B B B B B B B
Page 4

1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB
such that
3
5
AP
AB
.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 points (greater of 3 and 5)
1 2 3 4
, , , A A A A and
5
A such that
1 1 2 2 3 3 4 4 5
A A A A A A A A A A
Step 4: Join
5
. A B
Step 5: From
3
, A draw
3
A P parallel to
5
A B (draw an angle equal to
5
AA B ), meeting A B in
P.
Here, P is the point on AB such that
3
2
A P
P B
or
3
.
5
A P
A B
2. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7.6 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 8 13points
1 2 3 4 5 6 7 8 9 10 11 12
, , , , , , , , , , , A A A A A A A A A A A A and
13
A such that
1 1 2 2 3 3 4 4 5 6 7 8 9 9 10 10 11 11 12 12 13
. A A A A A A A A A A A A A A A A A A A A A A
Step 4: Join
13
. A B
Step 5: From
5
, A draw
5
A P parallel to
13
A B (draw an angle equal to
13
A A B ), meeting A B
in P.

Here, P is the point on A B which divides it in the ratio 5 :8.
Length of 2.9 AP c m (Approx)
Length of 4.7 B P c m (Approx)
3. Construct a P Q R , in which PQ = 6 cm, QR = 7 cm and PR =- 8 cm. Then, construct
another triangle whose sides  are
4
5
times the corresponding sides of P Q R
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . Q R c m
Step 2: With Q as center and radius 6 cm, draw an arc.
Step 3: With R as center and radius 8cm, draw an arc cutting the previous arc at P
Step 4: Join P Q and P R . Thus, P Q R is the required triangle.
Step 5: Below QR, draw an acute angle . R Q X
Step 6: Along OX, mark five points
1 2 3 4
, , , R R R R and
5
R such that
1 1 2 2 3 3 4 4 5
. Q R R R R R R R R R
Step 7: Join
5
. R R
Step 8: From
4
, R draw
4 5
'|| R R R R meeting Q R at R
Step 9: From R P R P R meeting P Q .

Here, ' ' P Q R is the required triangle, each of whose sides are
4
5
times the corresponding
sides of . P Q R
4. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides
are
7
5
of the corresponding sides of the first triangle.
Sol:
Steps of Construction :
Step 1: Draw a line segment 4 . BC c m
Step 2: With B as center, draw an angle of 90 .
Step 3: With B as center and radius equal to 3 cm, cut an arc at the right angle and name it
A.
Step 4: Join A B and A C.
Thus, ABC is obtained.
Step 5: Extend B C to D, such that
7
75 4 5.6 .
5
B D B C c m c m
Step 6: Draw || , D E C A cutting AB produced to E.
Thus, E B D is the required triangle, each of whose sides is
7
5
the corresponding sides of
. A B C
5. Construct a A B C with BC = 7 cm, 60 B and AB = 6 cm. Construct another triangle
whose sides are
3
4
times the corresponding sides of A B C
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . BC c m
Step 2: At B, draw 60 . XBC
Step 3: With B as center and radius 6 cm, draw an arc cutting the ray B X at A.
Step 4: Join A C. Thus, A B C is the required triangle.
Step 5: Below B C, draw an acute angle . Y B C
Step 6: Along B Y, mark four points
1 2 3
, , B B B and
4
B such that
1 1 2 2 3 3 4
. B B B B B B B B

Step 7: Join
4
. C B
Step 8: From
3
, B draw
3 4
'|| B C C B meeting B C at ' C
' '|| A C A C meeting A B
Here. ' ' A B C is the required triangle whose sides are
3
4
times the corresponding sides of
. A B C
6. Construct a A B C in which AB = 6 cm, 30 A and 60 A B . Construct another
' ' A B C similar to A B C
Sol:
Steps of Construction
Step 1: Draw a line segment 6 . AB c m
Step 2: At A, draw 30 . XA B
Step 3: At B, draw 60 . Y BA Suppose AX and BY intersect at C.
Thus, A B C is the required triangle.
Step 4: Produce AB to B such that ' 8 . AB c m
' '|| B C B C
. A B C
7. Construct a A B C in which BC = 8 cm, 45 B and 60 C . Construct another
triangle similar to A B C such that its sides are
3
5
of the corresponding sides of A B C .
Sol:
Steps of Construction
Step 1: Draw a line segment 8 . B C c m
Step 2: At B, draw 45 . XBC
Page 5

1. Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB
such that
3
5
AP
AB
.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 points (greater of 3 and 5)
1 2 3 4
, , , A A A A and
5
A such that
1 1 2 2 3 3 4 4 5
A A A A A A A A A A
Step 4: Join
5
. A B
Step 5: From
3
, A draw
3
A P parallel to
5
A B (draw an angle equal to
5
AA B ), meeting A B in
P.
Here, P is the point on AB such that
3
2
A P
P B
or
3
.
5
A P
A B
2. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Sol:
Steps of Construction:
Step 1: Draw a line segment 7.6 A B c m
Step 2: Draw a ray A X, making an acute angle . BA X
Step 3: Along A X, mark 5 8 13points
1 2 3 4 5 6 7 8 9 10 11 12
, , , , , , , , , , , A A A A A A A A A A A A and
13
A such that
1 1 2 2 3 3 4 4 5 6 7 8 9 9 10 10 11 11 12 12 13
. A A A A A A A A A A A A A A A A A A A A A A
Step 4: Join
13
. A B
Step 5: From
5
, A draw
5
A P parallel to
13
A B (draw an angle equal to
13
A A B ), meeting A B
in P.

Here, P is the point on A B which divides it in the ratio 5 :8.
Length of 2.9 AP c m (Approx)
Length of 4.7 B P c m (Approx)
3. Construct a P Q R , in which PQ = 6 cm, QR = 7 cm and PR =- 8 cm. Then, construct
another triangle whose sides  are
4
5
times the corresponding sides of P Q R
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . Q R c m
Step 2: With Q as center and radius 6 cm, draw an arc.
Step 3: With R as center and radius 8cm, draw an arc cutting the previous arc at P
Step 4: Join P Q and P R . Thus, P Q R is the required triangle.
Step 5: Below QR, draw an acute angle . R Q X
Step 6: Along OX, mark five points
1 2 3 4
, , , R R R R and
5
R such that
1 1 2 2 3 3 4 4 5
. Q R R R R R R R R R
Step 7: Join
5
. R R
Step 8: From
4
, R draw
4 5
'|| R R R R meeting Q R at R
Step 9: From R P R P R meeting P Q .

Here, ' ' P Q R is the required triangle, each of whose sides are
4
5
times the corresponding
sides of . P Q R
4. Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides
are
7
5
of the corresponding sides of the first triangle.
Sol:
Steps of Construction :
Step 1: Draw a line segment 4 . BC c m
Step 2: With B as center, draw an angle of 90 .
Step 3: With B as center and radius equal to 3 cm, cut an arc at the right angle and name it
A.
Step 4: Join A B and A C.
Thus, ABC is obtained.
Step 5: Extend B C to D, such that
7
75 4 5.6 .
5
B D B C c m c m
Step 6: Draw || , D E C A cutting AB produced to E.
Thus, E B D is the required triangle, each of whose sides is
7
5
the corresponding sides of
. A B C
5. Construct a A B C with BC = 7 cm, 60 B and AB = 6 cm. Construct another triangle
whose sides are
3
4
times the corresponding sides of A B C
Sol:
Steps of Construction
Step 1: Draw a line segment 7 . BC c m
Step 2: At B, draw 60 . XBC
Step 3: With B as center and radius 6 cm, draw an arc cutting the ray B X at A.
Step 4: Join A C. Thus, A B C is the required triangle.
Step 5: Below B C, draw an acute angle . Y B C
Step 6: Along B Y, mark four points
1 2 3
, , B B B and
4
B such that
1 1 2 2 3 3 4
. B B B B B B B B

Step 7: Join
4
. C B
Step 8: From
3
, B draw
3 4
'|| B C C B meeting B C at ' C
' '|| A C A C meeting A B
Here. ' ' A B C is the required triangle whose sides are
3
4
times the corresponding sides of
. A B C
6. Construct a A B C in which AB = 6 cm, 30 A and 60 A B . Construct another
' ' A B C similar to A B C
Sol:
Steps of Construction
Step 1: Draw a line segment 6 . AB c m
Step 2: At A, draw 30 . XA B
Step 3: At B, draw 60 . Y BA Suppose AX and BY intersect at C.
Thus, A B C is the required triangle.
Step 4: Produce AB to B such that ' 8 . AB c m
' '|| B C B C
. A B C
7. Construct a A B C in which BC = 8 cm, 45 B and 60 C . Construct another
triangle similar to A B C such that its sides are
3
5
of the corresponding sides of A B C .
Sol:
Steps of Construction
Step 1: Draw a line segment 8 . B C c m
Step 2: At B, draw 45 . XBC

Step 3: At C, draw 60 . Y C B Suppose B X and C Y intersect at A.
Thus, A B C is the required triangle
Step 4: Below B C, draw an acute angle . Z B C
Step 5: Along B Z, mark five points
1 2 3 4 5
, , , Z Z Z Z a n d Z such that
1 1 2 2 3 3 4 4 5.
B Z Z Z Z Z Z Z Z Z
Step 6: Join
5
. C Z
Step 7: From
3
, Z draw
3 5
'|| Z C C Z meeting B C
Step 8: From C ' '|| A C A C meeting AB
Here, ' ' A B C is the required triangle whose sides are
3
5
of the corresponding sides of
. A B C
8. To construct a triangle similar to A B C in which BC = 4.5 cm, 45 B and 60 C ,
using a scale factor of
3
7
, BC will be divided in the ratio
(a) 3 : 4 (b) 4 : 7 (c) 3 : 10 (d) 3 : 7
Answer: (a)  3 : 4
Sol:
To construct a triangle similar to A B C in which 4.5 , 45 60 , B C c m B a n d C
using a scale factor of
3
,
7
B C will be divided in the ratio 3: 4.
Here, ~ ' ' AB C A BC
': ' 3: 4 BC C C
or ': 3: 7 BC BC
Hence, the correct answer is option A.
```
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## Mathematics (Maths) Class 10

51 videos|346 docs|103 tests

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