Home Notes MCQ's Qestions NCERT Qestions Worksheets Blogs

Chapter 1:Real Numbers

Chapter 2:Polynomials

Chapter 3:Pair of Linear Equations in Two Variables

Chapter 4:Quadratic Equations

Chapter 5: Arithmetic Progression

Chapter 6: Triangles

Chapter 7:Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11:Constructions

Chapter 12:Area Related to Circles

Chapter 13:Surface Areas and Volumes

Chapter 14:Statistics

Chapter 15:Probability

NCERT Solutions class 10 maths chapter 11

Constructions

NCERT Solutions class 10 maths chapter 11 Exercise 11.1

In each of the following, give the justification of the construction also:

Question 1

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts

Answer

Step 1:Draw line AB of 7.6cm

Step 2:Draw a line AX with angle BAX

Step 3:Mark point A1,A2,A3,A4............A13 On AX such that AA1 = A1A2 = A2A3

Step 4:Join a line B and A13

Step 5:Draw a parallel line A5 to BA13 abd makes an angle ∠AA13B ,join a C

Step 6:now AP :PB = 5:8 and AP=2.9cm ,PB=4.7cm

Justification:

By construction, we have CA5 || B13 .

AC/CB =AA5/AA5AA13 = 5/8

AC/CB = 5/ 8

NCERT Solution Class 10 Ex 11.1 class 10

Question 2

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer

Step 1: Draw a line AB of 4 cm

Step 2: Using a point A draw an arc of radius 5 cm

Step 3: Similarly, Using a point B as its centre, and draw an arc of radius 6 cm.

Step 4: Now join C to A and B and form ΔABC triangle.

Step 5:Draw a line AX with angle BAX

Step 6: Make 3 points such as A1, A2, A3 on line AX such that AA1= A1A2 = A2A3.

Step 8: Join the BA3 through A2 which is parallel to the line BA3 meeting AB at point B’.

Step 9: Using point B’, draw a line parallel to the line BC meeting AC at C’.

Justification:
Here
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
By construction we get B’C’ || BC
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (by AA similarity )
So AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB,
∠A2AB’ =∠A3AB (Common)
∠AA2B’ =∠AA3B (Corresponding angles)
∴ ΔAAB’ ∼ ΔAAB (by AA similarity )
AA2B’ and AA3B
So, AB’/AB = AA2/AA3
AB’/AB = 2/3 ……. (2)
From the eq (1) and (2)
AB’/AB=B’C’/BC = AC’/ AC = 2/3
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC

NCERT Solution Class 10 Ex 11.1 class 10

Question 3

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

Answer

Construction Step

Step 1: Draw a line AB =5 cm.

Step 2: Take point A and B as centre, and draw the arcs of radius 6 cm and 5 cm and join at C.

Step 3: Draw a line AX with angle BAX

Step 5: Mark 7 points such as A1, A2, A3, A4, A5, A6, A7 , on line AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7

Step 6: Join the points B to A5

Step 7: A7B' is draw a parallel to A5B.

Step 8: Now,from B’ draw a line with point C’ that is parallel to the line BC.

Step 9: ΔAB’C’ is the triangle.
NCERT Solution Class 10 Ex 11.1 class 10

Question 4

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1
1 / 2
times the corresponding sides of the isosceles triangle.

Answer

Step 1: Draw a line of 8cm

Step 2:Make a perpendicular bisector of BC' at point D

Step 3: D as center deaw a arc of 4 cm intersect the perpendicular bisector at the point A

Step 4:meeting AB and AC

Step 5:Draw a line BX with angle line BC

Step 6:Mark 3 points B1, B2 and B3 on the line BX such that BB1 = B1B2 = B2B3

Step 7:Join the points B2C and draw a line from B3 parallel to the line B2C,to intersects the extended line BC at point C’.

Step 8:Draw a line from C’ parallel to the line AC to intersect extended line AB at A’

NCERT Solution Class 10 Ex 11.1 class 10

Question 5

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are
3 / 4
of the corresponding sides of the triangle ABC.

Answer

Step 1:Draw a line BC = 6 cm,

Step 2:At point B draw ∠ABC = 60°.

Step 3: B as center make arc of 5cm and mark point A

Step 4:Join AB and AC

Step 5: Draw a line BX makes an acute angle with BC

Step 6:Mark 4 points such as B1, B2, B3, B4, on line BX.

Step 7:Join the points B4C ,draw a line through B3, parallel to B4C intersecting the line BC at C’

Step 8: From C’ draw a line that is parallel to the line AC and intersects the line AB at A’.

NCERT Solution Class 10 Ex 11.1 class 10

Question 6

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are
4 / 3
times the corresponding sides of ∆ABC.

Answer

∠B = 45°, ∠A = 105°

We know that

Sum of angles in a triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

Step 1:Draw a line BC = 7 cm

Step 2:Draw ∠B =45 and ∠C =30

Step 3:Make a point A and join AB and AC

Step 4: Draw a line BX with acute angle with BC

Step 5: Mark 4 points such as B1, B2, B3, B4, on the ray BX.

Step 6: Join the points B3C

Step 7: from B4 draw a line parallel to B3C to intersects extended line BC at C’.

Step 8: From C’, draw a line parallel to the line AC to intersects extended line at C’.

NCERT Solution Class 10 Ex 11.1 class 10

Question 7

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are
5 / 3f
times the corresponding sides of the given triangle. Solution:

Answer

Step 1: Draw a line BC of 7cm

Step 2:At B draw ∠B =90

Step 3:B as center draw arc with 4 cm radius

Step 4:join AC

Step 5:ABC Triangle

Step 6: Draw a line BX with acute angle with BC .

Step 7: Mark 5 point B1, B2, B3, B4, on the line BX such that BB1 = B1B2 = B2B3= B3B4 = B4B5

Step 8: Join the points B3C.

Step 9: Draw a line from B5 parallel to B3C intersects the extended line BC at C’.

9. From C’, draw a line parallel to AC intersects the extended line AB at A’.

NCERT Solution Class 10 Ex 11.1 class 10