NCERT Solutions Class 10 Maths Chapter 11 Constructions

Question

Our NCERT Solutions Class 10 Maths Chapter 11 Constructions is the best way to understand the basics of the construction of the geometrical figure. These solutions are created by subject matter experts. All the concepts are explained in a very concise method it is a very precise and explained point-wise method. Our experts at Sarthaks have prepared the NCERT solution with utmost concern to make it understandable for every student and all the tough concepts are broken down into simple language in easy-to-understand for students. We have improvised the method of explaining with the help of shortcuts, tips, and tricks, flow charts, diagrams to make the studies student-friendly. We have covered all the topics and provided solutions for all the questions including intext questions and exercise questions.

In this NCERT Solutions Class 10 Maths, we have discussed different topics related to construction related to geometrical figures. We have provided solutions for all kinds of topics such as:

• Dividing a line segment in a given ratio.
• Construction of triangle with a scale > 1
• Construction of triangle with a scale < 1
• Construction of a tangent on the circle from any point lying outside of it.
• Construction of an Angle with the help of different construction instruments such as Protector, Compasses, and Ruler.
• Construction of Triangles where with the help of properties of triangles such as
  • The sum of all interior angles of a triangle is equal to 180°
  • The sum of the exterior angle of a triangle is equal to the sum of the opposite interior angle.
  • The length of the third side is always greater than the sum of the lengths of the two sides of the triangle.
  • Pythagoras Theorem: In a right-angled triangle, the square of the length of the hypotenuse of the triangle is equal to the sum of the square of the other two sides of the triangle.
  • Construction of a triangle with the help of SSS criteria.
  • Construction of a triangle with the help of SAS criteria.
  • Construction of a triangle with the help of ASA criteria.

Our NCERT Solutions Class 10 Maths will help students develop a practical approach to geometry. This will also help students learn about the properties of a different geometrical figures. It is suggested by our experts on the subject matter to have in hand practice of all kinds of questions and their solutions mentioned here. We have discussed all the theories, intext questions, and exercises in such a way that it helps students get clarity of all the concepts. A complete understanding of the chapter will help students score good marks in exams.

Get started now to ace your exam.

Answer 1

NCERT Solutions Class 10 Maths Chapter 11 Constructions

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

Answer:

A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows. 

Step 1: Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB. 

Step 2: Locate 13 (= 5 + 8) points, A₁, A₂, A₃, A₄ …….. A₁₃, on AX such that AA₁ = A₁A₂ = A₂A₃ and so on. 

Step 3: Join BA₁₃. 

Step 4: Through the point A₅, draw a line parallel to BA₁₃ (by making an angle equal to ∠AA₁₃B) at A₅ intersecting AB at point C. 

C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8. 

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

Justification:

The construction can be justified by proving that

\(\frac{AC}{CB}= \frac{5}{8}\)

By construction, we have A₅C || A₁₃B. 

By applying Basic proportionality theorem for the triangle AA₁₃B, we obtain

From the figure, it can be observed that AA₅ and A₅A₁₃ contain 5 and 8 equal divisions of line segments respectively.

On comparing equations (1) and (2), we obtain

\(\frac{AC}{CB}=\frac{5}{8}\)

This justifies the construction.

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

Answer:

Step 1: Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the required triangle.

Step 2: Draw a ray AX making an acute angle with line AB on the opposite side of vertex C. 

Step 3: Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃. 

Step 4: Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B'. 

Step 5: Draw a line through B' parallel to the line BC to intersect AC at C'.

∆AB'C' is the required triangle.

Justification:

The construction can be justified by proving that

By construction, we have B’C’ || BC 

∴ ∠A = ∠ABC (Corresponding angles) 

In ∆AB'C' and ∆ABC, 

∠AB'C' = ∠ABC (Proved above) 

∠B'AC' = ∠BAC (Common) 

∴∆AB'C' ∼ ∆ABC (AA similarity criterion)

In ∆AA₂B' and ∆AA₃B, 

∠A₂AB' = ∠A₃AB (Common) 

∠AA₂B' = ∠AA₃B (Corresponding angles) 

∴ ∆AA₂B' ∼ ∆AA₃B (AA similarity criterion)

From equations (1) and (2), we obtain

This justifies the construction.

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. Give the justification of the construction.

Answer:

Step 1: Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. ∆ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively. 

Step 2: Draw a ray AX making acute angle with line AB on the opposite side of vertex C. 

Step 3: Locate 7 points, A₁, A₂, A₃, A₄ A₅, A₆, A₇ (as 7 is greater between 5 and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇. 

Step 4: Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B'. 

Step 5: Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. 

∆AB'C' is the required triangle.

Justification:

The construction can be justified by proving that 

In ∆ABC and ∆AB'C', 

∠ABC = ∠AB'C' (Corresponding angles) 

∠BAC = ∠B'AC' (Common) 

∴ ∆ABC ∼ ∆AB'C' (AA similarity criterion)

In ∆AA₅B and ∆AA₇B', 

∠A₅AB = ∠A₇AB' (Common) 

∠AA₅B = ∠AA₇B' (Corresponding angles) 

∴ ∆AA₅B ∼ ∆AA₇B' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

This justifies the construction.

Answer 2

4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle. 

Give the justification of the construction.

Answer:

Let us assume that ∆ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm. 

A ∆AB'C' whose sides are 3/2 times of ∆ABC can be drawn as follows. 

Step 1: Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D. 

Step 2: Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ∆ABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm. 

Step 3: Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C. 

Step 4: Locate 3 points (as 3 is greater between 3 and 2) A₁, A₂, and A₃ on AX such that AA₁= A₁A₂ = A₂A₃. 

Step 5: Join BA₂ and draw a line through A₃ parallel to BA₂ to intersect extended line segment AB at point B'. 

Step 6: Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. 

∆AB'C' is the required triangle.

Justification: 

The construction can be justified by proving that 

In ∆ABC and ∆AB'C', 

∠ABC = ∠AB'C' (Corresponding angles) 

∠BAC = ∠B'AC' (Common) 

∴ ∆ABC ∼ ∆AB'C' (AA similarity criterion)

In ∆AA₂B and ∆AA₃B', 

∠A₂AB = ∠A₃AB' (Common) 

∠AA₂B = ∠AA₃B' (Corresponding angles) 

∴ ∆AA₂B ∼ ∆AA₃B' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

This justifies the construction.

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. 

Give the justification of the construction.

Answer:

∆A'BC' whose sides are 3/4 of the corresponding sides of ∆ABC can be drawn as follows. 

Step 1: Draw a ∆ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. 

Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A. 

Step 3: Locate 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX. 

Step 4: Join B₄C and draw a line through B₃, parallel to B4C intersecting BC at C'.

Step 5: Draw a line through C' parallel to AC intersecting AB at A'. 

∆A'BC' is the required triangle.

Justification:

The construction can be justified by proving 

In ∆A'BC' and ∆ABC, 

∠A'C'B = ∠ACB (Corresponding angles) 

∠A'BC' = ∠ABC (Common) 

∴ ∆A'BC' ∼ ∆ABC (AA similarity criterion)

In ∆BB₃C' and ∆BB₄C, 

∠B₃BC' = ∠B₄BC (Common) 

∠BB₃C' = ∠BB₄C (Corresponding angles)

∴ ∆BB₃C' ∼ ∆BB₄C (AA similarity criterion)

From equations (1) and (2), we obtain

This justifies the construction.

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 side of ∆ABC. Give the justification of the construction.

Answer:

∠B = 45°, ∠A = 105° 

Sum of all interior angles in a triangle is 180°. 

∠A + ∠B + ∠C = 180° 

105° + 45° + ∠C = 180° 

∠C = 180° − 150° 

∠C = 30° 

The required triangle can be drawn as follows. 

Step 1: Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°. 

Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3: Locate 4 points (as 4 is greater in 4 and 3), B₁, B₂, B₃, B₄, on B_X. 

Step 4: Join B₃C. Draw a line through B₄ parallel to B₃C intersecting extended BC at C'. 

Step 5: Through C', draw a line parallel to AC intersecting extended line segment at C'. 

∆A'BC' is the required triangle.

Justification:

The construction can be justified by proving that 

In ∆ABC and ∆A'BC',

∠ABC = ∠A'BC' (Common) 

∠ACB = ∠A'C'B (Corresponding angles) 

∴ ∆ABC ∼ ∆A'BC' (AA similarity criterion)

In ∆BB₃C and ∆BB₄C',

∠B₃BC = ∠B₄BC' (Common) 

∠BB₃C = ∠BB₄C' (Corresponding angles)

∴ ∆BB₃C ∼ ∆BB₄C' (AA similarity criterion) 

On comparing equations (1) and (2), we obtain 

This justifies the construction.

Answer 3

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Give the justification of the construction.

Answer:

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. 

Clearly, these will be perpendicular to each other. 

The required triangle can be drawn as follows. 

Step 1: Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it. 

Step 2: Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ∆ABC is the required triangle. 

Step 3: Draw a ray AX making an acute angle with AB, opposite to vertex C.

Step 4: Locate 5 points (as 5 is greater in 5 and 3), A₁, A₂, A₃, A₄, A₅, on line segment AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅. 

Step 5: Join A₃B. Draw a line through A₅ parallel to A3B intersecting extended line segment AB at B'. 

Step 6: Through B', draw a line parallel to BC intersecting extended line segment AC at C'. 

∆AB'C' is the required triangle.

Justification:

The construction can be justified by proving that 

In ∆ABC and ∆AB'C',

∠ABC = ∠AB'C' (Corresponding angles) 

∠BAC = ∠B'AC' (Common) 

∴ ∆ABC ∼ ∆AB'C' (AA similarity criterion)

In ∆AA₃B and ∆AA₅B',

∠A₃AB = ∠A₅AB' (Common) 

∠AA₃B = ∠AA₅B' (Corresponding angles) 

∴ ∆AA₃B ∼ ∆AA₅B' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

This justifies the construction.

8. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Give the justification of the construction.

Answer:

A pair of tangents to the given circle can be constructed as follows. 

Step 1: Taking any point O of the given plane as centre, draw a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP. 

Step 2: Bisect OP. Let M be the mid-point of PO. 

Step 3: Taking M as centre and MO as radius, draw a circle. 

Step 4: Let this circle intersect the previous circle at point Q and R. 

Step 5: Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each. 

Justification:

The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR.

∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle. 

∴ ∠PQO = 90°

⇒ OQ ⊥ PQ 

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. 

Similarly, PR is a tangent of the circle.

9. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. Give the justification of the construction.

Answer:

Tangents on the given circle can be drawn as follows. 

Step 1: Draw a circle of 4 cm radius with centre as O on the given plane. 

Step 2: Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle and join OP. 

Step 3: Bisect OP. Let M be the mid-point of PO. 

Step 4: Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R.

Step 5: Join PQ and PR. PQ and PR are the required tangents.

It can be observed that PQ and PR are of length 4.47 cm each. 

In ∆PQO, 

Since PQ is a tangent, 

∠PQO = 90°

PO = 6 cm 

QO = 4 cm 

Applying Pythagoras theorem in ∆PQO, we obtain 

PQ² + QO² = PQ² 

PQ² + (4)² = (6)² 

PQ² + 16 = 36 

PQ² = 36 − 16 

PQ² = 20 PQ 

PQ = 4.47 cm

Justification:

The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.

∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle. 

∴ ∠PQO = 90° 

⇒ OQ ⊥ PQ 

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. 

Similarly, PR is a tangent of the circle

10. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Give the justification of the construction.

Answer:

The tangent can be constructed on the given circle as follows. 

Step 1: Taking any point O on the given plane as centre, draw a circle of 3 cm radius. 

Step 2: Take one of its diameters, PQ, and extend it on both sides. Locate two points on this diameter such that OR = OS = 7 cm 

Step 3: Bisect OR and OS. Let T and U be the mid-points of OR and OS respectively. 

Step 4: Taking T and U as its centre and with TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. Join RV, RW, SX, and SY. These are the required tangents.

Justification:

The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is O and radius is 3 cm). For this, join OV, OW, OX, and OY.

∠RVO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle. 

∴ ∠RVO = 90° 

⇒ OV ⊥ RV 

Since OV is the radius of the circle, RV has to be a tangent of the circle. 

Similarly, OW, OX, and OY are the tangents of the circle

Answer 4

11. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. Give the justification of the construction.

Answer:

The tangents can be constructed in the following manner: 

Step 1: Draw a circle of radius 5 cm and with centre as O. 

Step 2: Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A. 

Step 3: Draw a radius OB, making an angle of 120° (180° − 60°) with OA. 

Step 4: Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.

Justification:

The construction can be justified by proving that ∠APB = 60° 

By our construction 

∠OAP = 90° 

∠OBP = 90° 

And ∠AOB = 120°

We know that the sum of all interior angles of a quadrilateral = 360° 

∠OAP + ∠AOB + ∠OBP + ∠APB = 360°

90° + 120° + 90° + ∠APB = 360° 

∠APB = 60° 

This justifies the construction.

12. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Give the justification of the construction.

Answer:

The tangents can be constructed on the given circles as follows. 

Step 1: Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius. 

Step 2: Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, and AR. These are the required tangents.

Justification:

The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B and radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm). For this, join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. 

∴ ∠ASB = 90° 

⇒ BS ⊥ AS 

Since BS is the radius of the circle, AS has to be a tangent of the circle. 

Similarly, AR, BP, and BQ are the tangents.

13. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle. Give the justification of the construction.

Answer:

Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as ∠BDC is of measure 90°. The centre E of this circle will be the mid-point of BC.

The required tangents can be constructed on the given circle as follows. 

Step 1: Join AE and bisect it. Let F be the mid-point of AE. 

Step 2: Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. Join AG. 

AB and AG are the required tangents.

Justification:

The construction can be justified by proving that AG and AB are the tangents to the circle. For this, join EG.

∠AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. 

∴ ∠AGE = 90° 

⇒ EG ⊥ AG 

Since EG is the radius of the circle, AG has to be a tangent of the circle. 

Already, ∠B = 90°

 ⇒ AB ⊥ BE 

Since BE is the radius of the circle, AB has to be a tangent of the circle.

14. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circles. Give the justification of the construction.

Answer:

The required tangents can be constructed on the given circle as follows. 

Step 1: Draw a circle with the help of a bangle. 

Step 2: Take a point P outside this circle and take two chords QR and ST. 

Step 3: Draw perpendicular bisectors of these chords. Let them intersect each other at point O. 

Step 4: Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre, draw a circle of radius OU, which will intersect the circle at V and W. Join PV and PW. 

PV and PW are the required tangents.

Justification:

The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW.

We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. ∠PVO is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. 

∴ ∠PVO = 90° 

⇒ OV ⊥ PV 

Since OV is the radius of the circle, PV has to be a tangent of the circle. 

Similarly, PW is a tangent of the circle.