NCERT Solutions for Class 10 Maths Chapter 11 Constructions

NCERT Solutions Class 10 Maths Chapter 11 Constructions resource was created by GFG Team to help students with any queries they might have as they go through problems from the NCERT textbook. It lessens the frustration of spending a long time working on a problem. The NCERT Solutions for Class 10 Maths address every issue in this chapter’s exercise from the NCERT textbook.

In Chapter 11 Constructions, students will discover how to construct various figures using implements like a compass and ruler. They discover several ways to create tangents to circles and how to divide a line segment into the appropriate ratios

These Solutions cover all the two exercises of the NCERT Class 10 Maths Chapter 11, which are as follows:

Class 10 Maths NCERT Solutions Chapter 11 Exercises
NCERT Maths Solutions Class 10 Exercise 11.1 – 7 Questions (7 Long Answers)
NCERT Maths Solutions Class 10 Exercise 11.2 – 7 Questions (7 Long Answers)

Constructions: 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.

Solution:

Steps of construction:

To divide the line segment of 7.6 cm in the ratio of 5 : 8.

Step 1. Draw a line segment AB of length 7.6 cm.

Step 2. Draw a ray AC which forms an acute angle with the line segment AB.

Step 3. Mark the points = 13 as (5+8=13) points, such as A₁, A₂, A₃, A₄ …….. A₁₃, on the ray AC such that it becomes AA₁ = A₁A₂ = A₂A₃ and such like this.

Step 4. Now join the line segment and the ray, BA₁₃.

Step 5. Hence, the point A₅, construct a line parallel to BA₁₃ which makes an angle equal to ∠AA₁₃B.

Step 6. Point A₅ intersects the line AB at point X.

Step 7. X is that point which divides line segment AB into the ratio of 5:8.

Step 8. Thus, measure the lengths of the line AX and XB. Hence, it measures 2.9 cm and 4.7 cm respectively.

Justification:

The construction can be justified by proving that

\frac{AX}{XB} = \frac{5}{ 8}

From construction, we have A₅X || A₁₃B. By the Basic proportionality theorem for the triangle AA₁₃B, we will get

\frac{AX}{XB} =\frac{AA_5}{A_5A_{13}}         ….. (1)

By the figure we have constructed, it can be seen that AA₅ and A₅A₁₃ contains 5 and 8 equal divisions of line segments respectively.

Thus,

\frac{AA_5}{A_5A_{13}}=\frac{5}{8}         … (2)

Comparing the equations (1) and (2), we get

\frac{AX}{XB} = \frac{5}{ 8}

Thus, Justified.

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.

Solution:

Steps of Construction:

Step 1. Draw a line segment XY which measures 4 cm, So XY = 4 cm.

Step 2. Taking point X as centre, and construct an arc of radius 5 cm.

Step 3. Similarly, from the point Y as centre, and draw an arc of radius 6 cm.

Step 4. Thus, the following arcs drawn will intersect each other at point Z.

Step 5. Now, we have XZ = 5 cm and YZ = 6 cm and therefore ΔXYZ is the required triangle.

Step 6. Draw a ray XA which will make an acute angle along the line segment XY on the opposite side of vertex Z.

Step 7. Mark the 3 points such as X₁, X₂, X₃ (as 3 is greater between 2 and 3) on line XA such that it becomes XX₁ = X₁X₂ = X₂X₃.

Step 8. Join the point YX₃ and construct a line through X₂ which is parallel to the line YX₃ that intersect XY at point Y’.

Step 9. From the point Y’, construct a line parallel to the line YZ that intersect the line XZ at Z’.

Step 10. Hence, ΔXY’Z’ is the required triangle.

Justification:

The construction can be justified by proving that

XY’ = \frac{2}{3}XY\\ Y’Z’ = \frac{2}{3}YZ\\ XZ’= \frac{2}{3}XZ

From the construction, we get Y’Z’ || YZ

∴ ∠XY’Z’ = ∠XYZ (Corresponding angles)

In ΔXY’Z’ and ΔXYZ,

∠XYZ = ∠XY’Z (Proved above)

∠YXZ = ∠Y’XZ’ (Common)

∴ ΔXY’Z’ ∼ ΔXYZ (From AA similarity criterion)

Therefore, 

\frac{XY’}{XY} = \frac{Y’Z’}{YZ}= \frac{XZ’}{XZ}          …. (1)

In ΔXXY’ and ΔXXY,

∠X₂XY’ =∠X₃XY (Common)

From the corresponding angles, we get,

∠AA₂B’ =∠AA₃B

Thus, by the AA similarity criterion, we get

ΔXX₂Y’ and XX₃Y

So, \frac{XY’}{XY} = \frac{XX_2}{XX_3}

Therefore, \frac{XY’}{XY} = \frac{2}{3}          ……. (2)

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

\frac{XY’}{XY}=\frac{Y’Z’}{YZ} = \frac{XZ’}{ XZ} = \frac{2}{3}

It is written as

XY’ = \frac{2}{3}XY

Y’Z’ = \frac{2}{3}YZ

XZ’=  \frac{2}{3}XZ

Therefore, justified.

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

Solution:

Steps of construction:

Step 1. Construct a line segment XY =5 cm.

Step 2. By taking X and Y as centre, and construct the arcs of radius 6 cm and 5 cm respectively.

Step 3. These two arcs will intersect each other at point Z and hence ΔXYZ is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 4. Construct a ray XA which will make an acute angle with the line segment XY on the opposite side of vertex Z.

Step 5. Pinpoint the 7 points such as X₁, X₂, X₃, X₄, X₅, X₆, X₇ (as 7 is greater between 5 and 7), on the line XA such that it becomes XX₁ = X₁X₂ = X₂X₃ = X₃X₄ = X₄X₅ = X₅X₆ = X₆X₇

Step 6. Join the points YX₅ and construct a line from X₇ to YX₅ that is parallel to the line YX₅ where it intersects the extended line segment XY at point Y’.

Step 7. Now, construct a line from Y’ the extended line segment XZ at Z’ that is parallel to the line YZ, and it intersects to make a triangle.

Step 8. Hence, ΔXY’Z’ is the needed triangle.

Justification:

The construction can be justified by proving that

XY’ = \frac{7}{5}XY

Y’Z’ = \frac{7}{5}YZ

XZ’= \frac{7}{5}XZ

By the construction, we have Y’Z’ || YZ

Therefore,

∠XY’Z’ = ∠XYZ {Corresponding angles}

In ΔXY’Z’ and ΔXYZ,

∠XYZ = ∠XY’Z   {As shown above}

∠YXZ = ∠Y’XZ’   {Common}

Therefore,

ΔXY’Z’ ∼ ΔXYZ   { By AA similarity criterion}

Therefore, 

\frac{XY’}{XY} = \frac{Y’Z’}{YZ}= \frac{XZ’}{XZ}        …. (1)

In ΔXX₇Y’ and ΔXX₅Y,

∠X₇XY’=∠X₅XY (Common)

From the corresponding angles, we will get,

∠XX₇Y’=∠XX₅Y

Hence, By the AA similarity criterion, we will get

ΔXX₂Y’ and XX₃Y

Thus, \frac{XY’}{XY} = \frac{XX_5}{XX_7}

Hence, \frac{XY }{XY’} = \frac{5}{7}        ……. (2)

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

\frac{XY’}{XY} = \frac{Y’Z’}{YZ} = \frac{XZ’}{ XZ} = \frac{7}{5}

It can be also shown as

XY’ = \frac{7}{5}XY

Y’Z’ =  \frac{7}{5}YZ

XZ’= \frac{7}{5}XZ

Thus, justified.

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

Solution:

Steps of construction:

Step 1. Construct a line segment YZ of 8 cm.

Step 2. Now construct the perpendicular bisector of the line segment YZ and intersect at the point A.

Step 3. Taking the point A as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point X.

Step 4. Join the lines XY and XZ and the triangle is the required triangle.

Step 5. Construct a ray YB which makes an acute angle with the line YZ on the side opposite to the vertex X.

Step 6. Mark the 3 points Y₁, Y₂ and Y₃ on the ray YB such that YY₁ = Y₁Y₂ = Y₂Y₃

Step 7. Join the points Y₂Z and construct a line from Y₃ which is parallel to the line Y₂Z where it intersects the extended line segment YZ at point Z’.

Step 8. Now, draw a line from Z’ the extended line segment XZ at X’, that is parallel to the line XZ, and it intersects to make a triangle.

Step 9. Hence, ΔX’YZ’ is the required triangle. 

Justification:

The construction can be justified by proving that

X’Y = \frac{3}{2}XY

YZ’ = \frac{3}{2}YZ

X’Z’= \frac{3}{2}XZ

By the construction, we will obtain X’Z’ || XZ

Therefore, 

∠ X’Z’Y = ∠XZY {Corresponding angles}

In ΔX’YZ’ and ΔXYZ,

∠Y = ∠Y (common)

∠X’YZ’ = ∠XZY

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Hence, 

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z}{XZ}

Thus, the corresponding sides of the similar triangle are in the same ratio, we get

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ} = \frac{3}{2}

Thus, justified.

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.

Solution:

Steps of construction:

Step 1. Construct a ΔXYZ with base side YZ = 6 cm, and XY = 5 cm and ∠XYZ = 60°.

Step 2. Construct a ray YA that makes an acute angle with YZ on the opposite side of vertex X.

Step 3. Mark 4 points (as 4 is greater in 3 and 4), such as Y₁, Y₂, Y₃, Y₄, on line segment YA.

Step 4. Join the points Y₄Z and construct a line through Y₃, parallel to Y₄Z intersecting the line segment YZ at Z’.

Step 5. Construct a line through Z’ parallel to the line XZ which intersects the line XY at X’.

Step 6. Therefore, ΔX’YZ’ is the required triangle.

Justification:

The construction can be justified by proving that

Since here the scale factor is \frac{3}{4}        , 

We need to prove

X’Y = \frac{3}{4}XY

YZ’ = \frac{3}{4}YZ

X’Z’= \frac{3}{4}XZ

From the construction, we will obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore,

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Thus, the corresponding sides of the similar triangle are in the same ratio, we get

Therefore,

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ}

Thus, it becomes 

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ} = \frac{3}{4}

Hence, justified.

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.

Solution:

To find ∠Z:

Given:

∠Y = 45°, ∠X = 105°

∠X+∠Y +∠Z = 180° {Sum of all interior angles in a triangle is 180°}

105°+45°+∠Z = 180°

∠Z = 180° − 150°

∠Z = 30°

Thus, from the property of triangle, we get ∠Z = 30°

Steps of construction:

Step 1. Construct a ΔXYZ with side measures of base YZ = 7 cm, ∠Y = 45°, and ∠Z = 30°.

Step 2. Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.

Step 3. Mark 4 points (as 4 is greater in 4 and 3), such as Y₁, Y₂, Y₃, Y₄, on the ray YA.

Step 4. Join the points Y₃Z.

Step 5. Construct a line through Y₄ parallel to Y₃Z which intersects the extended line YZ at Z’.

Step 6. Through Z’, construct a line parallel to the line YZ that intersects the extended line segment at Z’.

Step 7. Hence, ΔX’YZ’ is the required triangle.

Justification:

The construction can be justified by proving that

Here the scale factor is \frac{4}{3}     , we have to prove

X’Y = \frac{4}{3}XY

YZ’ = \frac{4}{3}YZ

X’Z’= \frac{4}{3}XZ

From the construction, we obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore. 

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore, 

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, 

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ}

We get,

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ} = \frac{4}{3}       

Thus, justified.

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 5/3 times the corresponding sides of the given triangle.

Solution:

Given:

The sides other than hypotenuse are of lengths 4cm and 3cm. Hence, the sides are perpendicular to each other.

Step of construction:

Step 1. Construct a line segment YZ =3 cm.

Step 2. Now measure and draw ∠= 90°

Step 3. Now taking Y as centre and draw an arc with the radius of 4 cm and intersects the ray at the point Y.

Step 4. Join the lines XZ and the triangle XYZ is the required triangle.

Step 5. Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.

Step 6. Mark 5 such as Y₁, Y₂, Y₃, Y₄, on the ray YA such that YY₁ = Y₁Y₂ = Y₂Y₃= Y₃Y₄ = Y₄Y₅

Step 7. Join the points Y₃Z.

Step 8. Construct a line through Y₅ parallel to Y₃Z which intersects the extended line YZ at Z’.

Step 9. Through Z’, draw a line parallel to the line XZ that intersects the extended line XY at X’.

Step 10. Therefore, ΔX’YZ’ is the required triangle.

Justification:

The construction can be justified by proving that

Here the scale factor is \frac{5}{3}     , we need to prove

X’Y = \frac{5}{3}XY

YZ’ = \frac{5}{3}YZ

X’Z’= \frac{5}{3}XZ

From the construction, we obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore, 

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore, 

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, 

\frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ}

So, it becomes \frac{X’Y}{XY} = \frac{YZ’}{YZ}= \frac{X’Z’}{XZ} = \frac{5}{3}

Therefore, justified.

Constructions: Exercise 11.2

Question 1. 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.

Solution:

Construction Procedure:

The construction to draw a pair of tangents to the given circle is as follows.

Step 1. Draw a circle with radius = 6 cm with centre O.

Step 2. A point P can be constructed 10 cm away from centre O.

Step 3. The points O and P are then joined to form a line

Step 4. Draw the perpendicular bisector of the line OP.

Step 5. Construct M as the mid-point of the line segment PO.

Step 6. Using M as the centre, measure the length of the line segment MO

Step 7. Now using MO as the radius, draw a circle.

Step 8. The circle drawn with the radius of MO, intersect the previous circle at point Q and R.

Step 9. Join the line segments PQ and PR.

Step 10. Now, PQ and PR are the required tangents.

Justification:

The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 6cm with centre O. This can be proved by joining OQ and OR which are represented in dotted lines.

From the construction, we can see that,

∠PQO is an angle in the semi-circle.

Every angle in a semi-circle is a right angle, therefore,

∴ ∠PQO = 90°

Now,

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle with a radius of 6 cm, PQ must be a tangent of the circle. Similarly, we can also prove that PR is a tangent of the circle. Hence, justified.

Question 2. 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.

Solution:

Construction Procedure:

For the specified circle, the tangent can be drawn as follows.

Step 1. Draw a circle of 4 cm radius with centre O.

Step 2. Taking O as the centre draw another circle of radius 6 cm.

Step 3. Locate a point P on this circle

Step 4. Join the points O and P to form a line segment OP.

Step 5. Construct the perpendicular bisector to the line OP, where M is the mid-point

Step 6. Taking M as its centre, draw a circle with MO as its radius

Step 7. The circle drawn with the radius OM, intersect the given circle at the points Q and R.

Step 8. Join the line segments PQ and PR.

Step 9. PQ and PR are the required tangents of the circle.

We can see that PQ and PR are of length 4.47 cm each.

Now, In ∆PQO,

Since PQ is a tangent,

∠PQO = 90°, PO = 6cm and 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 = 2√5

PQ = 4.47 cm

Therefore, the tangent length PQ = 4.47

Justification:

It can be proved that PQ and PR are the tangents to the circle of radius 4 cm with centre O.

Proof, 

Join OQ and OR represented in dotted lines. Now,

∠PQO is an angle in the semi-circle.

Every angle in a semi-circle is a right angle, therefore, ∠PQO = 90° s.t

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle with a radius of 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle.

Question 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q?

Solution:

Construction Procedure:

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

Step 1. Construct a circle with a radius of 3cm with centre O.

Step 2. Draw a diameter of a circle that extends 7 cm from the centre O of the circle and mark the endpoints as P and Q.

Step 3. Draw the perpendicular bisector of the constructed line segment PO.

Step 4. Mark the midpoint of PO as M.

Step 5. Draw another circle with M as centre and MO as its radius

Step 6. Now join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm.

Step 7. PA and PB are the required tangents of the circle.

Step 8. From that, QC and QD are the required tangents from point Q.

Justification:

The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 3 cm with centre O.

Proof, 

Join OA and OB. Now,

∠PAO is an angle in the semi-circle, which is equal to 90 degrees.

∴ ∠PAO = 90° s.t

⇒ OA ⊥ PA

Since OA is the radius of the circle with a radius of 3 cm, PA must be a tangent of the circle.

PB, QC, and QD are the tangent of the circle [by similar proof]. Hence, justified.

Question 4. Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°?

Solution:

Construction Procedure:

The tangents can be constructed in the following manner:

Step 1. Draw a circle with a centre O of the radius of 5 cm.

Step 2. Construct any arbitrary point Q on the circumference of the circle and join the line segment OQ.

Step 3. Also, draw a perpendicular to QP at point Q.

Step 4. Draw a radius OR, making an angle of 120° i.e(180°−60°) with OQ.

Step 5. Draw a perpendicular to the line RP at point R.

Step 6. Both the perpendicular bisectors intersect at P.

Step 7. Therefore, PQ and PR are the required tangents at an angle of 60°.

Justification:

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

We have,

∠OQP = 90°, ∠ORP = 90°

Also ∠QOR = 120°

We know that, the summation of the interior angles of a quadrilateral = 360°

Substituting values,

∠OQP + ∠QOR + ∠ORP + ∠QPR = 360^o

=> 90° + 120° + 90° + ∠QPR = 360°

Calculating, we get, ∠QPR = 60°

Hence, justified.

Question 5. Draw a line segment AB of length 8 cm. Taking A as the centre, draw a circle of radius 4 cm, and taking B as the centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle?

Solution:

Construction Procedure:

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

Step 1. Construct a line segment named AB = 8 cm.

Step 2. Taking A as the centre and draw a circle of a radius of 4 cm.

Step 3. Taking B as centre, draw another circle of radius 3 cm.

Step 4. Draw the perpendicular bisector of the line AB with M as the midpoint.

Step 5. Taking M as the centre, draw another circle with the radius of MA or MB which intersects the circle at the points P, Q, R, and S.

Step 6. Join the line segments AR, AS, BP, and BQ respectively.

Step 7. The required tangents are AR, AS, BP, and BQ.

Justification:

The construction can be justified by proving that AS and AR are the tangents of the circle with centre B and BP and BQ are the tangents of the circle with a circle centered at A.

Proof, 

Join AP, AQ, BS, and BR.

∠ASB is an angle in the semi-circle.

∴ ∠ASB = 90°

⇒ BS ⊥ AS

Now, BS is the radius of the circle. Therefore, AS must be a tangent of the circle. Similarly, AR, BP, and BQ are the required tangents of the given circle.

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

Solution:

Construction Procedure:

The tangent for the given circle can be constructed as follows

Step 1. Draw a line segment with base BC = 8cm

Step 2. Construct an angle of 90° at point B, s.t ∠ B = 90°.

Step 3. Taking B as centre, draw an arc of 6cm.

Step 4. Mark the point of intersection as A.

Step 5. Join the line segment AC.

Step 6. Now, we have ABC as the required triangle.

Step 7. Now, construct the perpendicular bisector to the line BC with the midpoint as E.

Step 8. Taking E as the centre, draw a circle with BE or EC as the radius.

Step 9. Join A and E to form a line segment.

Step 10. Now, again draw the perpendicular bisector to the line AE and the midpoint is taken as M

Step 11. Taking M as the centre, draw a circle with AM or ME as the radius.

Step 12. Both the circles intersect at points B and Q.

Step 13. Join points A and Q to form a line segment.

Step 14. Therefore, AB and AQ are the required tangents

Justification:

The construction can be justified by proving that AG and AB are the tangents to the circle.

Proof,

Join EQ.

∠AQE is an angle in the semi-circle.

∴ ∠AQE = 90°

Now, ⇒ EQ⊥ AQ

Since EQ is the radius of the circle, AQ will be a tangent of the circle. Also, ∠B = 90°

⇒ AB ⊥ BE

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

Question 7. 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 circle?

Solution:

Construction Procedure:

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

Step 1. Draw an arbitrary circle. Mark its centre as O.

Step 2. Draw two non-parallel chords such as AB and CD. 

Step 3. Draw the perpendicular bisector of AB and CD

Step 4. Taking O as the centre where both the perpendicular bisectors AB and CD intersect.

Step 5. Take a point P outside the circle.

Step 6. Join the points O and P to form a line segment.

Step 7. Now draw the perpendicular bisector of the line PO and mark its midpoint as M.

Step 8. Taking M as centre and MO as the radius draw a circle.

Step 9.  Both the circles intersect at the points Q and R.

Step 10. Now join PQ and PR. 

Step 11. Therefore, PQ and PR are the required tangents.

Justification:

The construction can be justified by proving that PQ and PR are the tangents to the circle. The perpendicular bisector of any chord of the circle passes through the centre. Now, join the points OQ and OR.

Both the perpendicular bisectors intersect at the centre of the circle. Since ∠PQO is an angle in the semi-circle. 

∴ ∠PQO = 90°

⇒ OQ⊥ PQ

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

∴ ∠PRO = 90°

⇒ OR ⊥ PO

Similarly, since OR is the radius of the circle, PR has to be a tangent of the circle. Therefore, PQ and PR are the tangents of a circle.

Important Points to Remember:

• These NCERT solutions are developed by the GfG team, with a focus on student benefit.
• These solutions are accurate and can be used by students to prepare for their board exams. 
• Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

FAQs on NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Q1: Why is it important to learn constructions?

Answer:

With the help of geometric construction, we can create angles, bisect lines, draw line segments, and all the geometric shapes and practise of these constructions help us in future roles.

Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 11 Constructions?

Answer:

NCERT Solutions for Class 10 Maths Chapter 11 Constructions covers topics such construction of figures with the help of ruler and compass , how to divide line into particular ratio etc.

Q3: How can NCERT Solutions for Class 10 Maths Chapter 11 Constructions help me?

Answer:

NCERT Solutions for Class 10 Maths Chapter 11 Constructions can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

Q4: How many exercises are there in Class 10 Maths Chapter 11 Quadratic Equations?

Answer:

There are 2 exercises in the Class 10 Maths Chapter 11 Constructions which covers all the important topics and sub-topics.

Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 11 Constructions?

Answer:

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.