NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Construction according to the required data is done as follows:

Step1: Draw a line segment AB=7.6 cm and at point A draw a ray AC making acute angle with AB.

Step 2: As shown in figure, start locating 13 marks (5+8) as \(A_1,A_2,A_3,.....A_{13} \), such that \(AA_1=AA_2=AA_3 \)..and so on.

Step 3: Now join the points \(A_{13} \) and B.

Step 4: From point \(A_5 \), draw a line parallel to \(A_{13}B\) and mark the intersection point as P.

Step 5: Measure AP and PB. we will find AP= 2.9cm and PB =4.7cm(approx) which are in approx ration of 5:8.

**Justification:**

In \(ΔAA_5P and △AA_13B \) we have,

\(A_5P ∥ A_{13}B\)

\(\frac{AP}{BP}=\frac{AA_5}{A_5A_{13}} \)

(By the Basic proportionality theorem)

\(\frac{AP}{BP}=\frac{5}8 \) {∵ \(\frac{AA_5}{A_5A_{13}}=\frac{5}8 \)}

\(\therefore \) , AP:BP=5:8

Q2 ) Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3} \) of the corresponding sides of the first trianngle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of Construction from the given data is done in the following steps :

Step 1. Draw a line segment BC=6 cm.

Step 2. With B as a centre draw an arc of radius equal to 5 cm.

Step 3. With C as centre draw an arc of radius equal to 4 cm, such that this arc intersects the previously drawn arc at A.

Step 4. Join both AB and AC,such that we get ΔABC, which is our required triangle.

Step 5. Below the line segment BC, make an acute angle CBX.

Step 6. Along BX, locate three points: \(B_1,B_2 and B_3 \)such that \(BB_1=B_1B_2=B_2B_3 \)

Step 7. Join \(B_3C \)
Step 8. From \(B_2\), draw a line \(B_2D\) ∥ \(B_3C \)such that it meets BC at D.

Step 9. From D, draw ED ∥AC such that it meets BA at E. Then,
EBD is the required triangle whose sides are \(\frac{2}3\) rd of the corresponding sides of ΔABC.

**JUSTIFICATION:**

By construction we have

\(\frac{BD}{DC}=\frac{2}1\)

Therefore, \(\frac{BC}{BD}=\frac{BD+DC}{BD}=1+\frac{DC}{BD}\)

\(\frac{BC}{BD}=\frac{3}2 so \frac{BD}{BC}=\frac{2}3\)

Also we have DE and AC as parallel

Therefore, ΔABC~ΔEBD(By AA property)

{\(\angle \) D=\(\angle \) C (by construction) and \(\angle \) A is common}

So \(\frac{EB}{AN}=\frac{BD}{BC}=\frac{DE}{CA}=\frac{2}3\)

Hence, we get the new triangle whose sides are equal to
\(\frac{2}{3} \) rd of the corresponding sides of \(\triangle \) ABC

Q3 ) Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5} \) of the corresponding sides of the first triangle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction by the given data are:

Step 1. Draw a \(\triangle \) ABC with sides as AB=5 cm, BC= 7 cm, AC= 6cm.

Step 2. At point B, draw an acute angle CBX below BC.

Step 4. Join \(B_5\) and C.

Step 5.Draw \(B_7C'\) parallel to \(B_5C\), where C` is a point on extended line BC.

Step 6. Draw A'C' parallel to AC, where A' is a point on extended line BA.

Hence,A’BC’ is the required triangle.

**JUSTIFICATION:**

In \(\triangle \) ABC and \(\triangle \) A’BC’,

AC || A’C’

\(\frac{AB}{A’B}=\frac{BC}{BC’}\)

(By the basic Proportionality Theorem)...(i)

In \(ΔBB_5C \)and \(ΔBB_7C’\),

\(B_5C || B_7C’\)

\(\frac{BC}{BC’}=\frac{BB_5}{BB_7}=\frac{5}7\) (By the basic Proportionality Theorem)...(ii)

Equating (i) and (ii) \(\frac{AB}{A’B}=\frac{BB_5}{BB_7} \Rightarrow \frac{AB}{A’B}=\frac{5}7 \)

\(\therefore A’B=\frac{7}5AB\)

\(\therefore \) Sides of new triangle is \(\frac{7}5 \)times the corresponding sides of the first triangle.

Q4 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct figure by given data are as follows:

Step 1: Construct an isosceles triangle ABC with BC=8cm and altitude AD=4cm.

Step 2:Draw a ray BX, making an acute angle with BC.

Step 3: Locate 3 points on BX, such that BP=PQ=QR.

Step 4: Join QC.

Step 5: Through R, draw a line RC parallel to OC, meeting produced line BC at C’.

Step 6: Through C, draw a line CA parallel to CA, meeting the produced line BA at A’.

Thus, \(\triangle \) A’BC’ is the required isosceles triangle

**JUSTIFICATION:**

In \(\triangle \) ABC and \(\triangle \) A’BC’, we have

\(\triangle ACB=\triangle A’C’B \)(corresponding angles)

\(∠B=∠ B \)(common)

\(∴ ΔABC∼ΔA’BC’ \) (By AA similarity)

\(∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’} \)

But, \(\frac{BC}{BC’}=\frac{BQ}{BR}=\frac{2}3 ∴ \frac{BC}{BC’}=\frac{2}3 \)

\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{B’C}{BC}=\frac{3}2.\)

Hence, we get the required triangle whose sides are \(\frac{3}{2} \)
, i.e., \(1\frac{1}{2}\) times of the corresponding sides of the isosceles \(\therefore \) ABC.

Q5 ) Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \(\angle \)ABC=60° ;. Then construct a triangle whose sides are \(\frac{3}4\) of the corresponding sides of the triangle ABC.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Step 1: Draw a line segment BC=6cm and at point B draw an angle ABC=60°;.

Step 2: Cut AB=5cm and join AC so we obtain \(\angle \) ABC.

Step 3: Draw a ray BX making an acute angle with BC on the side opposite to the A.

Step 4: Locate 4 points \(A_1,A_2,A_3 \)and \(A_4\) on the ray BX so that \(BA_1=A_1A_2=A_2A_3=A_3A_4\).

Step 5: Join \(A_4\) to C.

Step 6: At \(A_3\), draw \(A_3C’ || A_4C\), where C’ is a point on the line segment BC.

Step 7: At C’, draw C’A’ || CA, where A’ is a point on the line segment BA.

\(\therefore \triangle \) A’BC’ is the required triangle.

**JUSTIFICATION:**

In \(ΔA’BC’ and ΔABC\),

\(A’C’ ∥AC\)

\(\frac{A’B}{AB}=\frac{BC’}{BC} \) (By the basic proportionality Theorem)...(i)

In \(ΔBA_3C’ and ΔA_4C\),

\(A_3C’ ∥A_4C \)

\(\frac{BC’}{BC}=\frac{BA_3}{BA_4}=\frac{3}4 \) (By the Basic Proportionality Theorem)

\(∴\frac{BC’}{BC}=\frac{3}4 \) …........(ii)

From(i) and (ii), we get

\(\frac{A’B}{AB}=\frac{3}4 \Rightarrow A’B=\frac{3}4 AB\)

\(\therefore \) Sides of the new triangle formed are \(\frac{3}{4} \) times the corresponding sides of the first triangle.

Q6 ) Draw a triangle ABC with side BC = 7 cm, \(\angle \) B=45°, \(\angle \) A=105°;. Then construct a triangle whose sides are \(\frac{4}3 \) times the corresponding sides of \(\triangle \) ABC.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the figure by given data:

Step 1: Draw a line segment BC=7 cm.

Step 2: Draw \(\angle \) ABC=45° ; and \(\angle ACB=30° \) ; ,i.e, \(\angle BAC=105° \) ;

Step 3: We get \(\triangle \) ABC.

Step 4: Now, draw a ray BX making an acute angle with BC.

Step 5: Mark 4 points \(B_1,B_2,B_3 \) and \(B_4\) on the ray BX so that \(BB_1=B_1B_2=B_2B_3=B_3B_4 \) and Join \(B_3C\).

Step 6: Through \(B_4\), draw a line \(B_4C’\) parallel to \(B_3C\), intersecting the extended line segment BC at C’.

Step 7: Through C’, draw a line A’C’ parallel to CA, intersecting the extended line segment BA at A’.

Thus, ΔA’BC’, is the required triangle.

**JUSTIFICATION:**

In \(\triangle A’BC’ and \triangle \) ABC,

\(\angle ABC=\angle A’BC’ \)(common)

\(\angle ACB= \angle A’C’B\) (corresponding angles)

\(\therefore \triangle ABC ~ A’BC’ \) (By AA similarity)

\(\therefore \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’}\)

But, \(\frac{BC}{BC’}=\frac{BB_3}{BB_4}=\frac{3}4 ∴ \frac{BC’}{BC}=\frac{4}3 \)

\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{BC’}{BC}=\frac{4}3 .\)

\(\therefore \) Sides of the new triangle formed are \(\frac{4}{3} \) times the corresponding sides of the first triangle.

Q7 ) 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 \(\frac{5}3\)times the corresponding sides of the given triangle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of Construction the figure by the given data:

Step 1. Construct a \(\triangle \) ABC, with BC = 4 cm, CA = 3 cm and ∠BCA = 90°.

Step 2. By making an acute angle with BC, draw a ray BX .

Step 3. Mark five points \(B_1, B_2, B_3, B_4 \)and \(B_5 \)on BX, such that \(BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5\) and join \(B_3C\).

Step 6. Through \(B_5\), draw \(B_5C’\) parallel to \(B_3C\) intersecting BC produced at C’.

Step 7. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.

Thus, \(\triangle \) A’BC’ is the required right triangle.

**JUSTIFICATION:**

In \(ΔA’BC’ and ΔABC\),

\(∠ABC=∠ A’BC’ \)(common)

\(∠ACB=∠ A’C’B\) (corresponding angles)

\(∴ ΔABC∼ΔA’BC’ \) (By AA similarity)

\(∴ \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’}\)

But, \(\frac{BC}{BC’}=\frac{BB_3}{BB_5}=\frac{3}5 ∴ \frac{BC’}{BC}=\frac{5}3 \)

\(⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{BC’}{BC}=\frac{5}3 .\)

\(\therefore \) Sides of the new triangle formed are \(\frac{5}{3} \) times the corresponding sides of the first triangle.

Q.1In each of the following, give the justification of the construction also:

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

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

(2) Draw an acute angle BAX on base AB. Mark the ray AX.

(3) Locate 13 point A1, A2, A3........A13 on the ray AX so that AA1=A1A2=A2A3........

(4) join A13 with B and A5 draw a line || to BA3 that is A5C . The line intersects AB at C .

(5) On measure AC = 2.9 cm and BC = 4.7 cm

A5C || A13B

\(\frac{AC}{BC} = \frac{AA5}{A5A13} = \frac{5}{8} \)

\(\therefore \) AC : BC = 5:8

Q.2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3} \) of the corresponding sides of the first triangle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

2. Draw two arc 5cm from A and 4 cm from C are intersecting each other on B . Join BA and BC .

3. Draw AX making an acute angle with AC.

4. Locate 3 point P,Q and are on AY such that AP = PQ= QR

5. Join CR and darw a line QC' which is || to CR

6. Draw an angle on AC'B' which is equal to angle ACB

7. Therefore, \(\triangle \)AB’C’ is the required triangle.

Q.3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5} \) of the corresponding sides of the first triangle

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Step of construction:
1. Draw a triangle which has AB = 5 cm, BC = 6cm , and CA = 7 cm

2. Draw an acute angle CBX below BC at point B .

3. Mark the ray BX as B1, B2 B3,B4 ....B7 such that BB1= B1B2................B6=B7

4.join B5 to C

5. Draw B7 C' || to B5C , where C' is extended line BC.

6. Draw A'C'|| AC where A' is a point on extended line BA.

7. A'BC' is required triangle

Q.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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Step of construction

1. Draw a line segment BC with the measure of 8 cm.

2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D

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

4. Now join the lines AB and AC and the triangle is the required triangle.

5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

6. Locate the 3 points P, Q and R on the ray BX such that BP= PQ = QR

7. Join the points B2C and draw a line from B3 which is parallel to the line B2C where it intersects the extended line segment BC at point C’.

8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.

9. Therefore, \(\triangle \) A’BC’ is the required triangle.

Q.5 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \(\angle \)ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4} \) of the corresponding sides of the triangle ABC.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

1. Draw a \(\triangle \)ABC with base side BC = 6 cm, and AB = 5 cm and \(\angle \)ABC = 60°.

2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 3 and 4), such as A1, A2, A3, A4, on line segment BX.

4. Join the points A4C and also draw a line through A3, parallel to A4C intersecting the line segment BC at C’.

5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.

6. Therefore, \(\triangle \) A’BC’ is the required triangle.

Q.6 Draw a triangle ABC with side BC = 7 cm, \(\angle \) B = 45°, \(\angle \) A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3} \)times the corresponding sides of \(\triangle \)ABC.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

To find \(\angle \)C:

Given:

\(\angle \)B = 45°, \(\angle \)A = 105°

We know that,

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

\(\angle A+\angle B +\angle C = 180° \)

105°+45°+\(\angle \)C = 180°

\(\angle \)C = 180° ? 150°

\(\angle \) C = 30°

So, from the property of triangle, we get \(\angle \)C = 30°

The required triangle can be drawn as follows.

1. Draw a \(\triangle \)ABC with side measures of base BC = 7 cm, \(\angle \)B = 45°, and \(\angle \)C = 30°.

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

3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.

4. Join the points B3C.

5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.

6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.

7. Therefore, \(\triangle \)A’BC’ is the required triangle.

Q.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 \(\frac{5}{3} \) times the corresponding sides of the given triangle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of Construction:

1. Construct a \(\triangle \)ABC, such that BC = 4 cm, CA = 3 cm and \(\angle \) BCA = 90°

2. Draw a ray BX making an acute angle with BC.

3. Mark five points B1, B2, B3, B4 and B5 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.

4. Join B3C.

5. Through B5, draw B5C’ parallel to B3C intersecting BC produced at C’.

6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.

Thus, \(\triangle \) A’BC’ is the required right triangle.

Q1 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1: Draw a circle with center O and radius =6cm and draw a point P such that OP=10cm.

Step 2: Draw a perpendicular bisector of OP and mark a point M such that M is the mid- point of OP.

Step 3: With center as M and radius PM=MO, draw a circle which cuts the given circle at S and T.

Step 4: Now join PS and PT.

Thus, we have the required tangents as PS and PT and also the lengths of each tangent is 8cm.

**JUSTIFICATION:**

Join OS

Now in the triangle PSO,we have

\(∠PSO=90°\)

\(∴ PS= \sqrt{{OP}^2 -{OS}^2} \)

[by pythagoras’ theorem]

\(=\sqrt{{(10)}^2-{(6)}^2}=\sqrt{100-36} \)

\(=\sqrt{64}=8cm \)

Q2 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1: Draw concentric circles C and C’ of radius 4cm and 6cm respectively and mark the center as O.

Step 2: Mark two points as A and P such that O,A and P lie on the same line.

Step 3:Draw perpendicular bisector of OP which intersects OP at O’.

Step 4: By taking O’ as the center, draw a circle of radius OO’ which intersects the circle C at point T and Q.

Step 5: Now join PT and PQ.

Step 6: PT and PQ are our required tangents whose length is 4.5cm approx.

**JUSTIFICATION:**

Join OT and OQ,

We have OT ⊥ PT [Radius ⊥ to tangent ]

In the right angled triangle OTP,

\({OP}^2={OT}^2+{PT}^2 \)

\({(6)}^2={(4)}^2+{(PT)}^2 \)

\({PT}^2=36-16=20 \)

\(PT=\sqrt{20}=2\sqrt{5} cm\)

Similarly \(PQ=2\sqrt{5} cm \)

A pair of tangents can be drawn from an external point outside the circle and these tangents have equal length.

\(\therefore \) PT=PQ

Q3 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1:Draw a circle of radius 3cm and mark it’s center as O.

Step 2:Extend the diameter in both the direction as shown in the figure and mark the points as P and Q such that OP=OQ=7 cm.

Step 3: Mark the midpoint of OP and OQ as M and M’ respectively.

Step 4:With these points as M and M’, draw circles of radius MP and MO respectively.

Step 5: Circles with the center M intersects the first circle at R and S and the circle with center M’ intersects the first circle at T and U.

Step 6: Join PR,PS,QT and QU

Thus we have got the pair of tangents as PR and PS from point P and OT and QU as another pair of tangent from point Q drawn to the circle.

**JUSTIFICATION:**

Join OR and we have in triangle PRO,

∠PRO=90° [ Angle in a semicircle]

Also OR is the radius of the circle with center O.

\(\therefore \)
Line PR\(\perp \) OR

We know that a line drawn through the end of a radius and perpendicular to it is a tangent to the circle. Hence, we have PR as the tangent to the point R and similarly ,PS,QT and QU as the tangents at the points S,T and U respectively.

Q4 ) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1:Draw a circle of radius 5 cm and mark the center as O.

Step 2: As we have angle between the tangents as 60°

\(\therefore \) By quadrilateral property we have angle between the radii of circle is 120°

Step 3:Draw two radius OA and OB such that they have angle of 120°

Step 4:At points A and B, draw 90°angles and mark the point P where the arms of these angles intersect.

Step 5:we get PA and PB as our required tangents.

**JUSTIFICATION:**

Since, OA is the radius , so PA has to be a tangent to the circle.

Similarly, PB is the tangent to the circle.

\(\angle APB=360° - \angle OAP - \angle OBP - \angle AOB \)

= 360° - 90° - 90° - (180° - 60°)

= 360° – 360° +60° = 60°

Thus the tangents PA and PB are inclined to each other at an angle of 60°

Q5 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1: Draw a line segment AB of 8 cm.

Step 2. Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.

Step 3. With M as centre which is the center of line AB, draw a circle of radius MA or MB, intersecting circle with centre B at R and S, circle with center as A at P and Q.

Step 4. Join AR,AS, BP and BQ and hence we got the required tangents.

** JUSTIFICATION:**

On joining BP, we get \(\angle \) BPA=90° ; [\(\therefore \) BPA is the angle in the semi-circle]

Therefore, AP \(\perp \) BP

So,AP is the tangent to the circle because BP is the radius of the given circle.

Similarly, AR,BQ and AS are the tangents.

Q6 ) Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \(\angle B=90° \) ;. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure by the given data:

Step 1: Draw a \(\triangle \) ABC, in which AB=6cm, BC=8cm and \(\angle B=90° \) ;

Step 2: Draw BD perpen AC and also draw the perpendicular bisectors of BC and BD.They meet at point O’.

Step 3:Taking O’ as the center, draw a circle of radius O’B. This circle passes through B,C and D.

Step 4:Join O’A and draw a perpendicular bisector of O’A. This bisector meets O’A at K.

Step 5:Taking K as centre, draw an arc of radius KO’ intersects the previous circle C’ at T.

Step 6:Join AP and AQ and thus we got our required tangents.

Q7 ) 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps to construct the required figure:

Step 1: Draw a circle with a bangle.

Step 2:Take two non- parallel chords as AB and CD of the circle.

Step 3: Draw the perpendicular bisector of these chords intersecting each other at O,which is the center of the circle made by bangle.

Step 4:Take a point P outside the circle and join OP.

Step 5: Mark the midpoint M of OP.

Step 6: Draw a circle with center as M and radius as MP=OM,which will intersect the first circle at R and Q.

Step 7:Join PQ and PR and hence we got the required tangents.

Q.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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:
1. Draw a circle with radius = 6 cm with centre O.

2. Locate a point P, which is 10 cm away from O.

3. Join the points O and P through line

4. Draw the perpendicular bisector of the line OP.

5. Let M be the mid-point of the line PO.

6. Take M as centre and measure the length of MO

7. The length MO is taken as radius and draw arc on circle on S and T .

8. Join PS and PT.

10. Therefore, PQ and PR are the required tangents.

Q.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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:
1. Draw concentric circle of radius 4 cm and 6 cm having same centre .

2. Locate a point P on the circle whose radius is 6 cm.

3.Join the points O and P through lines such that it becomes OP.

4. Draw the perpendicular bisector to the line OP

5. Let O' be the mid-point of PO.

6. Draw arc with centre M as on the small circle on R and Q

7. Join PQ and PR.

10. PQ and PR are the required tangents.

Q.3 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

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Step of construction:

1. Draw a circle with centre O and radius 3 cm

2. Produce the diameter of the circle to both the ends up to P and Q such that OP = OQ = 7cm

3. Draw the perpendicular bisector of the line PO and QO
and mark the midpoint as M and M' respectively

4. Draw arc with centre M on circle on R and S

5. Draw arc with centre M' on circle on T and U

6. PR , PS , QT and QU are the required tangents.

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

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:
1. Draw a circle of radius 5cm .

2. As tangents are inclined to each other at an angle 60 °

\(\therefore \) angle between the radii of circle is 120° ( \(\because \) the sum of all angle if quadrilateral is 360° )

3. draw radii OA and OB inclined to each other at angle 120°
4. At point A and B draw 90° angle. The arm of these angle interest at point P.

5. PA and PB are the required tangents.

Q.5 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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Draw a line segment AB = 8 cm.

2. Take A as centre and draw a circle of radius 4 cm

3. Take B as centre, draw a circle of radius 3 cm

4. Draw the perpendicular bisector of the line AB and the midpoint is taken as M.

5. Now, take M as centre draw a arc with the radius of MA or MB which the intersects the circle at the points P, Q, R and S.

6. Now join AR, AS, BP and BQ

7. Therefore, the required tangents are AR, AS, BP and BQ

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

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Dwar a right triangle ABC with AB = 6 cm, BC = 8 cm and \(\angle \) B = 90°

2. From B draw BD perpendicular to AC.

3. Draw perpendicular bisector of BC at point O'.

4. Take O' as a centre and OB as radius as radius, draw a circle passes through point B and C and D

5. Join O'A and draw perpendicular bisector O'A which intersect O'A.

6. Take that mid point centre, draw an arc on circle on point T.

7. Join AT, AT is required tengent.

Q.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.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Answer :

Steps of construction:

1. Draw a circle with bangle.

2. Take two non-parallel chords AB and CD of the circle.

3. Draw perpendicular bisector of these chord intersecting each other at O which is the centre of the circle.

4. Take a point P outside of the circle.

5. Join OP and mark the mid point M of OP.

6. With M as a centre and radius MP draw arc on the circle at the point Q and R

7. Join PQ and PR

There are total 28 questions present in ncert solutions for class 10 maths chapter 11 constructions

There are total 4 long question/answers in ncert solutions for class 10 maths chapter 11 constructions

There are total 2 exercise present in ncert solutions for class 10 maths chapter 11 constructions