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Construction Class 10 NCERT Solutions for Maths Chapter 11 are made by our team of subject experts at Toppr in a detailed manner. NCERT Solutions for class 10 maths chapter 11 Constructions are curated strictly in accordance with the CBSE Curriculum and the exam pattern. The NCERT textbook questions are answered in a way to provide you with a better understanding of the concepts in a systematic and step-by-step manner. It also contains the appropriate diagrams to explain the concepts to the students. As Class 10 exams are Board exams, NCERT Solutions provided by Toppr are the best study material to excel in the exams. These solutions will not only help the students in preparing for the board exams but also for the Olympiads. Also, the MCQs and long and short questions are all answered according to the weightage and the exam pattern. With the help of NCERT Solutions for Class 10 Maths Chapter 11 Constructions you can also analyze your shortcomings and work on them before the exams. These are the best resources designed after proper research and study to assist the students in scoring good marks.

Table of Content

Exercise 11.1

Question 1

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

Solution

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$Dividealinesegmentof$$7.6cm,$$lengthintheratio$ $5:8$

$andmeasure$.

$m:n=5:8$

$m+n=5+8=13$

$AC:CB=5:8$

$bymeasurement$ $:AC=3cm,CB=4.6cm$.

$Stepsofconstruction$:

$1.Drawanyray$ $AB$, $suchthat$ $AB=7.6cm$

$2.Draw$$A$$X$$rayatpoint$ $A$ $makingacuteangle$.

$3$.$Locatethepoints$ $A_{1},A_{2},A_{3},........A_{13}$ $from$ $A$.

$4$. $Join$ $BA_{13}$.$Draw$ $BA_{13}∣∣A_{5}C$.

$Now$ , $AC:CB=5:8$

$Ifmeasured$ ,$AC=3cm,CB=4.6cm$.

$m:n=5:8$

$m+n=5+8=13$

$AC:CB=5:8$

$bymeasurement$ $:AC=3cm,CB=4.6cm$.

$Stepsofconstruction$:

$1.Drawanyray$ $AB$, $suchthat$ $AB=7.6cm$

$2.Draw$$A$$X$$rayatpoint$ $A$ $makingacuteangle$.

$3$.$Locatethepoints$ $A_{1},A_{2},A_{3},........A_{13}$ $from$ $A$.

$4$. $Join$ $BA_{13}$.$Draw$ $BA_{13}∣∣A_{5}C$.

$Now$ , $AC:CB=5:8$

$Ifmeasured$ ,$AC=3cm,CB=4.6cm$.

Question 2

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $53 $ time of the corresponding sides of the given triangle.

Solution

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1) Draw$BC=4cm$

2) Cut arc of $5$ cm from B and arc of $6$ cm from C. Mark point of intersection as A.

3) Draw BD making an acute angle with BC. Mark 5 equal arcs on BD.

4) Join $B_{5}$ to C, draw a line $B_{3}C_{′}$ parallel to $B_{5}C$.

5) Draw a line $C_{′}A_{′}$ parallel to $CA$.

$A_{′}BC_{′}$ ia the required triangle.

Question 3

Construct a triangle of sides $5cm,6cm$ and $7cm$, then construct a triangle similar to it, whose sides are $32 $ of corresponding sides of the first triangle.

Solution

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Making the triangle

Draw a line segment AB = $5$ cm

Draw an arc at $6$ cm from point A.

Draw an arc at $7$ cm from point B so that it intersects the previous arc.

Joint the point of intersection from A and B.

This gives the required triangle ABC.

Dividing the base:

Draw a ray AX at an acute angle from AB

Plot three points on AX so that; $AA_{1}$ = $A_{1}A_{2}$ = $A_{2}A_{3}$

Join $A_{3}$ to B.

Draw a line from point $A_{2}$ so that this line is parallel to $A_{3}$B and intersects AB at point B'

Draw a line from point B' parallel to BC so that this line intersects AC at point C'

$⇒△AB_{′}C_{′}$ is the required triangle

Draw a line segment AB = $5$ cm

Draw an arc at $6$ cm from point A.

Draw an arc at $7$ cm from point B so that it intersects the previous arc.

Joint the point of intersection from A and B.

This gives the required triangle ABC.

Dividing the base:

Draw a ray AX at an acute angle from AB

Plot three points on AX so that; $AA_{1}$ = $A_{1}A_{2}$ = $A_{2}A_{3}$

Join $A_{3}$ to B.

Draw a line from point $A_{2}$ so that this line is parallel to $A_{3}$B and intersects AB at point B'

Draw a line from point B' parallel to BC so that this line intersects AC at point C'

$⇒△AB_{′}C_{′}$ is the required triangle

Question 4

Draw a line segment PQ= 8 cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meets PQ at point R. Measure the length of PR and QR. Is PR=QR?

Solution

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Steps of Construction :

$1.$ With $P$ and $Q$ as centers, draw arcs on both sides of $PQ$ with equal radii. The radius should be more than half the length of $PQ$.

$2.$ Let these arcs cut each other at points $R$ and $RS$

$3.$ Join $RS$ which cuts $PQ$ at $D$. Then $RS=PQ$. Also $∠POR=90_{∘}$.

Hence, the line segment $RS$ is the perpendicular bisector of $PQ$ as it bisects $PQ$ at $P$ and is also perpendicular to $PQ$. On measuring the lengths of $PR=4$cm, $QR=4$ cm Since $PR=QR$, both are $4$cm each

$∴PR=QR$.

Question 5

Draw a triangle ABC with $BC=6$cm, AB$=5$cm and $∠$ABC $=60_{o}$. Then construct a triangle whose sides are $43 $ of the corresponding sides of the $Δ$ ABC.

Solution

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The $ΔA_{′}BC_{′}$ whose sides are $43 $ of the corresponding sides of $ΔABC$ can be drawn as follows:

**Step 1:** Draw a $ΔABC$ with side $BC=6cm,AB=5cm,∠ABC=60_{∘}$

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

**Step 3:** Locate $4$ points, $B_{1},B_{2},B_{3},B_{4}$ on line segment $BX$.

**Step 4:** Join $B_{4}C$ and draw a line through $B_{3}$, parallel to $B_{4}C$ intersecting $BC$ at $C_{′}$.

**Step 5:** Draw a line through $C_{′}$ parallel to $AC$ intersecting $AB$ at $A_{′}$.

The triangle $A_{′}BC_{′}$ is the required triangle.

Question 6

Draw a triangle $ABC$ with side $BC=7cm,∠B=45_{o},∠A=105_{o}.$ Then construct a triangle whose side are $34 $ times the corresponding sides of $△ABC.$ Write the construction.

Solution

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In order to construct $△ABC$, we follow the following steps :

$STEPI$ Draw $BC=7cm$.

$STEPII$ At $B$ construct $∠CBX=45_{o}$ and at $C$ construct

$∠BCY=180_{o}−(45_{o}−105_{o})=30_{o}$

$∠BCY=180_{o}−(45_{o}−105_{o})=30_{o}$

Suppose $BX$ and $CY$ intersect at $A.△ABC$ so obtained is the given triangle.

To construct a triangle similar to $△ABC$, we follow the following steps.

$STEPI$ Construct an acute angle $∠CBZ$ at $B$ on opposite side of vertex $A$ of $△ABC$.

$STEPII$ Mark-off four (greater $4$ of and $3$ in $34 $) points $B_{1},B_{2},B_{3},B_{4}$ on $BZ$ such that

$BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}$.

$BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}$.

$STEPIII$ Join $B_{3}$ (the third point) to $C$ and draw a line through $B_{4}C$ parallel to $B_{3}C$, intersecting the extended line segment $BC$ at $C$.

$STEPIV$ Draw a line through $C$ parallel to $CA$ intersecting the extended line segment $BA$ at $A$.

Triangle $ABC$ so obtained is the required triangle such that

$ABAB =BCBC =ACAC =34 $

Question 7

Draw a right triangle which the sides (other than hypotenuse) are of lengths $4cm$ and $3cm$. Then construct another triangle whose sides are $35 $ times the corresponding sides of the given triangle.

Solution

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Construct $△ABC$, the given right angled triangle, as follows:

Step 1: Draw base $BC$ of length $4cm$.

Step 2: Draw a right angle at point $B$.

Step 3: Draw an arc of radius $3cm$ on the right angle drawn from $B$ and label that point as $A$

Step 4: Join $A−C$.

$△ABC$ is the given triangle.

To construct $△EBD∼△ABC$:

Step 1: Draw a ray $BX$ at an acute angle to line $BC$ on the opposite side of $A$.

Step 2: Mark $5$ equidistant points $B_{1},B_{2},B_{3},B_{4},B_{5}$ on ray $BX$.

Step 3: Join $B_{3}$ and $C$.

Step 4: Draw $B_{5}D$ parallel to $B_{3}C$ and label the intersection on the extension of $BC$ as $D$.

Step 5: Draw $DE$ parallel to $BA$ and label the intersection with the extension of $BA$ as $E$

$△EBD$ is the required triangle.

Exercise 11.2

Question 1

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

Solution

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Steps of Construction:

1. A circle with radius $6cm$ is drawn taking $O$ as centre

2. Point $P$ is marked at $10cm$ away from centre of circle.

3. With the half of compass mark $M$ which is the midpoint of $OP$.

4. Draw a circle with centre $M$, taking radius $MO$ or $MP$ which intersects the given circle at $Q$ and $R$.

5. Now join $PQ$ and $PR$. These are the tangents of the circle.

1. A circle with radius $6cm$ is drawn taking $O$ as centre

2. Point $P$ is marked at $10cm$ away from centre of circle.

3. With the half of compass mark $M$ which is the midpoint of $OP$.

4. Draw a circle with centre $M$, taking radius $MO$ or $MP$ which intersects the given circle at $Q$ and $R$.

5. Now join $PQ$ and $PR$. These are the tangents of the circle.

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

$∴$In $△OPQ,OQ⊥QP$

and in $△OPR,OR⊥PR$

Hence, both $△OPQ$ and $△OPR$ are right angle triangles.

Applying Pythagoras theorem to both $△s$, we get:

$OP_{2}=OQ_{2}+PQ_{2}$

and $OP_{2}=OR_{2}+PR_{2}$

$OQ=OR=radius=6cm$ and $OP=10cm$

$∴10_{2}=6_{2}+PQ_{2}$

and $10_{2}=6_{2}+PR_{2}$

$⇒PQ_{2}=100−36=64$

and $PR_{2}=100−36=64$

$∴PQ=PR=8cm$

Hence, the length of the tangents to a circle of radius $6cm$, from a point $10cm$ away from the centre of the circle, is $8cm$.

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

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$(1)$Draw two concentric circle $C_{1}$ and $C_{2}$ with common center

0 and radius $4cm$ and $6cm$

$(2)$ Take a point P on the outer circle $C_{2}$ and join OP.

$(3)$ Draw the bisector of OP which bisect OP at M'.

$(4)$ Taking M' as center and OM' as radius draw a dotted circle which

cut the inner circle $C_{1}$ at two point M and P.

$(5)$ Join PM and PP'. Thus, PM and PP' are required tangent.

On measuring PM and PP'.

$PM=PP_{′}=4.4cmBycalculation:InΔOMP,∠PMO=90_{0}PM_{2}=OP_{2}−OM_{2}(bypythagorastheorem)PM_{2}=(6)_{2}−(4)_{2}=36−16=20PM_{2}=20cmPM=20 =4.4cmHence,thelengthofthetangentis4.4cm $

Question 3

Draw a circle of radius $3cm$. Take two points $P$ and $Q$ on one of its extended on both sides of its centre, each at a distance of $7cm$ on opposites of its centre. Draw tangents to the circle from each of these two points $P$ and $Q$

Solution

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Step 1: Draw a circle with centre $O$ and radius $3cm$ using a compass.

Step 2: Draw a secant passing through the centre. Mark points $P$ and $Q$ on opposite sides of the centre at a distance of $7cm$ from $O$.

Step 3: Place the compass on $P$ and draw two arcs on opposite sides of $OP$. Now place the compass on $O$ and draw two arcs intersecting the arcs drawn from point $P$.

Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of $OP$. Mark the mid point of $OP$ as $M_{1}$

Step 5: From $M_{1}$ draw a circle with radius $=M_{1}P=M_{1}O$

Step 6: Mark the intersection points of the circle drawn from $M_{1}$ with the circle drawn from $O$ as $A$ and $B$.

Step 7: Join $P−A$ and $P−B$

Step 8: Repeat steps $3$ to $7$ for point $Q$ and obtain tangents $QC$ and $QD$

$PA,PB,QC$ and $QD$ are the required tangents.

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Practice more questions

- The locus of a point that moves at an identical distance from two points, is normal to the line joining both the points.
- Perpendicular or Normal means right angles.
- A bisector cuts a line segment in two half.

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Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Question 1. What is meant by the scale factor in the constructions of triangles?

Answer. With respect to the construction of triangles, the scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Question 2. Write a short note on the Construction of Tangents to a Circle.

Answer. There cannot be a tangent to the circle through a point that lies inside a circle. But, if a point lies on the circle, then only one tangent to the circle is possible at this point. This tangent is perpendicular to the radius of the circle through this point. Thus, if we want to draw a tangent at a point on a circle, we can simply draw the radius through this point and draw a line perpendicular to this radius through this point. We will therefore get the required tangent at the point. However, if a point lies outside the circle, there will be two tangents to the circle from this point.

Question 3. What types of constructions are covered in Chapter 11 – Constructions?

Answer. Chapter 11 – Constructions covers the following constructions:

- Dividing a line segment in a given ratio.
- Constructing a triangle similar to a given triangle as per a given scale factor which may be less than 1 or greater than 1.
- Constructing the pair of tangents from an external point to a circle.

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 12 : Area Related to CirclesChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability