# NCERT Solutions Class 10 Maths Chapter 11

## NCERT Solutions for Class 10 Mathematics Chapter 11 Constructions

The NCERT Solutions for Class 10 Mathematics Chapter 11 are accurate and reliable. Students looking for step-by-step answers to all the NCERT Mathematics Class 10 Chapter 11 textbook questions can refer to the solutions by Extramarks. The subject matter experts have prepared the solutions as per the latest CBSE guidelines. The answers are written in a simple and easy-to-understand manner with examples, as well as diagrams, and have been used, wherever possible, to help students understand it in a better way.

The NCERT solutions for Class 10 Mathematics chapter 11 by Extramarks are the perfect resource for last-minute revision and preparation. The answers are explained thoroughly to make it easier for students to grasp the concepts. Additionally, by referring to the Extramarks Solutions, students will get a better idea of how to attempt the questions and score better marks in tests and board exams.

## Introduction

Chapter 11 Constructions helps students understand how to make geometrical figures like a triangle, circle and more. They will also learn how to bisect a line segment or draw perpendiculars and tangents as demanded by the question. Although the chapter is more practical, it demands theoretical knowledge as well. This is because, without a strong understanding of concepts like proportionality theorem, Pythagoras theorem, etc, students won’t be able to draw accurate constructions.

## NCERT Solutions for Class 10 Mathematics

Extramarks offers NCERT Solutions for Class 10 Mathematics for the following chapters:

Chapter 1 – Real Numbers

Chapter 2 – Polynomials

Chapter 3 – Pair of Linear Equations in Two Variables

Chapter 5 – Arithmetic Progressions

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 – Areas Related to Circles

Chapter 13 – Surface Areas and Volumes

Chapter 14 – Statistics

Chapter 15 – Probability

### Division of a Line Segment

For dividing a line segment internally in the ratio m:n,  follow the steps given below:

•       Step 1: Draw a line segment of a given length. Now name one of its points as A and the other one as B.
•       Step 2: Now, draw a ray that should be making an acute angle with AB. Name the ray as AX.
•       Step 3: Alongside AX, start marking off (m + n) points. This includes A1,  A2, ……., Am, Am+1,….., Am+n. To give you an example,          if the ratio is to be 2:3, then mark 5 (=2+ 3) points).
•       Step 4: Join BAAm+n.
•       Step 5: Draw a line through Am parallel to Am+n B . Make an angle equal to ∠AAm+n B.

Make this line meet AB at point P. It is the point that divides AB internally in the ratio m:n.

### Constructing a Triangle Similar to a Given Triangle

When constructing a triangle similar to the given triangle, the former can be smaller or bigger than the latter. The following term needs to be defined:

Scale factor – The ratio of the to-be-constructed figure’s sides with measurements of the given figure.

Let ABC be the given triangle.   Suppose you want to construct a triangle similar to ABC, each of its sides is (m/n)th of the corresponding sides of ABC.

### The Following are the Steps to Be Taken for the Construction of a Triangle When M<n:

Step 1: Draw ABC by using the given data.

Step 2: Make AB as the base of the given ABC.

Step 3: At one end, say A of AB draw an acute angle ∠BAX below the base AB.

Step 4: Along AX mark off n points A1, A2,A3, ……., Am, Am+1,….., An such that AA1= A1A2 = A2A3

= A3 = Am-1Am = …….. = An-1An.

Step 5: Join AnB.

Step 6: Start from A to reach point AnB which meets AB at B’

Step 7: From B’ draw B’C’ || CB meeting AC at C’.

AB’C’ is the required triangle each of whose sides is (m/n)th of the corresponding sides of ABC.

### When constructing a tangent to a circle, one of the two cases will be presented:

Case A: When the center of the circle is known,  then the  steps of construction are:

Step 1: Take point O as a centre and use a compass to draw the circle.

Step 2: Let there be a point A on the circle.

Step 3: Join OA

Step 4: Construct ∠OAB = 90°.

Step 5: Draw TP to T’ to obtain the line TPT’ as the required tangent.

Case B: When the centre of the circle is not known, then the steps of construction are:

Step 1: Through point A draw a chord AC.

Step 2: A&C are joined to a point R in the major arc of the circle.

Step 3: Construct ∠CAT equal to ∠CRP on the opposite side of the chord AC.

Step 4: Draw TP to T’ to obtain  TPT’ as the required tangent.

### Construction of Tangents to a Circle from an External Point

Two tangents can be drawn to a circle from an external point. But one of the following two things  can happen: :

Case A: When the centre of the circle is known, then the steps of construction are:

Step 1: Given external point P is joined to centre O of the circle.

Step 2: Draw a perpendicular bisector of OP, intersecting OP at Q.

Step 3: Draw a circle with Q as the centre and OQ = QP as the radius, intersecting the given

circle at T & T’.

Step 4: Join PT & PT’ – the two tangents to the circle drawn from the external point P.

Case B: When centre of the circle is not known, then the steps of construction are:

Step 1: From P draw a secant PAB intersecting the given circle at A & B.

Step 2: Draw AP to C, such that AP = PC.

Step 3: Locate the midpoint of BC as M, & draw a semicircle.

Step 4: Draw a perpendicular PD on BC intersecting the semicircle at D.

Step 5: With P as centre & radius PD draws arcs intersecting the given circle at T

& T’.

Step 6: Join PT & PT’ – the two required tangents drawn on the given circle from the external point P​​

## Exam Strategy to Boost Your Preparation

•      Always refer to reliable study material like NCERT solutions Class 10 Mathematics Chapter 11 by Extramarks.
•      No first-time preparation should be left until the minute. . Instead,  last few days before the exam should be spend solely revising.
•      Always cross-check your answers with solved examples in NCERT Solutions for Class 10 Mathematics Chapter 11. These solutions           are prepared as per CBSE guidelines.
•     Pay enough attention to the accuracy of the diagrams as well.

Advantages of using Extramarks NCERT Solutions

One of the obvious advantages of using Extramarks NCERT Solutions is increasing the chances of securing high grades. Other perks include:

•   The convenience of studying whenever and from wherever the students wish to. You may study online or offline anywhere.
•   In-depth explanations of complex basics along with illustrations.
•   Systematic and to-the-point answers of each and every question, of all the chapters of the Class 10 CBSE’s entire syllabus.

## Related Questions

1. 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. Then the steps of construction will be:

•       Using the bangle,  draw a circle.
•       Now draw two non-parallel chords namely AB and CD
•       Now perpendicularly bisect AB and CD
•       Make O the center when perpendicular bisectors intersect.
•       Take a Point P outside the circle.
•       Join the points O and P to draw tangents.
•       With M as the midpoint, draw the perpendicular bisector of the line PO.
•       Keeping M as the center, draw MO as the circle’s radius.
•       Let the circle intersect at the points Q and R.
•       Join the points PQ and PR.

Reason:

The construction is to prove PQ and PR are the tangents to the circle.

Since, O is the centre of a circle, the perpendicular bisector of the chords passes through here and becomes OQ and OR.

It is obvious that the intersection point of these perpendicular bisectors is the centre of the circle.

Since ∠PQO is an angle in the semi-circle, it is by default a right angle.

∴ ∠PQO = 90°

⇒ OQ⊥ PQ

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

∴ ∠PRO = 90°

⇒ OR ⊥ PO

Since OR is the radius, PR must be a tangent.

Therefore, PQ and PR are tangents of a circle.

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

Ans.

$\begin{array}{l}\mathrm{Following}\text{are the steps to divide a line segment of length}\\ \text{7.6 cm in the ratio of 5:8.}\\ \text{Step 1: Draw line segment AB of 7.6 cm and draw a ray AX}\\ \text{making an acute angle with line segment AB.}\\ \text{Step 2: Locate 13(}=5+8\right){\text{points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}...,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{13}\text{on AX}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} such that AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}=...={\mathrm{A}}_{12}{\mathrm{A}}_{13}.\\ {\text{Step 3: Join the points B and A}}_{13}.\\ {\text{Step 4: Through the point A}}_{5}{\text{, draw a line parallel to BA}}_{13}\text{at}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} A}}_{5}\text{intersecting AB at point C.}\\ \text{C is the point which divides line segment AB of length 7.6 cm}\\ \text{in the ratio 5}:8.\\ \mathrm{On}\text{measuring lengths of AC and BC, we get AC}=2.9\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\text{}\mathrm{and}\\ \mathrm{BC}=4.7\text{cm.}\end{array}$ Q.2

$\begin{array}{l}\text{Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are}\frac{\text{2}}{3}\text{of the}\\ \text{corresponding sides of the first triangle.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Step}1:\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{AB}=4 \mathrm{cm}.\mathrm{Taking}\mathrm{point}\mathrm{A}\mathrm{as}\\ \mathrm{centre},\mathrm{draw}\mathrm{an}\mathrm{arc}\mathrm{of}5\mathrm{cm}\mathrm{radius}.\mathrm{Again},\mathrm{taking}\\ \mathrm{point}\mathrm{B}\mathrm{as}\mathrm{centre},\mathrm{draw}\mathrm{an}\mathrm{arc}\mathrm{of}6\mathrm{cm}.\mathrm{These}\mathrm{arcs}\\ \mathrm{intersect}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{point}\mathrm{C}.\mathrm{So},\mathrm{we}\mathrm{have}\\ \mathrm{AC}=5\mathrm{ }\mathrm{cm}\mathrm{}\mathrm{and}\mathrm{BC}=6 \mathrm{cm}.\mathrm{}\mathrm{\Delta }\mathrm{ }\mathrm{ABC}\mathrm{is}\mathrm{the}\mathrm{required}\\ \mathrm{triangle}\mathrm{.}\\ \mathrm{Step}2:\mathrm{Draw}\mathrm{a}\mathrm{ray}\mathrm{AX}\mathrm{making}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{with}\mathrm{line}\mathrm{AB}\\ \mathrm{on}\mathrm{the}\mathrm{opposite}\mathrm{side}\mathrm{of}\mathrm{vertex}\mathrm{C}\mathrm{.}\\ \mathrm{Step}3:\mathrm{Locate}3\mathrm{points}{\mathrm{A}}_{1},{\mathrm{A}}_{2}, {\mathrm{A}}_{3}\mathrm{on}\mathrm{AX}\mathrm{such}\mathrm{that}\\ \mathrm{ }{\mathrm{AA}}_{1}={\mathrm{A}}_{1}{\mathrm{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}.\\ \mathrm{Step}4:\mathrm{Join}\mathrm{the}\mathrm{points}\mathrm{B}\mathrm{and}{\mathrm{A}}_{3}.\\ \mathrm{Step}5:\mathrm{Through}\mathrm{the}\mathrm{point}{\mathrm{A}}_{2},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}{\mathrm{BA}}_{3}\mathrm{}\\ \mathrm{intersecting}\mathrm{AB}\mathrm{at}\mathrm{point}\mathrm{B}‘\mathrm{.}\\ \mathrm{Step}6:\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{through}\mathrm{B}‘\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{BC}\mathrm{to}\\ \mathrm{intersect}\mathrm{AC}\mathrm{at}\mathrm{C}‘\mathrm{.}\\ \mathrm{The}\mathrm{required}\mathrm{triangle}\mathrm{is}\mathrm{\Delta AB}‘\mathrm{C}‘\mathrm{.}\end{array}$ Q.3

$\begin{array}{l}\text{Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are}\frac{\text{7}}{5}\text{of the}\\ \text{corresponding sides of the first triangle.}\end{array}$

Ans.

$\begin{array}{l}\text{Step 1: Draw a line segment AB}=5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{Taking point A as}\\ \text{centre, draw an arc of 6 cm radius. Again, taking}\\ \text{point B as centre, draw an arc of 7 cm. These arcs}\\ \text{​​​​​​​​​​​​​​​​​​​​ intersect each other at point C. So, we have}\\ \text{AC}=6\text{\hspace{0.17em}}\mathrm{cm}\text{}\mathrm{and}\text{BC}=7\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{}\mathrm{\Delta }\text{\hspace{0.17em}ABC is the required}\\ \text{triangle.}\\ \text{Step 2: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 3: Locate 7 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{,}...\text{,}{\mathrm{A}}_{7}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}=...={\mathrm{A}}_{6}{\mathrm{A}}_{7}.\\ {\text{Step 4: Join the points B and A}}_{5}.\\ {\text{Step 5: Through the point A}}_{7}{\text{, draw a line parallel to BA}}_{5}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B’.}\\ \text{Step 6: Draw a line through B’ parallel to the line BC to}\\ \text{intersect extended line segment AC at C’.}\\ \text{The required triangle is}\mathrm{\Delta }\text{AB ’C’.}\end{array}$ Q.4

$\begin{array}{l}\text{Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose}\\ \text{sides are}\mathrm{}\text{\hspace{0.17em}}1\frac{1}{2}\text{times the corresponding sides of the isosceles triangle}.\mathrm{}\end{array}$

Ans. $\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a line segment AB}=8\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{Draw arcs of same}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}radius on both sides of the line segment while taking}\\ \text{point A and B as its centre. Let these arcs intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} each other at O and O’. Join OO’. Let OO’ intersect}\\ \text{\hspace{0.17em}AB at D.}\\ \text{Step 2: Taking D as centre, draw an arc of 4 cm radius which}\\ \text{cuts the extended line segment OO’ at point C. An}\\ \text{isosceles}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}\text{is formed havind CD as 4 cm and AB}\\ \text{as 8 cm.}\\ \text{​​​​​​​​​​​​​​​​​Step 3: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 4: Locate 3 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}.\\ {\text{Step 5: Join the points B and A}}_{2}.\\ {\text{Step 6: Through the point A}}_{3}{\text{, draw a line parallel to BA}}_{2}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B’.}\\ \text{Step 7: Draw a line through B’ parallel to the line BC to}\\ \text{intersect extended line segment AC at C’.}\\ \text{The required triangle is}\mathrm{\Delta }\text{AB ’C’.}\end{array}$

Q.5 Draw a triangle ABC with side BC=6 cm, AB=5 cm and ABC=60°. Then construct a triangle whose sides are34  of the corresponding sides of the triangle ABC.

Ans. $\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a}\mathrm{\Delta }\text{\hspace{0.17em}ABC with sides AB}=5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm},\text{BC}=6\text{cm and}\\ \text{}\angle \text{ABC}=60°.\text{}\\ \text{​​​​​​​​​​​​​​​​​Step 2: Draw a ray BX making an acute angle with line BC}\\ \text{on the opposite side of vertex A.}\\ {\text{Step 3: Locate 4 points B}}_{1},{\text{B}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{B}}_{3}\text{, \hspace{0.17em}}{\mathrm{B}}_{4}\text{on BX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} BB}}_{1}={\text{B}}_{1}{\text{B}}_{2}={\mathrm{B}}_{2}{\mathrm{B}}_{3}={\mathrm{B}}_{3}{\mathrm{B}}_{4}.\\ {\text{Step 4: Join the points C and B}}_{4}.\\ {\text{Step 5: Through the point B}}_{3}{\text{, draw a line parallel to CB}}_{4}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting line segment BC at point C’.}\\ \text{Step 6: Draw a line through C’ parallel to the line AC to}\\ \text{intersect line segment AB at A’.}\\ \text{The required triangle is}\mathrm{\Delta }\text{A ’BC’.}\end{array}$

Q.6

$\begin{array}{l}\text{Draw a triangle ABC with sides BC=7 cm,}\angle =°,\angle =°,\text{}\\ \text{then construct a triangle whose sides are}\frac{4}{3}\text{times the}\\ \text{corresponding sides of}\mathrm{\Delta }\text{\hspace{0.17em}}\mathrm{ABC}.\end{array}$

Ans. $\begin{array}{l}\mathrm{It}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{that}\text{}\angle \text{B}=45°,\text{}\angle \mathrm{A}=105°.\\ \therefore \text{In}\mathrm{\Delta }\text{\hspace{0.17em}ABC, we have}\angle \mathrm{C}=180°-105°-45°=30°\\ \mathrm{StepsofConstruction}:\\ \text{Step 1: Draw a}\mathrm{\Delta }\text{\hspace{0.17em}ABC with side BC}=7\text{cm and}\\ \text{}\angle \text{B}=45°,\text{}\angle \mathrm{C}=30°.\text{}\\ \text{​​​​​​​​​​​​​​​​​Step 2: Draw a ray BX making an acute angle with line BC}\\ \text{on the opposite side of vertex A.}\\ {\text{Step 3: Locate 4 points B}}_{1},{\text{B}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{B}}_{3}\text{, \hspace{0.17em}}{\mathrm{B}}_{4}\text{on BX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} BB}}_{1}={\text{B}}_{1}{\text{B}}_{2}={\mathrm{B}}_{2}{\mathrm{B}}_{3}={\mathrm{B}}_{3}{\mathrm{B}}_{4}.\\ {\text{Step 4: Join the points C and B}}_{3}.\\ {\text{Step 5: Through the point B}}_{4}{\text{, draw a line parallel to CB}}_{3}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment BC at point C’.}\\ \text{Step 6: Draw a line through C’ parallel to the line AC to}\\ \text{intersect extended line segment BA at A’.}\\ \text{The required triangle is}\mathrm{\Delta }\text{A ’BC’.}\end{array}$

Q.7

$\begin{array}{l}\text{Draw a right triangle in which the sides (other than hypotenuse}\right)\text{are of length 4 cm and 3 cm}.\text{Then}\\ \text{construct another triangle whose sides are}\frac{\text{5}}{3}\text{times the corresponding sides of the given triangle}.\mathrm{}\end{array}$

Ans. $\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a right angle}\mathrm{\Delta }\text{\hspace{0.17em}ABC with base AB}=4\text{cm,}\\ \text{AC}=3\text{cm and}\angle \mathrm{A}=90°.\text{}\\ \text{​​​​​​​​​​​​​​​​​Step 2: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 3: Locate 5 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{, \hspace{0.17em}}{\mathrm{A}}_{4}{\text{, A}}_{5}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\mathrm{A}}_{1}{\mathrm{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}={\mathrm{A}}_{3}{\mathrm{A}}_{4}={\mathrm{A}}_{4}{\mathrm{A}}_{5}.\\ {\text{Step 4: Join the points B and A}}_{3}.\\ {\text{Step 5: Through the point A}}_{5}{\text{, draw a line parallel to BA}}_{3}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B’.}\\ \text{Step 6: Draw a line through B’ parallel to the line BC to}\\ \text{intersect extended line segment AC at C’.}\\ \text{The required triangle is}\mathrm{\Delta }\text{AB’ }\mathrm{C}‘\text{.}\end{array}$

Q.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.

Ans. $\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a circle of radius 6 cm with centre at point O.}\\ \text{Locate a point P, 10 cm away from O, and join O and P.}\\ \text{Step 2:}\mathrm{Bisect}\text{OP. Let M be the mid-point of OP.}\\ \text{Step 3:}\mathrm{Draw}\text{a circle with centre at M and MO as radius. Q and}\\ \text{R are points of intersections of this circle with the circle}\\ \text{having centre at O.}\\ \text{Step 4: Join PQ and PR. PQ and PR are the required tangents.}\\ \mathrm{The}\text{lengths of tangents PQ and PR are 8 cm each.}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a circle of radius 4 cm with centre at point O.}\\ \text{Draw anothe circle of radius 6 cm with centre at}\\ \text{point O. Locate a point P on this circle and join OP.}\\ \text{}\\ \text{Step 2:}\mathrm{Bisect}\text{OP. Let M be the mid-point of OP.}\\ \text{Step 3:}\mathrm{Draw}\text{a circle with centre at M and MO as radius. Q and}\\ \text{R are points of intersections of this circle with the circle}\\ \text{having centre at O.}\\ \text{Step 4: Join PQ and PR. PQ and PR are the required tangents.}\\ \mathrm{On}\mathrm{measuring}\mathrm{we}\mathrm{get}t\mathrm{he}\text{lengths of tangents PQ and PR are 4.5 cm each.}\end{array}$ $\begin{array}{l}\mathrm{In}\text{}\mathrm{\Delta }\text{\hspace{0.17em}PQO, we have}\\ \angle \text{PQO}=90\mathrm{°}\text{[PQ is a tangent]}\\ \text{PO}=\text{\hspace{0.17em}}6\text{cm [Radius of the circle]}\\ \text{QO}=4\text{cm [Radius of the circle]}\\ \mathrm{By}\text{applying Pythagoras theorem in}\mathrm{\Delta }\text{\hspace{0.17em}PQO},\text{we have}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{2}+{\mathrm{QO}}^{2}={\mathrm{OP}}^{2}\\ \mathrm{or}{\text{\hspace{0.17em}PQ}}^{2}+{4}^{2}={6}^{2}\\ \mathrm{or}{\text{\hspace{0.17em}PQ}}^{2}=36-16=20\\ \mathrm{or}\text{\hspace{0.17em}PQ}=\sqrt{20}=2\sqrt{5}=4.5\text{cm approx.}\\ \text{We know that PQ}=\mathrm{PR}\text{as these tangents are drawn from}\\ \text{a single point P.}\\ \mathrm{Therefore},\text{PQ}=\mathrm{PR}=4.5\text{cm approx.}\end{array}$

Q.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.

Ans.

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a circle of radius 3 cm with centre at point O.}\\ \text{Step 2: Take one of its diametre, RS, and extend it on both}\\ \text{sides. Locate two points P and Q on this diametre}\\ \text{such that OP}=\mathrm{OQ}=7\text{cm}.\\ \text{Step 3:}\mathrm{Bisect}\text{OP and OQ. Let T and U be the mid-points of}\\ \text{OP and OQ respectively.}\\ \text{Step 4: Taking T and U as centres and OT and OU as radii,}\\ \text{d}\mathrm{raw}\text{two circles. These two circles will intersect}\\ \text{the circle at points V, W, X, Y. Join PV, PW, QX and QY.}\\ \text{These are the required tangents.}\end{array}$ Q.11 Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Ans.

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a circle of radius 5 cm with centre at point O.}\\ \text{Step 2: Take a point A on circumference of the circle and join}\\ \text{OA. Draw a perpendicular to OA at point A.}\\ \text{Step 3: Draw a radius OB, making an angle of 120° with OA.}\\ \text{Step 4:}\mathrm{Draw}\text{a perpendicular to OB at point B. Let both the}\\ \text{perpendiculars intersect at point P. PA and PB are}\\ \text{the required tangents at an angle of 60°.}\end{array}$ Q.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.

Ans.

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a line segment AB of 8 cm. Taking A and B as}\\ \text{centre, Draw a circle of 4 cm radius centred at A}\\ \text{and another circle of radius 3 cm centred at B.}\\ \text{Step 2: Bisect the line AB. Let the mid-point of AB be C.}\\ \text{Taking C as centre, draw a circle with radius AC.}\\ \text{This circle intersects the other two circles at points}\\ \text{P, Q, R and S. Join BP, BQ, AS and AR. These are}\\ \text{the required tangents.}\end{array}$ Q.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, D is drawn. Construct the tangents from A to this circle.

Ans.

As per given information, we have the following figure. $\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Join AE and bisect it. Let F be the mid-point of AE.}\\ \text{Step 2: Taking F as centre and FE as its radius, draw a circle}\\ \text{which will intersect the other circle at points B and G.}\\ \text{Join AG. AB and AG are the required tangents.}\end{array}$ Q.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 circle.

Ans.

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a circle with the help of a bangle.}\\ \text{Step 2: Take a point P outside this circle and take two}\\ \text{chords QR and ST.}\\ \text{Step 3: Draw perpendicular bisectors of QR and ST. Let}\\ \text{the perpendicular bisectors intersect each other}\\ \text{at point O.}\\ \text{Step 4: Join OP and bisect it. Let U be the mid-point of OP.}\\ \text{Taking U as centre, draw a circle of radius OU,}\\ \text{which intersects the previous circle at V and W.}\\ \text{Join PV and PW.}\\ \text{PV and PW are the required tangents.}\end{array}$ 