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.
Access NCERT Solutions for Class 10 Mathematics Chapter 11 – Constructions
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 4 – Quadratic Equations
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
- 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.