NCERT Solutions for Exercise 11.2 Class 10 Maths Chapter 11 - Constructions

NCERT Solutions for Exercise 11.2 Class 10 Maths Chapter 11 - Constructions

Edited By Ramraj Saini | Updated on Nov 27, 2023 04:47 PM IST | #CBSE Class 10th
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NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2

NCERT Solutions for Exercise 11.2 Class 10 Maths Chapter 11 Constructions are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 Maths ex 11.2 consists of descriptive points of creating circles in a logical way, with justification. Also, some properties and technical terms of special conic curves. This chapter contains important concepts to understand in two-dimensional geometry. starts with the construction of a circle under certain given conditions ie measurement of radius, a distance of a point from the start of tracing tangents and chords briefly covered in this chapter and mentioned question somehow is like the foundation of the examination, are indeed a major element to cover thoroughly.

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  1. NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.2
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 11 exercise 11.2
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  4. More About NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.2
  5. NCERT Solutions Subject Wise
NCERT Solutions for Exercise 11.2 Class 10 Maths Chapter 11 - Constructions
NCERT Solutions for Exercise 11.2 Class 10 Maths Chapter 11 - Constructions

10th class Maths exercise 11.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Constructions Class 10 Chapter 11 Exercise: 11.2

Q1 In each of the following, give also the justification of the construction:
Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.

Answer:

Steps of construction:-

(i) Taking point O as center draw a circle of radius 6 cm.

(ii) Now, name a point P which is 10 cm away from point O. Join OP.

(iii) Draw a perpendicular bisector of OP name the intersection point of bisector and OP as O'.

(iv) Now draw a circle considering O' as center and O'P as the radius.

(v) Name the intersection point of circles as Q and R.

(vi) Join PQ and PR. These are the required tangents.

(vii) Measure lengths of PQ = 8cm and PR = 8cm

1636090832534

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.

Answer:

Steps of constructions:-

(i) Taking point O as a center draw a circle of radius 4 cm.

(ii) Now taking O as center draw a concentric circle of radius 6 cm.

(iii) Taking any point P on the outer circle, join OP.

(iv) Draw a perpendicular bisector of OP.

(v) Name the intersection of bisector and OP as O'.

(vi) Now, draw a circle taking O' as center and O'P as the radius.

(vii) Name the intersection point of two circles as R and Q.

(viii) Join PR and PQ. These are the required tangents.

(ix) Measure the lengths of the tangents. PR = 4.47 cm and PQ = 4.47 cm.

1636090849002

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 center. Draw tangents to the circle from these two points P and Q.

Answer:

Steps of construction:-

(i) Taking O as a center draw a circle of radius 3 cm.

(ii) Now draw a diameter PQ of this circle and extend it.

(iii) Mark two points R and S on the extended diameter such that OR = OS = 7 cm.

(iv) Draw the perpendicular bisector of both the lines and name their mid-points as T and U.

(v) Now, taking T and U as center draw circles of radius TR and QS.

(vi) Name the intersecting points of the circles with the first circles as V, W, X, Y.

(vii) Join the lines. These are the required tangents.

1636090862749

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

Answer:

Steps of construction:-

(i) Draw a circle with center O and radius 5 cm.

(ii) Now mark a point A on the circumference of the circle. And draw a line AP perpendicular to the radius OA.

(iii) Mark a point B on the circumference of the circle such that \angle AOB = 120 o . (As we know, the angle at the center is double that of the angle made by tangents).

(iv) Join B to point P.

(v) AP and BP are the required tangents.

1636090901571

Q5 Draw a line segment AB of length 8 cm. Taking A as a center, draw a circle of radius 4 cm and taking B as center, draw another circle of radius 3 cm. Construct tangents to each circle from the center of the other circle.

Answer:

Steps of construction:-

(i) Draw a line segment AB having a length of 8 cm.

(ii) Now, taking A as a center draw a circle of radius 4 cm. And taking B as a center draw a circle of radius 3 cm.

(iii) Bisect the line AB and name the mid-point as C.

(iv) Taking C as a center and AC as radius draw a circle.

(v) Name the intersection points of the circle as P, Q, R, S.

(vi) Join the lines and these are our required tangents.

1636090914889

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

Answer:

Steps of construction:-

(i) Draw a line segment BC of length 8 cm.

(ii) Construct a right angle at point B. Now draw a line of length 6 cm. Name the other point as A.

(iii) Join AC. \Delta ABC is the required triangle.

(iv) Now construct a line BD on the line segment AC such that BD is perpendicular to AC.

(v) Now draw a circle taking E as a center (E is the midpoint of line BC) and BE as the radius.

(vi) Join AE. And draw a perpendicular bisector of this line.

(vii) Name the midpoint of AE as F.

(viii) Now, draw a circle with F as center and AF as the radius.

(ix) Name the intersection point of both the circles as G.

(x) Join AG. Thus AB and AG are the required tangents.

1636090928158

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.

Answer:

Steps of construction:-

(i) Draw a circle using a bangle.

(ii) Now draw 2 chords of this circle as QR and ST.

(iii) Take a point P outside the circle.

(iv) Draw perpendicular bisector of both the chords and let them meet at point O.

(v) Joinpoint PO.

(vi) Draw bisector of PO and name the midpoint as U.

(vii) Now, taking U as a center and UP as radius draw a circle.

(viii) Name the intersection point of both the circles as V and W.

(ix) Join PV and PW. These are the required tangents.

1636090941880

More About NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.2

The problems from the theory point of view are covered in exercise 11.2 Class 10 Maths. The Initial questions of NCERT solutions for Class 10 Maths chapter 11 exercise 11.2 is to construct figures using protractors and dividers. And later on questions of Class 10 Maths chapter 11 exercise, 11.2 is to prove based on fundamental properties of Circle and tangents attached to it. In NCERT syllabus Class 10 Maths chapter 11 exercise 11.2 also covers some exclusive problems on triple circle scenarios with chords.

Also Read| Constructions Class 10 Notes

Also, see-

NCERT Solutions Subject Wise

Frequently Asked Questions (FAQs)

1. Minimum number of Points required to construct a circle?

The Minimum of 3 points is required to construct a circle from scratch.

2. What is the measurement of the sum of all interior angles of the Quadrilateral inscribed in a circle?

The Sum of all the angles for the above-mentioned condition is 360 degrees.

3. Do we have to draw diagrams using a pencil or pen?

Ans: We have to draw/construct diagrams of curves using dark pointed pencils in order to get good marks.

4. Can we prove theorems using construction?

Ans: Yes we can prove theorems with constructions, in fact, students will find some questions related to it.

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

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According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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