NCERT Solutions for Exercise 11.2 Class 9 Maths Chapter 11

NCERT Solutions for Exercise 11.2 Class 9 Maths Chapter 11 - Constructions

Edited By Sumit Saini | Updated on Dec 29, 2021 06:17 PM IST

NCERT Solutions for Class 9 Maths chapter 11 exercise 11.2 introduces us with the construction of triangles till now we have learned about how to construct the bisector of a given angle, and the perpendicular bisector of a given line segment must be constructed. and to create a 60-degree angle at the starting point of a given ray.NCERT solutions for Class 9 Maths chapter 11 exercise 11.2 is based on the construction of triangles using the prior knowledge of exercise 11.1 Class 9 Maths and the congruence of triangles as covered in chapter seventh Class 9. All the construction in the NCERT solutions for Class 9 Maths chapter 11 exercise 11.2 is carried out with the help of compass and ruler but we can use a protector to re-check if construction is accurate. Actually, it is very helpful as it cross-examines the whole construction. Along with Class 9 Maths chapter 11 exercise 11.2, the following exercises are also present.

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This Story also Contains

Constructions Class 9 Maths Chapter 11 Excercise: 11.2
Q1 Construct a triangle ABC in which , and .
More About NCERT Solutions for Class 9 Maths Exercise 11.2
Benefits of NCERT Solutions for Class 9 Maths Exercise 11.2
NCERT Solutions of Class 10 Subject Wise
NCERT Solutions for Class 9 Maths

Constructions Exercise 11.1

Constructions Class 9 Maths Chapter 11 Excercise: 11.2

Q1 Construct a triangle ABC in which $BC = 7cm$ , $\angle B = 75\degree$ and $AB + AC = 13 cm$ .

Answer:

The steps of construction are as follows:

Step 1: Draw a line segment BC of 7cm length. Taking the help of protractor make an $\angle XBC = 75^{\circ}$ .

1640601458338

Step 2: Now, cut a line segment BD having 13 cm on BX which is $(AB+AC).$

1640601468104

Step 3: Now, join CD.

1640601479666

Step 4: Draw a perpendicular bisector of CD to intersect BD at a point A. Join AC. Then ABC is the required triangle

Hence, the required triangle is ABC.

Q2 Construct a triangle ABC in which $BC = 8cm$ , $\angle B = 45\degree$ ° and $AB - AC = 3.5cm$

Answer:

The steps of construction to be followed:

Step 1: Draw a line segment BC = 8cm and make an angle of $45^{\circ}$ at point B i.e., $\angle XBC.$

1640601500495

Step 2: Now, cut the line segment BD = 3.5 cm on ray BX. i.e. $(AB-AC)$ .

Step 3: Join CD and draw a perpendicular bisector of CD i.e., PQ.

1640601513380

Step 4: Let the perpendicular bisector of CD intersects BX at point A. Then,

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Step 5: Join AC, to get the required triangle $\triangle ABC.$

Q3 Construct a triangle PQR in which $QR = 6cm$ , $\angle Q = 60 \degree$ and $PR -PQ = 2cm$ .

Answer:

The steps of construction to be followed:

Step 1: Draw a ray QX and cut off a line segment QR which is equal to 6 cm in length.

1640601586225

Step 2: With an angle of $60^{\circ}$ with QR, construct a ray QY and extend it to form a line YQY'.

Step 3: Now, cut off a line segment QS equal to 2 cm from QY' and join RS.

1640601600248

Step 4: Draw a perpendicular bisector of RS which intersects QY at a point P.

Step 5: Join PR to get the required triangle $\triangle PQR.$

1640601611912

Q4 Construct a triangle XYZ in which $\angle Y = 30\degree$ , $\angle Z = 90\degree$ ° and $XY + YZ + ZX = 11 cm$ .

Answer:

The steps of construction to be followed:

Step 1: For given $XY+YZ+ZX = 11 cm$ , a line segment $PQ =11 cm$ is drawn.

Step 2: At points, P and Q angles of $\angle RPQ = 30^{\circ}$ and $\angle SQP =90^{\circ}$ are constructed respectively.

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Step 3: Now, bisects the angle RPQ and SQP. The bisectors of these angles intersect each other at a point X.

1640601651930

Step 4: Construct the perpendicular bisector of PX and QX, name them as TU and WV respectively.

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Step 5: Let the bisector TU intersect PQ at Y and bisector WV intersect PQ at Z. Then XY and ZY are joined.

1640601685870

Therefore, $\triangle XYZ$ is the required triangle.

Q5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Answer:

The steps of construction to follow:

Step 1: Draw a ray BX and Cut off a line segment $BC=12cm$ from the ray.

1640601714690

Step 2: Now, construct an angle $\angle XBY = 90^{\circ}$ .

1640601724612

Step 3: Cut off a line segment BD of length 18 cm on BY. Then join the CD.

1640601736897

Step 4: Now, construct a perpendicular bisector of CD which intersects BD at A and AC is joined.

1640601751004

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Thus, the constructed triangle is ABC.

Benefits of NCERT Solutions for Class 9 Maths Exercise 11.2

Class 9 Maths chapter 11 exercise 11.2 covers the topic of construction of triangle in three different way as provided in Class 9 Maths chapter 11 exercise 11.2
NCERT Class 9 Maths chapter 11 exercise 11.2 , will be helpful in Class 11 Maths chapter 11 construction.
Construction is a very important topic of an entrance exam JEE Main (joint entrance exam - 2) known as B-arch
Construction is base engineering in physics and mechatronic

Also See:

NCERT Solutions of Class 10 Subject Wise

Frequently Asked Questions (FAQs)

1. What should be used in construction of problems in NCERT solutions for Class 9 Maths exercise 11.2 ?

We should only use ruler and compass to construct the figures

2. Can we use a protractor to construct figures ?

No we cannot use protractor to construct figures we can only use to recheck the figure

3. When is a triangle unique ?

(i) two sides and the included angle is given,

(ii) three sides are given,

(iii) two angles and the included side is given and,

(iv) in a right triangle, hypotenuse and one side is given.

4. How many new constructions are introduced before exercise 11.2 Class 9 Maths ?

three new constructions are introduced before Exercise 11.2 Class 9 Maths

5. Which rules of congruence are used in NCERT solutions for Class 9 Maths exercise 11.2 ?

SAS (side angle side)

Side by side, side by side SSS

ASA (angle side angle)

RHS (right hand side)

6. How many questions are answered in the NCERT answers for Class 9 Mathematics chapter 11 exercise 11.2?

Class 9 Mathematics chapter exercise 11.2 has 5 questions.

7. How many solved examples are there before the NCERT answers for Class 9 Mathematics chapter 11 exercise 11.2?

Before NCERT solution for Class 9 Mathematics chapter 11 exercise 11.2 there are 1 solved examples

8. What kinds of questions do the NCERT answers for Class 9 Mathematics chapter 11 exercise 11.2 cover?

All the questions covered in the Class 9 Maths chapter exercise 11.2 was about construction. It included the bases of the construction covered before exercise in theory part.

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