# NCERT Solutions for Class 9 Maths Chapter 11 Constructions

NCERT solutions for class 9 maths chapter 11 Constructions: The chapter deals with the construction of certain angles and also triangles in certain conditions. In the previous chapters, you learnt to draw an approximate figure while writing the solutions. In this chapter, the scenario is different because the main purpose to introduce this chapter in the curriculum is to develop an approach to create an accurate geometric figure using some hints. CBSE NCERT solutions for class 9 maths chapter 11 Constructions has designed to provide you help in creating those accurate figures. Let's understand the kinds of questions which appear in this particular chapter. For example, you are asked to give a sketch of a roof of a triangular shape with a given perimeter and two base angles. The concepts learnt in this particular chapter can be used in the designing such cases. CBSE NCERT solutions for class 9 maths chapter 11 Constructions has point by point solutions to every constructional problem. When there is any constructional project to start, then the first step to initiate that project is the planning and drawing. Geometrical constructions have their own line of study. If you want to pursue your career in this domain then the higher study related to this can be civil engineering and architecture. In this chapter, there are 2 exercises having total of 14 questions. Solutions of NCERT for class 9 maths chapter 11 Constructions have an in-depth solution to each and every question present in the practice exercises. For 360 degree help in your school examination preparation, you can use NCERT solutions for other chapters, classes, and subjects.

### NCERT solutions for class 9 maths chapter 11 Constructions Excercise: 11.1

The steps of construction to follow:

Step 1: Draw a ray OP.

Then, take O as the centre and any radius draw an arc cutting OP at Q.

Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.

Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.

Hence, $\angle POA = 90^{\circ }$.

Justification:

We need to justify, $\angle POA = 90^{\circ }$

So, join OR and OS and RQ. we obtain

By construction OQ = OS = QR.

So, $\triangle ROQ$ is an equilateral triangle. Similarly $\triangle SOR$ is an equilateral triangle.

So, $\angle SOR = 60^{\circ}$

Now, $\angle ROQ = 60^{\circ}$ that means $\angle ROP = 60^{\circ}$.

Then, join AS and AR:

Now, in triangles OSA and ORA:

$SR = SR$  (common)

$AS = AR$  (Radii of same arcs)

$OS = OR$  (radii of the same arcs)

So, $\angle SOA = \angle ROA = \frac{1}{2}\angle SOR$

Therefore, $\angle ROA = 30^{\circ}$

and $\angle POA = \angle ROA+\angle POR = 30^{\circ} +60^{\circ} =90^{\circ}$

Hence, justified.

The steps of construction to follow:

Step 1: Draw a ray OY.

Then, take O as the centre and any radius, mark a point A on the arc ABC.

Step 2: Now, taking A as the centre and the same radius, mark a point B on the arc ABC.

Step 3: Take B as a centre and the same radius, mark a point C on the arc ABC.

Step 4: Now, taking C and B as centre one by one, draw an arc from each centre intersecting each other at a point X.

Step 5: X and O are joined and a ray making an angle $90^{\circ}$ with OY is formed.

Let the arc AC touches OX at E

Step 6: With A and E as centres, 2 arcs are marked intersecting each other at D and the bisector of angle XOY is drawn.

Justification:

By construction we have,

$\angle XOY = 90^{\circ}$

We constructed the bisector of  $\angle XOY$ as $\angle DOY$

Thus,

$\angle DOY = \frac{1}{2}\angle XOY = \frac{1}{2}\times90^{\circ} = 45^{\circ}$

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a center and take any radius, then draw an arc AB cutting OY at A.

Step 3: Take A and B as centres, draw 2 arcs are marked intersecting each other at X and hence, the bisector of $30^{\circ}$ is constructed.

Thus, $\angle XOY$ is the required angle.

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a centre and take any radius, then mark a point B on the arc

Step 3: Take B as a centre with the same radius, mark a point C on the arc ABC.

Step 4: Now, taking B and C as centres simultaneously, then draw an arc from each centre intersecting each other at a point X. Then join X and O and a ray making an angle with OY is formed.

Let the arc AC touches OX at E

Step 5: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$.

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, take A as a centre and take any radius, draw an arc AB which cuts OY at A.

Step 3: Take A and B as a centre, then mark 2 arcs which intersect each other at X and Hence, the bisector is constructed of $30^{\circ}$.

Step 4: Now, with A and E as centres, mark 2 arcs which intersect each other at D and we obtain the bisector of the angle $XOY$.

Thus, the angle of $15^{\circ}$ is obtained which is $\angle EOY.$

Steps to construction to follow:

Step 1: Draw a ray OY.

Step 2: Now, taking O as the centre draw an arc ABC.

Step 3: On taking A as a centre, draw two arcs B and C on the arc ABC.

Step 4: Now, taking B and C as centres, arcs are made to intersect at point E and the $\angle EOY = 90^{\circ}$ is constructed.

Step 5: Taking A and C as centres, arcs are made to intersect at D.

Step 6: Now, join OD and hence, $\angle DOY = 75^{\circ}$ is constructed.

Hence, $\angle DOY$ is the required angle.

The steps of construction to be followed:

Step 1: Draw a ray OY.

Step 2: Then, taking O as a centre, draw an arc ABC.

Step 3:  Now, with A as a centre, draw two arcs B and C which are made on the arc ABC.

Step 4: Taking B and C as centres simulataneously, arcs are made to intersect at E and $\angle EOY =90^{\circ}$ is constructed.

Step 5: With B and C as centres, arcs are made to intersect at X.

Step 6: Join the OX and we get $\angle XOY =105^{\circ}$ is constructed.

Thus, the angle $XOY$ is $105^{\circ}.$

The steps of construction to be followed:

Step 1: Draw a ray DY.

Step 2: Draw an arc ACD with O as a center.

Step 3: Now, with A as a centre, draw two arcs B and C on the arc ACD.

Step 4: Taking B and C as centres, arcs are made to intersect at E and the angle formed is $\angle EOY = 90^{\circ}$.

Step 5: Take F and D as centres, draw arcs to intersect at point X or the bisector of angle EOD is made.

Step 6: Join OX and the  $\angle XOY = 135^{\circ}$ is made.

Hence, the angle required $\angle XOY$ is $135^{\circ}$.

The following steps to make an equilateral triangle:

Step 1: Draw a line segment AB = 4 cm.

Step 2: With A and B as centres, make two arcs in the line segment AB. Mark it as D and E respectively.

Step 3: Now, with D and E as centres, make the two arcs cutting the previous arcs respectively, and forming an angle of $60^{\circ }$ each.

Step 4: Extend the lines of A and B until they intersect each other at point C.

Hence, triangle constructed is ABC which is equilateral.

Now, Justification;

Since the angles constructed are of $60^{\circ }$ each, so the third angle will also be $60^{\circ }$.

$\left [ \because The\ sum\ of\ all\ angles\ of\ triangle = 180^{\circ} \right ]$

## NCERT solutions for class 9 maths chapter 11 Constructions Excercise: 11.2

The steps of construction are as follows:

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

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

Step 3: Now, join CD.

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.

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., $\angle XBC.$

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.

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

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

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.

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.

Y

Y'

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

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

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.

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

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

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

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

The steps of construction to follow:

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

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

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

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

Thus, the constructed triangle is ABC.

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 11 Constructions

• Understand the basic construction of angles using the previous class book.

• Learn the constructional process of triangles and rectangles.

• Go through some solved examples to understand the solutions pattern.

• Apply the concepts learned in the practice exercises.

• If you feel trouble in solving any problem then take the assistance of NCERT solutions for class 9 maths chapter 11 Constructions.