{"id":100554,"date":"2021-02-09T05:13:37","date_gmt":"2021-02-09T05:13:37","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=100554"},"modified":"2023-09-18T02:54:32","modified_gmt":"2023-09-18T02:54:32","slug":"ncert-solutions-for-9th-class-maths-chapter-11-constructions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-9th-class-maths-chapter-11-constructions\/","title":{"rendered":"NCERT Solutions for 9th class Maths : Chapter 11 Constructions"},"content":{"rendered":"\n

Class 9: Maths Chapter 11 solutions. Complete Class 9 Maths Chapter 11 Notes.<\/p>\n\n\n\n

NCERT Solutions for 9th class Maths : Chapter 11 Constructions<\/strong><\/h2>\n\n\n\n

NCERT 9th Maths Chapter 11, class 9 Maths Chapter 11 solutions<\/p>\n\n\n\n

Page No: 191<\/p>\n\n\n\n

Exercise 11.1<\/strong><\/p>\n\n\n\n

1. Construct an angle of 90\u00b0 at the initial point of a given ray and justify the construction.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of construction:<\/p>\n\n\n\n

Step 1: A ray YZ is drawn.
Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.
Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.
Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray XY making an angle 90\u00b0 with YZ is formed.<\/p>\n\n\n\n

Justification for construction:
We constructed \u2220BYZ = 60\u00b0 and also \u2220AYB = 60\u00b0.
Thus, \u2220AYZ = 120\u00b0.
Also, bisector of \u2220AYB is constructed such that:
\u2220AYB = \u2220XYA +<\/p>\n\n\n\n

\u2220XYB
\u21d2 \u2220XYB = 1\/2\u2220AYB
\u21d2 \u2220XYB = 1\/2\u00d760\u00b0
\u21d2 \u2220XYB = 30\u00b0
Now,
\u2220XYZ = \u2220BYZ + \u2220XYB = 60\u00b0 + 30\u00b0 = 90\u00b0<\/p>\n\n\n\n

2. Construct an angle of 45\u00b0 at the initial point of a given ray and justify the construction.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of construction:<\/p>\n\n\n\n

Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.
Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.
Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.
Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.
Step 6: X and Y are joined and a ray making an angle 90\u00b0 with YZ is formed.
Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of \u2220XOY is drawn.<\/p>\n\n\n\n

Justification for construction:
By construction,
\u2220XOY = 90\u00b0
We constructed the bisector of \u2220XOY as DOY.
Thus,
\u2220DOY = 1\/2 \u2220XOY
\u2220DOY = 1\/2\u00d790\u00b0 = 45\u00b0<\/p>\n\n\n\n

3. Construct the angles of the following measurements:<\/strong>
(i) 30\u00b0        (ii) 22.5\u00b0        (iii) 15\u00b0<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 30\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:
Step 1: A ray OY is drawn.
Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.
Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.
Thus, \u2220XOY is the required angle making 30\u00b0 with OY.<\/p>\n\n\n\n

(ii) 22.5\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:
Step 1: An angle \u2220XOY = 90\u00b0 is drawn.
Step 2: Bisector of \u2220XOY is drawn such that \u2220BOY = 45\u00b0 is constructed.
Step 3: Again, \u2220BOY is bisected such that \u2220AOY is formed.
Thus, \u2220AOY is the required angle making 22.5\u00b0 with OY.<\/p>\n\n\n\n

(iii) 15\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:
Step 1: An angle \u2220AOY = 60\u00b0 is drawn.
Step 2: Bisector of \u2220AOY is drawn such that \u2220BOY = 30\u00b0 is constructed.
Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of \u2220BOY is drawn.
Thus, \u2220XOY is the required angle making 15\u00b0 with OY.<\/p>\n\n\n\n

NCERT 10th Maths Chapter 11, class 10 Maths Chapter 11 solutions<\/p>\n\n\n\n

4. Construct the following angles and verify by measuring them by a protractor:<\/strong>
(i) 75\u00b0         (ii) 105\u00b0        (iii) 135\u00b0<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 75\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:<\/p>\n\n\n\n

Step 1: A ray OY is drawn.
Step 2: An arc BAE is drawn with O as a centre.
Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.
Step 4: With A and B as centres, arcs are made to intersect at X and \u2220XOY = 90\u00b0 is made.
Step 5: With A and C as centres, arcs are made to intersect at D
Step 6: OD is joined and and \u2220DOY = 75\u00b0 is constructed.
Thus, \u2220DOY is the required angle making 75\u00b0 with OY.<\/p>\n\n\n\n

(ii) 105\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:<\/p>\n\n\n\n

Step 1: A ray OY is drawn.
Step 2: An arc ABC is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.
Step 4: With B and C as centres, arcs are made to intersect at E and \u2220EOY = 90\u00b0 is made.
Step 5: With B and C as centres, arcs are made to intersect at X
Step 6: OX is joined and and \u2220XOY = 105\u00b0 is constructed.
Thus, \u2220XOY is the required angle making 105\u00b0 with OY.<\/p>\n\n\n\n

(iii) 135\u00b0<\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:Step 1: A ray DY is drawn.
Step 2: An arc ACD is drawn with O as a centre.
Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.
Step 4: With B and C as centres, arcs are made to intersect at E and \u2220EOY = 90\u00b0 is made.
Step 5: With F and D as centres, arcs are made to intersect at X or bisector of \u2220EOD is constructed.
Step 6: OX is joined and and \u2220XOY = 135\u00b0 is constructed.
Thus, \u2220XOY is the required angle making 135\u00b0 with DY.<\/p>\n\n\n\n

5. Construct an equilateral triangle, given its side and justify the construction.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of constructions:
Step 1: A line segment AB=4 cm is drawn.
Step 2: With A and B as centres, two arcs are made.
Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60\u00b0 each.
Step 5: Lines from A and B are extended to meet each other at C.
Thus, ABC is the required triangle formed.<\/p>\n\n\n\n

Justification:
By construction,
AB = 4 cm, \u2220A = 60\u00b0 and \u2220B = 60\u00b0
We know that,
\u2220A +<\/p>\n\n\n\n

\u2220B +<\/p>\n\n\n\n

\u2220C = 180\u00b0 (Sum of the angles of a triangle)
\u21d2 60\u00b0 + 60\u00b0 + \u2220C = 180\u00b0
\u21d2 120\u00b0 + \u2220C = 180\u00b0
\u21d2 \u2220C = 60\u00b0
BC = CA = 4 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 4 cm
\u2220A = \u2220B = \u2220C = 60\u00b0<\/p>\n\n\n\n

Page No: 195<\/p>\n\n\n\n

NCERT 9th Maths Chapter 11, class 9 Maths Chapter 11 solutions<\/p>\n\n\n\n

Exercise 11.2<\/strong><\/p>\n\n\n\n

1. Construct a triangle ABC in which BC = 7cm, \u2220B = 75\u00b0 and AB + AC = 13 cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of Construction:
Step 1: A line segment BC of 7 cm is drawn.
Step 2: At point B, an angle \u2220XBC is constructed such that it is equal to 75\u00b0.
Step 3: A line segment BD = 13 cm is cut on BX (which is equal to AB+AC).
Step 3: DC is joined and \u2220DCY = \u2220BDC is made.
Step 4: Let CY intersect BX at A.
Thus, \u0394ABC is the required triangle.<\/p>\n\n\n\n

2. Construct a triangle ABC in which BC = 8cm, \u2220B = 45\u00b0 and AB – AC = 3.5 cm.<\/p>\n\n\n\n

Answer<\/strong>

Steps of Construction:
Step 1: A line segment BC = 8 cm is drawn and at point B, make an angle of 45\u00b0 i.e. \u2220XBC.
Step 2: Cut the line segment BD = 3.5 cm (equal to AB – AC) on ray BX.
Step 3: Join DC and draw the perpendicular bisector PQ of DC.
Step 4: Let it intersect BX at point A. Join AC.
Thus, \u0394ABC is the required triangle.<\/p>\n\n\n\n

3. Construct a triangle PQR in which QR = 6cm, \u2220Q = 60\u00b0 and PR \u2013 PQ = 2cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of Construction:
Step 1: A ray QX is drawn and cut off a line segment QR = 6 cm from it.
Step 2:. A ray QY is constructed making an angle of 60\u00ba with QR and YQ is produced to form a line YQY’
Step 3: Cut off  a line segment QS = 2cm  from QY’. RS is joined.
Step 5: Draw perpendicular bisector  of RS intersecting QY at a point  P. PR is joined.
Thus, \u0394PQR is the required triangle.<\/p>\n\n\n\n

NCERT 9th Maths Chapter 11, class 9 Maths Chapter 11 solutions<\/p>\n\n\n\n

4. Construct a triangle XYZ in which \u2220Y = 30\u00b0, \u2220Z = 90\u00b0 and XY + YZ + ZX = 11 cm.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of Construction:
Step 1: A line segment PQ = 11 cm is drawn. (XY + YZ + ZX = 11 cm)
Step 2: An angle, \u2220RPQ = 30\u00b0 is constructed at point A and an angle \u2220SQP = 90\u00b0 at point B.
Step 3: \u2220RPQ and \u2220SQP are bisected . The bisectors of these angles intersect each other at point X.
Step 4: Perpendicular bisectors TU of PX and WV of QX are constructed.
Step V: Let TU intersect PQ at Y and WV intersect PQ at Z. XY and XZ are joined.
Thus, \u0394XYZ is the required triangle.<\/p>\n\n\n\n

5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 c<\/strong>m.<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

\"NCERT<\/figure>\n\n\n\n

Steps of Construction: <\/p>\n\n\n\n

Step 1: A ray BX is drawn and a cut off a line segment BC = 12 cm is made on it.<\/p>\n\n\n\n

Step 2: \u2220XBY = 90\u00b0 is constructed.
Step 3: Cut off a line segment BD = 18 cm is made on BY. CD is joined.
Step 4: Perpendicular bisector of CD is constructed intersecting BD at A. AC is joined.
Thus, \u0394ABC is the required triangle.<\/p>\n\n\n\n

NCERT Solutions of Chapter 11 Constructions provided here are detailed and accurate that will be give in depth study of concepts. Through this chapter you will learn some basic constructions and then after construct certain kinds of triangles.<\/p>\n\n\n\n

\u2022 Basic Constructions: We will learn:<\/p>\n\n\n\n

1. To construct the bisector of a given angle.<\/p>\n\n\n\n

2. To construct the perpendicular bisector of a given line segment.<\/p>\n\n\n\n

3. To construct an angle of 60\u00b0 at the initial point of a given ray.<\/p>\n\n\n\n

\u2022 Some Constructions of Triangles: We will learn:<\/p>\n\n\n\n

1. To construct a triangle, given its base, a base angle and sum of other two sides.<\/p>\n\n\n\n

2. To construct a triangle given its base, a base angle and the difference of the other two sides.<\/p>\n\n\n\n

3. To construct a triangle, given its perimeter and its two base angles.<\/p>\n\n\n\n

There are only two exercises in the whole chapter which will let you learn variety of constructions with given measures. You can always find exercisewise NCERT Solutions for Chapter 11<\/strong> just by clicking on the links given below.<\/p>\n\n\n\n

While preparing these NCERT Solutions for Class 9 Maths, Indcareer Schools experts have taken every care and provided a detailed step by step solutions of every question which will help you in learning the concepts embedded in the question.<\/p>\n\n\n\n

NCERT Solutions for Class 9 Maths Chapters:<\/h4>\n\n\n\n

FAQ on <\/strong>Chapter <\/strong>11 Constructions<\/strong><\/p>\n\n\n\n

Why we should solve NCERT Solutions for Chapter 11 Constructions Class 9?<\/h4>\n\n\n\n

These NCERT Solutions will make you equipped with variety of concepts which can be helpful in solving questions in the examinations. Also, our subject matter experts have prepared these Chapter 11 NCERT Solutions as per the latest marking scheme which let you score more marks in tests.<\/p>\n\n\n\n

What are the instruments required for drawing geometrical figures?<\/h4>\n\n\n\n

A protractor, a pair of compasses, a pair of set squares, a pair of dividers, a graduated scale are required  instruments for drawing geometrical figures.<\/p>\n\n\n\n

How many exercises in Chapter 11 Constructions?<\/h4>\n\n\n\n

Chapter 11 Constructions consists of two exercises. In the first one, you will learn to draw basic constructions such as constructing a bisector while in the second one you will have to draw triangles with given measures.<\/p>\n\n\n\n

How can I understand the topics given in Chapter 11 Class 9 Maths?<\/h4>\n\n\n\n

It is very necessary to understand the topics present in Chapter 11 Class 9 Maths and NCERT Solutions is one of the best way of doing. By practicing questions, you will get to know the concepts behind them and will make you ready for supplementary Maths Books of Class 9.<\/p>\n\n\n\n

NCERT 9th Maths Chapter 11, class 9 Maths Chapter 11 solutions<\/p>\n\n\n\n

NCERT Solutions for 9th class Maths : Chapter 11: Download PDF<\/strong><\/h2>\n\n\n\n

NCERT Solutions for 9th class Maths : Chapter 11 Constructions<\/p>\n\n\n\n

Download PDF<\/strong>: NCERT Solutions for 9th class Maths : Chapter 11 Constructions PDF<\/a><\/p>\n\n\n\n

Chapterwise NCERT Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n