CBSE Important Questions Class 9 Maths Chapter 11

Important Questions Class 9 Mathematics Chapter 11 – Constructions

In Chapter 11 of Class 9, students will learn some basic constructions. The method will subsequently be utilised to construct different sorts of triangles. This chapter discusses how to construct an angle or its bisector as well as how to construct triangles when different parameters are provided. Students will learn how to create the perpendicular bisector of a given line segment, the bisector of a given angle, the sum of a triangle, and much more.

Students can more easily and successfully prepare for all the concepts contained in the CBSE Syllabus with the help of Extramarks’ important questions Class 9 Mathematics Chapter 11. Students are also provided with revision notes, which contain explanations, important formulas, and time-saving suggestions to help them quickly revise all the topics. Students can prepare for exams better by practising the important questions Class 9 Mathematics Chapter 11.

You may prepare for the next board exams and improve your marks by using the important questions Class 9 Mathematics Chapter 11. Extramarks’ team has decided to get you ready for the Class 9 exam in accordance with the CBSE curriculum. Solving and practising important questions Class 9 Mathematics Chapter 11 will also help you grasp the topic better.

These important questions from Mathematics Class 9 Chapter 11 provide a sample of the kinds of questions that are frequently asked in board exams. Learning about these can also give you more assurance as you take the examinations. To get impressive scores in their final exams, students should practise these important questions Class 9 Mathematics Chapter 11. 

Important Questions Class 9 Mathematics Chapter 11 – With Solutions

Construction is a crucial topic that is included in Class 9 Math board exams and helps students understand how various shapes are constructed. Discover how to create a perpendicular bisector, a bisector, and much more. By solving important questions Class 9 Mathematics Chapter 11, you can learn more about Construction and how to tackle different types of questions. 

A few Important Questions Class 9 Mathematics Chapter 11 are provided here, along with their answers:

Question 1: With the help of a ruler and compass, it is not possible to construct an angle of

(a) 22.5°

(b) 40°

(c) 37.5°

(d) 67.5°

Solution 1:

Question 2:

The construction of ΔABC, if given that BC = 6 cm, ∠B = 45° is not possible when the difference between AB and AC is equal to

(a) 6.9 cm

(b) 4.0 cm

(c) 5.0 cm

(d) 5.2 cm

Solution 2:
We know the construction of a triangle is not possible when the sum of the two sides of a triangle is less than or equal to the third side of a triangle.

i.e., AB + BC < AC

= BC < AC – AB

i.e., 6 < AC-AB

Thus, if AC – AB= 6.9 cm, then the construction of ΔABC with the given conditions is not possible.

Question 3:

The construction of a ΔABC, given that BC = 3 cm, ∠C = 60° is possible when the difference between AB and AC is equal to

(a) 3 cm            

(b) 3.1 cm

(c) 3.2 cm 

(d) 2.8 cm

Solution 3:
We know the construction of a triangle is possible when the sum of any two sides of the triangle is greater than the third side.

 i.e.,

AB+ BC > AC

=> BC > AC – AB

=> 3 > AC – AB

Thus, if AC – AB = 2.8 cm, then the construction of ΔABC with the given conditions is possible.

Write True or False and justify the answer.

Question 4:

An angle of 52.5° can be constructed.

Solution 4:
For constructing an angle of 52.5° firstly, we will construct an angle of 90°, then we construct an angle of measure 120° and then plot the angle bisector of measurement 120° and 90° to get an angle of 105° (90° + 15°). Now,  we bisect this angle to get an angle of 52.5°.

Question 5:

It is possible to construct an angle of 42.5° with the ruler and compass.

Solution 5:
We know 42.5° = ½ x 85° and in general, construction of an angle of 85° with the ruler and compass is not possible.

Question 6:

ΔABC can be constructed where AB = 5 cm, ∠A = 45° and BC + AC = 5 cm.

Solution 6:
We know a triangle can be constructed if only the sum of the two sides is greater than the third side.

Here, BC + AC = AB = 5 cm

So, ΔABC cannot be constructed.

Question 7:

ΔABC can be constructed where ∠C = 30°, BC = 6 cm, and AC – AB =4 cm.

Solution 7:
We know a triangle can be constructed if the sum of its two sides is greater than the third side.

i.e., in ΔABC,  AB + BC > AC

=> BC > AC – AB

i.e., 6 > 4, which is true, so ΔABC with the given conditions can be constructed.

Question 8:

A ΔABC can be constructed where AB +BC + CA = 10 cm and ∠B =105°, ∠C = 90°.

Solution 8:
Given, ∠B = 105°, ∠C = 90°

and AB + BC + CA = 10cm

We know the sum of the angles of a triangle is 180°.

∠A + ∠B + ∠C = 180°

Here, ∠B + ∠C = 105°+90°

= 195° >180°, which is not true.

So, ΔABC with the given conditions cannot be constructed.

Question 9:

A ΔABC can be constructed where AB + BC + CA = 12 cm and ∠B = 60°, ∠C =45° .

Solution 9:
We know that the sum of the angles of a triangle is 180°.

∠A + ∠B + ∠C = 180°

Here, ∠B + ∠C = 60°+ 45° = 105°< 180°,

So, ΔABC with the given conditions can be constructed.

Question 10: Construct an angle of measure 90° at the initial point of a given ray and justify the construction.

Solution 10:
Step I: Draw AB.

Step II: Taking O as the centre and having any suitable radius, we draw a semicircle, which cuts OA at B.

Step III: Keeping the radius the same, we divide the said semicircle into three equal parts in such a way that BC=CD=DE

Step IV: We draw OC and OD.

Step V: We draw OF, the bisector of  ∠COD.

Thus, ∠AOF = 90°

Justification:

∵ O is the centre of the semicircle and is divided into 3 equal parts.

∴ BC=CD=DE

⇒ ∠BOC = ∠COD = ∠DOE [Since equal chords subtend the equal angles at the centre]

And, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC + ∠BOC + ∠BOC = 180°

⇒ 3∠BOC = 180°

⇒ ∠BOC = 60°

In a similar manner, ∠COD = 60° and ∠DOE = 60°

∵ OF  is the bisector of ∠COD

∴ ∠COF = 12 ∠COD = 12(60°) = 30°

Now, ∠BOC + ∠COF = 60° + 30°

⇒ ∠BOF = 90° or ∠AOF = 90°

Question 11: Construct an angle of measure 45° at the initial point of a given ray and justify the construction.

Solution 11:
Step I: Draw OA.

Step II: Taking O as the centre and with a suitable radius, we draw a semicircle such that it intersects 

OA at B.

Step III: Taking B as the centre and keeping the same radius, cut the semicircle at C. Now, taking C as the centre and keeping the same radius, we cut the semicircle at D and E, such that BC=CD=DE.

Step IV: We draw OC and OD.

Step V: We draw OF, the angle bisector of ∠BOC.

Step VI: We draw OG, the angle bisector of ∠FOC.

Thus, ∠BOG = 45° or ∠AOG = 45°

Justification:

∵ BC=CD=DE

∴ ∠BOC = ∠COD = ∠DOE [Since equal chords subtend the equal angles at the centre]

Since, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC = 60°

∵ OF  is the bisector of ∠BOC.

∴ ∠COF = 12 ∠BOC = 12 (60°) = 30° …equation (1)

Also, OG is the bisector of ∠COF.

∠FOG = 12 ∠COF = 12 (30°) = 15° …equation (2)

Adding (1) and (2), we get

∠COF + ∠FOG = 30° + 15° = 45°

⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF]

⇒ ∠BOG = 45°

Question 12. Construct the angles of the given measurements

(i) 30°

(ii) 22 12∘

(iii) 15°

Solution 12:
Steps of Construction:

Step I: Draw OA.

Step II: With O as the centre and having a suitable radius, draw an arc-cutting OA at B.

Step III: With the centre at B and the same radius as above, we draw an arc which cuts the previous arc at C.

Step IV: We join OC, which measures  ∠BOC = 60°.

Step V: We draw OD, the bisector of ∠BOC, in such a way that ∠BOD = 12∠BOC = 12(60°) = 30°

Hence, ∠BOD = 30° or ∠AOD = 30°

(ii) Angle of 22 12∘

Steps of Construction:

Step I: We draw OA.

Step II: We construct ∠AOB = 90°

Step III: We draw OC, the bisector of ∠AOB, in such a way that

∠AOC = 12∠AOB = 12(90°) = 45°

Step IV : Now, we draw OD, the bisector of ∠AOC, so that ∠AOD = 12∠AOC = 12(45°) = 22 12∘

Hence, ∠AOD = 22 12∘

(iii) Angle of 15°

Steps of Construction:

Step I: We draw OA.

Step II: We construct ∠AOB = 60°.

Step III : We draw OC, the bisector of ∠AOB, so that ∠AOC = 12∠AOB =  12(60°) = 30° i.e., ∠AOC = 30°

Step IV : We draw OD, the bisector of ∠AOC so that ∠AOD = 12 ∠AOC = 12(30°) = 15°

Thus, ∠AOD = 15°

Question 13: Construct the following angles and verify by measuring them with a protractor

(i) 75°

(ii) 105°

(iii) 135°

Solution 13:
Step II: With O as the centre and a radius, we draw an arc cutting  OA at B.

Step III: With centre B and taking the same radius, we mark point C on the previous arc.

Step IV: Taking centre C and having the same radius, we mark another point, D, on the arc of step II.

Step V: We join OC  and OD, which measures ∠COD = 60° = ∠BOC.

Step VI : We draw OP, the bisector of ∠COD, so that ∠COP = 12∠COD = 12(60°) = 30°.

Step VII: We draw OQ, the bisector of ∠COP, so that ∠COQ = 12∠COP = 12(30°) = 15°.

Hence, ∠BOQ = 60° + 15° = 75°∠AOQ = 75°

(ii) Steps of Construction:

Step I: We draw OA.

Step II: Taking centre O and a suitable radius, we draw an arc which cuts OA  at B.

Step III: Taking centre B and the same radius, we mark point C on the previous arc.

Step IV: Taking centre C and the same radius, we mark another point D on the arc drawn in step II.

Step V: Draw OP, the bisector of CD, which cuts CD at E such that ∠BOP = 90°.

Step VI: Draw  OQ, the bisector of BC, such that ∠POQ = 15°

Thus, ∠AOQ = 90° + 15° = 105°

(iii) Steps of Construction:

Step I: Draw OP.

Step II: With centre O and having a suitable radius, draw an arc which cuts OP at A

Step III: Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that AQ = QR = RS.

Step IV: Draw OL, the bisector of RS, which cuts the arc RS at T.

Step V: Draw OM, the bisector of RT.

Thus, ∠POQ = 135°

Question 14: Construct an equilateral triangle, given its side(say) and justify the construction.

Solution 14:
Steps of Construction:

Step I: We draw OA.

Step II: Taking O as the centre and radius equal to 3 cm, we draw an arc to cut OA at B so that OB = 3 cm.

Step III: Taking B as the centre and radius equal to OB, we draw an arc to intersect the previous arc at C.

Step IV: Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:

∵ The arcs OC and BC are drawn with the same radius.

∴ OC = BC

⇒ OC = BC [Chords corresponding to equal arcs are equal]

∵ OC = OB = BC

∴ OBC is an equilateral triangle.

Question 15: Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Solution 15:
Step I: We draw BX

Step II: Along BX, we cut off a line segment BC = 7 cm.

Step III: At B, we construct ∠CBY = 75°

Step IV : From BY, we cut off BD = 13 cm (= AB + AC)

Step V: We join DC.

Step VI: We draw a perpendicular bisector of CD, which meets BD at A.

Step VII: We join AC.

Thus, ∆ABC is the required triangle.

Question 16: Construct a ABC in which AB – AC = 35 cm and BC = 8 cm, ∠B = 45° .

Solution 16:
Step I: Draw BX

Step II: Along BX, cut off a line segment BC = 8 cm.

Step III: At B, construct ∠CBY = 45°

Step IV : From BX, cut off BD = 3.5 cm (= AB – AC)

Step V: Join DC.

Step VI: Draw PQ, the perpendicular bisector of DC, which intersects BY at A.

Step VII: Join AC.

Thus, ∆ABC is the required triangle.

Question 17: Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.

Solution 17:
Step I: Draw QX

Step II: Along QX, cut off a line segment QR = 6 cm.

Step III: Construct a line YQY’ such that ∠RQY = 60°.

Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’.

Step V: Join SR.

Step VI: Draw MN, the perpendicular bisector of SR, which intersects QY at P.

Step VII: Join PR.

Thus, ∆PQR is the required triangle.

Question 18: Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.

Solution 18:
Step I : Draw a line segment AB = 11 cm = (XY+YZ + ZX)

Step II : Construct ∠BAP = 30°

Step III: Construct ∠ABQ = 90°

Step IV: Draw AR, the bisector of ∠BAP.

Step V: Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X.

Step VI: Draw the perpendicular bisector of AX, which intersects AB at Y.

Step VII: Draw the perpendicular bisector of XB, which intersects AB at Z.

Step VIII: Join XY and XZ.

Thus, ∆XYZ is the required triangle.

Question 19: Construct a right triangle in which the base is 12 cm, and the sum of its hypotenuse and other side is 18 cm.

Solution 19:
Step I : Draw BC = 12 cm.

Step II: At B, construct ∠CBY = 90°.

Step III: Along BY, cut off a line segment BX = 18 cm.

Step IV: Join CX.

Step V: Draw PQ, the perpendicular bisector of CX, which meets BX at A.

Step VI: Join AC.

Thus, ∆ABC is the required triangle.

Question 20:

Draw an angle of measure 110° with the help of a protractor and bisect it. Measure each angle.

Solution 20:

1. We take X as the centre and any radius, and we draw an arc to intersect the rays XA and XB at E and D, respectively.
2. We take D and E as the centres, and with a radius of more than ½ DE, we draw arcs to intersect
3.  each other at F.
4. We draw the ray XF.
5. So, the ray XF is the bisector of the angle B X A. Measuring each angle, we get
6. ∠BXC = ∠AXC = 55°.
7. [∴ ∠BXC = ∠AXC = ½ ∠BXA = ½ x 110° = 55° ]

Question 21: Draw a line segment AB of measure 4 cm in length. Draw a line perpendicular to AB through points A and B, respectively. Are these lines parallel?

Solution 21:

• We draw the line segment AB = 4 cm.
• We take 4 as the centre and a radius of more than ½ AB (i.e., 2 cm), and we draw an arc such that it intersects AB at E.
• We take E as the centre, and with the same radius as above, we draw an arc which intersects the previous arc at F.
•  Again, we take F as the centre and, with the same radius as the above, we draw an arc intersecting the previous arc (obtained in step ii) at G.

1. We take G and F as the centres and draw arcs which intersect each other at H.
2. We then join AH. Hence, AX is perpendicular to AB at A. In a similar manner, we draw BY ⊥ AB at B.
3. Now, we know that if the two lines are parallel, then the angle between them will be 0° or
4.  180°.
5. i.e., , ∠XAB = 90° [∴ XA ⊥ AB]
6. and ∠YBA = 90° [ ∴ YB ⊥ AB]
7. ∠XAB+ ∠YBA = 90° + 90° = 180°
8. Thus, lines XA and YS are parallel.
9. [since the sum of the interior angle on the same side of the transversal is 180°, then the two lines are parallel]

Question 22: Construct a triangle in which the sides are 3.6 cm, 4. 8 cm and 3.0 cm. Bisect the smallest angle and measure each part.

Note: The angle opposite to the smallest side is the smallest.

Solution 22:

• We draw a line segment BC of the length 4.8 cm.
• From B, point A is at a distance of 3.6 cm. So, taking B as the centre, we draw an arc of radius 3.6 cm.
• From C, point A is at a distance of 3 cm. So, taking C as the centre, we draw an arc taking a radius 3 cm and intersecting the previous arc at A.
• Join AB and AC. Thus, ΔABC is the required triangle.

Here, angle B is the smallest, as AC is the smallest side. Now, to direct angle B, we use the following steps.

1. We take B as the centre, and we draw an arc intersecting AB and BC respectively at D and E.
2. Taking D and E as the centres, we draw arcs intersecting at P.
3. Joining BP, we obtain an angle bisector of ∠B.
4. Flere, ∠ABC=39°
5. Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

Question 23:

Construct a ΔABC where BC = 5 cm, ∠B = 60° and

Solution 23:

• We draw the base BC = 5 cm.
• At point B, we make an ∠XBC = 60°.
• We then cut a line segment BD which is equal to AB + AC = 7.5 cm from the ray BX.
• Join DC.
• Make a ∠DCY = ∠BDC.
•  Let CY intersect BX at A.
• Then, ΔABC is the required triangle.

Question 24: Construct a square of side 3 cm.

Thinking Process: Firstly, we draw a line segment of the given length. At both ends of the segment, draw an angle of 90° from these line segments and draw a line up to the given length. Further, we draw an angle of 90° from these lines, and thus we join the other line to get the required construction.

Solution 24:
We are to construct a square of side 3 cm using the following steps.

1. We draw a line segment AS of length 3 cm.
2. Now, we generate an angle of measure 90° at points A and B of the line segment, and then we plot the parallel lines AX and BY at these points.
3. We then cut AD and SC of length 3 cm from AX and BY, respectively.
4. We draw an angle of 90° at any one of the points C or D and then join both the points by the line segment CD of length 3 cm.
5. Therefore, ABCD is the required square of side, 3 cm.

Alternatively,

We are to construct a square ABCD of side 3 cm, using the following steps

1. We draw a line segment AB of length 3 cm.
2. Now, we draw an angle XAB = 90° at point A of the line segment AB.
3. We then cut the line segment AD = 3 cm from the ray AX and joined BD.
4. Now, from D, point C is at a distance of 3 cm. 
5. Thus, taking D as the centre, we draw an arc of radius 3 cm.
6. From B, point C is at the distance of 3 cm. So, taking 8 as the centre, we draw an arc of 3 cm radius and intersecting the previous arc (obtained in step iv) at C.
7. We join DC and BC. So, ABCD is the required square of side 3 cm.

Question 25: Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.

Solution 25:
We are to construct a rectangle whose measure of adjacent sides are of lengths 5 cm and 3.5 cm, using the following steps.

1. We draw a line segment BC of length 5 cm.
2. Now, we generate an angle measuring 90° at the points B and C of the line segment BC and then plot the parallel lines BX and CY at these points.

1. We then cut AB and CD measuring lengths 3.5 cm from BX and CY, respectively.
2. We draw an angle of 90° at one of the points A or D and then join both the points by a line segment AD of length 5 cm.
3. So, ABCD is a rectangle with
4. adjacent sides of length 5 cm and 3.5 cm.

Alternatively,

We are to construct a rectangle ABCD whose adjacent sides measure of lengths 5 cm and 3.5 cm using the following steps.

1. We draw a line segment SC of length 5 cm.
2. Now, we draw an ∠XBC = 90° at point B of the line segment SC.
3. We cut a line segment AB = 3.5 cm from the ray BX and joined AC.
4. Now, from A, point D is at a distance of 5 cm. So, taking A as the centre, we draw an arc of radius 5 cm.
5. From C, point D is at the distance of 3.5 cm. So, taking C as the centre, we draw an arc of radius 3.5 cm and intersect the previous arc (obtained in step iv) at D.
6. We join AD and CD.
7. Hence, ABCD is the required rectangle with adjacent sides measuring 5 cm and 3.5 cm.

Question 26:

Construct a rhombus where the side is of length 3.4 cm, and one of the angles is 45°.

Solution 26:

1. We draw a line segment AS of length 3.4 cm.

1. Now, we generate an angle of 45° at both the ends A and B of the line segment AB and then plot the parallel lines AX and BY.
2. We cut AD and SC measuring lengths 3.4 cm from AX and BY, respectively.
3. We draw an angle of 45° at one of the points D or C and then join both the points by the line segment DC of length 3.4 cm, which is parallel to AB.
4. Therefore, ABCD is the rhombus in which the side is of length 3.4 cm, and one of the angles is 45°.

Alternatively,

We are to construct a rhombus in which the side is of length 3.4 cm, and one of the angles is 45°, using the following steps.

1. We draw a line segment AB of length 3.4 cm.
2. Now, we draw an angle XAB = 45° at point A of the line segment AB.
3. We cut the line segment AD = 3.4 cm from the ray AX and followed by joining BD.
4. Now, from D, point C is at a distance of 3.4 cm. Thus, taking D as the centre, we draw an arc of radius 3.4 cm.
5. From B, point C is at a distance of 3.4 cm. So, having B as the centre draws an arc of radius 3.4 cm and intersecting the previous arc (obtained in step iv) at C.
6. We join CD and BC.
7. Thus, ABCD is the rhombus whose side measures 3.4 cm, and one of its angles is 45°.

Question 27: Construct a triangle if its perimeter is 10.4 cm and the two angles are 45° and 120°.

Solution 27:
perimeter = 10.4 cm, i.e., AB+ BC + CA = 10.4 cm 

and the two angles are 45° and 120°.

suppose ∠B = 45° and ∠C = 120°

Now, we are to construct the ΔABC using the following steps.

1. We draw a line segment XY  which is equal to the perimeter, i.e., AB+ BC + CA = 10.4 cm.

We make angle ∠LXY = ∠B = 45° and ∠MYX = ∠C = 120°.

1. And bisect ∠LXY and ∠MYX and let these bisectors intersect at point A (suppose).
2. We draw the perpendicular bisectors PQ and RS, respectively, of AX and AY.
3. We suppose PQ intersecting XY at B and RS intersecting XY at C. We then join AB and AC. Therefore, ΔABC is the required triangle.

Justification

Since B is lying on the perpendicular bisector PQ of AX.

So, AB+ BC + CA = XB+ BC + CY=XY

Again, ∠BAX = ∠AXB [∴ in ΔAXB, AB = XB] …equation(i)

Also, ∠ABC = ∠BAX + ∠AXB [ ∠ABC is the exterior angle of ΔAXB]

= ∠AXB + ∠AXB [from Equation (i)]

= 2 ∠AXB= ∠LXY [ AX is the bisector of ∠LXB]

And, ∠CAY = ∠AYC [∴ in A AYC, AC = CY]

∠ACB=∠CAY + ∠AYC [ ∠ACB is the exterior angle of ΔAYC]

= ∠CAY + ∠CAY

= 2 ∠CAY= ∠MYX [∴ AY is the bisector of ∠MYX]

Thus, our construction is justified.

Question 28: Proof that ΔPQR is a triangle., given that ∠PQR = 45° ,QR = 3 cm, and QP – PR =2 cm.

Solution 28:
and QP – PR = 2 cm

Since C is lying on the perpendicular bisector RS of AY.

We are to construct ΔPQR using the following steps.

1. We draw the base QR of length 3 cm.
2. We make an angle XQR = 45° at the point Q of base QR.
3. We cut the line segment QS =QP- PR = 2 cm from the ray QX.

1. We join SR and draw the perpendicular bisector of SR, i.e., AB.
2. Let the bisector AB intersect QX at P. We join PR. Thus, ΔPQR is the required triangle.

Justification

As given, the base QR and ∠QPR are drawn.

Here, point P is lying on the perpendicular bisector of SR.

PS = PR

Now, QS = PQ – PS

= PQ -PR

Thus, our construction is justified.

Question 29: Construct an equilateral triangle, if its altitude is 3.2 cm.

Solution 29:
Given the altitude of an equilateral triangle, ABC is 3.2 cm. We are to construct the ΔABC using the following steps.

1. We draw a line PQ.
2. We take a point D on PQ and then draw a ray DE ⊥ PQ.
3. We then cut the line segment AD measuring length 3.2 cm from DE.
4. We make the angles equal to 30° at A on both the sides of AD, suppose ∠CAD and ∠BAD, where B and C lies on PQ.
5. We cut the line segment DC from PQ so that DC = BD

We join AC

Thus,  ABC is a required triangle.

Justification

Here, ∠A = ∠BAD + ∠CAD

= 30°+30° =60°.

And AD ⊥ SC

∴ ∠ADS = 90°.

In ΔABD, ∠BAD + ∠DBA = 180° [by angle sum property]

30° + 90° + ∠DBA = 180° [since ∠BAD = 30°, by construction ]

∠DBA = 60°

Similarly, ∠DCA = 60°

Thus, ∠A = ∠B=∠C = 60°

Hence, ΔABC is an equilateral triangle.

Question 30: Construct a rhombus whose diagonals are 6 cm and 4 cm in length.

Solution 30:sing the following steps.

1. We draw the diagonal, say AC = 4 cm.
2. We take A and C as centres and radius of more than ½ AC, and we draw the arcs on both sides of the line segment AC intersecting each other.
3. We cut both the arcs so that they intersect each other at P and Q. We then join PQ.
4. Let PQ intersect AC at point O. Thus, PQ is the perpendicular bisector of AC.
5. We cut off 3 cm lengths from OP and OQ, and then we get points B and D.
6. Now, join AB, BC, CD, and DA.
7. Thus, ABCD is the required rhombus.

Justification

Since D and B are lying on the perpendicular bisector of AC.

DA = DC and BA = BC …equation(i)

[since every point on the perpendicular bisector of the line segment is equidistant from endpoints

of line segment]

Now, ∠DOC = 90°

Also, OD = OB = 3 cm

Thus, AC is a perpendicular bisector or BD.

CD = CB …equation (ii)

AB = BC =CD = DA

From Equations (i) and (ii)

ABCD is a rhombus.

Question 31. Draw a line segment measuring AB = 8 cm. Draw 13 parts of it. Measure the length of 13 the part of AB.

Solution 31:
Steps of Construction :

1. We draw a line segment AB = 8 cm.
2. We draw its perpendicular bisector, which intersects AB in M.
3. We draw the perpendicular bisector of MB and which intersects AB in N. Thus, AN = 13 of AB = 6 cm.

Question 32: Why it is not possible  to construct a ∆ABC, if AB = 6 cm, ∠A = 60°,  and AC + BC = 5 cm but construction of ∆ABC is possible if ∠A = 60°, AB = 6 cm and AC – BC = 5 cm ?

Solution 32: = 5 cm which is clearly less than AB (6 cm) Therefore, ∆ABC is not possible.

Also, by triangle inequality property, the construction of a triangle is possible if the difference between two sides of the triangle is less than the third side.

i.e., AC – BC = 5 cm, thus, less than AB (6 cm)

Therefore, ∆ABC is possible.

Question 33. Construct an angle of measure 90° at the initial point of the given ray.

Solution 33:
Steps of Construction :

1. We draw a ray OA.
2. Taking O as the centre and any convenient radius, we draw an arc, cutting OA at P.
3. With P as the centre and the same radius, we draw another arc cutting the arc drawn previously in step 2 at Q.
4. Taking Q as the centre and the same radius as in steps 2 and 3, we draw another arc, cutting the arc drawn previously in step 2 at R.
5. With Q and R as the centres and the same radius, we draw two arcs, cutting each other at S.
6. We join OS and produce it to B. Thus, ∠AOB is the required angle of 90°

Question 34: Draw a straight angle. Using a compass, bisect it. Name the angles obtained.

Solution 34:
Steps of Construction :

1. Draw any straight angle (say ∠AOC).
2. Bisect ∠AOC and join BO.
3. ∠AOB is the required bisector of straight-angle AOC.

Question 35: Draw any reflex angle. Bisect it using a compass. Name the angles so obtained.

Solution 35:
Steps of Construction :

1. Let ∠AOB be any reflex angle.
2. With O as the centre and any convenient radius, we draw an arc-cutting OA in P and OB in Q.
3. With P and Q as the centres, we draw the two arcs of radius little more than half of it and which intersects each other in C. Join OC. Thus, OC is the required bisector. Angles so obtained are ∠AOC and ∠COB.

Question 36: Construct a triangle whose sides are in the ratio of 2 : 3: 4 and the perimeter is 18 cm.

Solution 36:
Steps of Construction :

1. We draw a line segment AB =18 cm.
2. At A,  we construct an acute angle ∠BAX (< 90°).
3. We mark 9 points on AX, in such a way that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6
4. = A6A7 = A7A8 = A8A9.
5. We then join A9B.
6. From A2 and A5, we draw A2M || A5N || A9B, intersecting AB at M and N, respectively.
7. Taking M as the centre and radius AM, we draw an arc.
8. With N as the centre and radius NB, we draw another arc intersecting the previous arc at L.
9. We join LM and LN. Thus, ∆LMN is the required triangle.

Question 37: Construct a ∆ABC where ∠B = 45° BC = 8 cm, and AB – AC = 3.1 cm.

Solution 37:
Steps of Construction :

1. We draw any line segment BC = 8 cm.
2. At B,  we construct an angle ∠CBX = 45°.
3. From BX, we cut off BD = 3.1 cm.
4. We join DC.
5. We draw the perpendicular bisector ‘p’ of DC and which intersects BX in A.
6. We join AC. Thus, ∆ABC is the required triangle.

Question 38: Construct a ∆ABC so that  ∠B = 45°, BC = 3.2 cm,and AC – AB = 2.1 cm.

Solution 38:
Steps of Construction :

1. We draw a line segment BC = 3.2 cm.
2. At B, we construct an angle ∠CBX = 45° and produce it to the point X’.
3. We cut off BD = 2.1 cm and joined CD.
4. We draw the perpendicular bisector of CD and which intersects X’BX in A.
5. We join AC. Thus, ∆ABC is the required triangle.

Question 39: Draw a line segment QR = 5 cm. Construct perpendiculars at points Q and R to it. Name them QX and RY, respectively. Are they both parallel?

Solution 39:
Steps of Construction :

1. We draw a line segment QR = 5 cm.
2. With Q as the centre, we construct an angle of measure 90° and extend this line through Q and name it QX.
3. With R as the centre, we construct an angle of measure 90° and extend this line through R and name it RY. 
4. Thus, the perpendicular lines QX and RY are parallel.

Question 40: Construct an isosceles triangle whose two equal sides measures 6 cm each and whose measure of the base is 5 cm. Draw the perpendicular bisector of its base showing that it passes through the opposite vertex.

Solution 40:
Steps of Construction :

1. We draw a line segment AB = 5 cm.
2. With A and B as centres, we draw two arcs of radius 6 cm which intersect each other in C.
3. We join AC and BC to get ∆ABC.
4. With A and B as centres, we draw two arcs of radius little more than half of AB. We suppose they intersect each other in P and Q. We join PQ and produce it to pass through C.

Question 41: Construct a triangle ABC where BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.

Solution 41:
Given : In ∆ABC, AB + AC = 8.2 cm BC = 4.7 cm, and ∠C = 60°.

Required: To construct ∆ABC.

Steps of Construction :

1. We draw BC = 4.7 cm.
2. Draw
3. From ray CX, we cut off CD = 8.2 cm.
4.  We join BD.
5.  We draw the perpendicular bisector of BD meeting CD at A.
6. We join AB to obtain the required triangle ABC.

Justification :

∵ point A lies on the perpendicular bisector of BD. Thus, AB = AD

Now, CD = 8.2 cm

⇒ AC + AD = 8.2 cm

⇒ AC + AB = 8.2 cm.

Question 42: Construct ∆XYZ. If its perimeter is 14 cm, one side of the length is 5 cm and ∠X = 45°.

Solution 42:
Here, perimeter of the ∆XYZ = 14 cm and measure of one side XY = 5 cm

∴  YZ + XZ = 14 – 5 = 9 cm and ∠X = 45°.

Construction Steps:

1. We draw a line segment XY = 5 cm.
2. We construct a ∠YXA = 45° with the help of a compass and ruler.
3. From ray XA, we cut off XB = 9 cm.
4. We join BY.
5. We draw a perpendicular bisector of BY and which intersects XB in Z.
6. Join ZY. Thus, ∆XYZ is the required triangle.

Question 43: To construct a triangle with a perimeter of 10 cm and base angles of 60° and 45°.

Solution 43:
Given: In ∆ABC,

AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.

Required: We are to construct ∆ABC.

Steps of Construction :

1. We draw DE = 10 cm.
2. At D, we construct ∠EDP= 5 of 60°= 30° and at E, we construct DEQ = 1 of 45o = 22°
3. Let DP and EQ meet at A.
4. We draw a perpendicular bisector of AD, which meets DE at B.
5. We draw a perpendicular bisector of AE, which meets DE at C.
6. We join AB and AC. Thus, ABC is the required triangle.

Question 44: Construct an equilateral triangle in which the altitude is 6 cm long.

Solution 44:
Steps of Construction :

1. We draw a line PQ and consider any point S on it.
2. We construct the perpendicular SR on PQ.
3. From SR, we cut a line segment SA of measuring= 6 cm.
4. At the initial point A of the line segment AS we construct ∠SAB = 30° and ∠SAC = 30°.
5. The arms AB and AC of the angles ∠SAB and ∠SAC meeting PQ at B and C, respectively. Thus, ∆ABC is the required equilateral triangle with an altitude of length 6 cm.

Question 45: Construct a rhombus in which the diagonals are 8 cm and 6 cm long. Measure the length of each side of the rhombus.

Solution 45:
Steps of Construction :

1. We draw a line segment PR = 8 cm.
2. We draw the perpendicular bisector XY of the line segment PR. We suppose O be the point of intersection of PR and XY so that O is the mid-point of PR measuring 8 cm.
3. From OX, we cut a line segment OS = 3 cm, and then from OY, we cut a line segment OQ = 3 cm.
4. We join PS, SR, RQ and QP, and then PQRS is the required rhombus.
5. We measure the length of segments PQ, QR, RS and SP, and each is found to be 5 cm long.

Benefits of Solving Important Questions Class 9 Mathematics Chapter 11

A fundamental idea that has applications in numerous domains is geometry. As a result, familiarising yourself with the idea and its applications is essential. The Chapter 11 Class 9 Mathematics important questions are the finest resource for learning about the Construction chapter. To help students master their CBSE Class 9 Mathematics examination, important questions Class 9 Mathematics Chapter 11 were created by experts at Extramarks.

Below are a few benefits for students to practise from our ImportantQuestions Class 9 Mathematics Chapter 11:

• It provides students with a one-stop solution for all their questions. Consequently, they will save time during preparation by gathering questions and answers beforehand.
• Questions and answers are reviewed and verified by a team of expert Mathematics teachers. The Extramarks teaching resources have gained the trust of lakhs of students.
• Expert Mathematics teachers have created these solutions after reviewing the updated CBSE curriculum. The latest NCERT guidelines are followed by them.
• Exam-focused questions and answers enable students to overcome exam anxiety and approach exams confidently.

Be in touch with us at Extramarks to get the important exam questions for CBSE Class 9 Mathematics, all chapters, with solutions. Use the additional questions offered here by clicking on the links below to practise your revision.

• CBSE revision notes
• CBSE syllabus
• CBSE sample papers
• CBSE past years’ question papers
• CBSE extra questions and solutions