NCERT Exemplar Class 9 Maths Solutions Chapter 11 Constructions

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 11 Constructions for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 11 Constructions pdf, free NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 11 Constructions book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 11
Chapter NameConstructions
CategoryNCERT Exemplar

NCERT Exemplar Class 9 Maths Solutions Constructions

Exercise 11.1: Multiple Choice Questions (MCQs)

Question 1:
With the help of a ruler and a compass it is not possible to construct an angle of
(a) 37.5°
(b) 40°
(c) 22.5°
(d) 67.5°
Solution:
(b) With the help of a ruler and a compass, we can construct the angels, 90°, 60°, 45°, 22.5°, 30° etc., and its bisector of an angle. So, it is not possible to construct an angle of 40°.

Question 2:
The construction of ΔABC, given that BC = 6 cm, ∠B = 45° is not possible when difference of AB and AC is equal to
(a) 6.9 cm
(b) 5.2 cm
(c) 5.0 cm
(d) 4.0 cm
Solution:
(a) Given, BC = 6 cm and ∠B= 45°
We know that, the construction of a triangle is not possible, if sum of two sides is less than or equal to the third side of the triangle.
i.e., AB + BC < AC
=> BC < AC – AB
=> 6 < AC-AB
So, if AC – AB= 6.9 cm, then construction of ΔABC with given conditions is not possible.

Question 3:
The construction of a ΔABC, given that BC = 3 cm, ∠C = 60° is possible when difference of AB and AC is equal to
(a) 3.2 cm                
(b)
3.1 cm
(c) 3 cm
(d) 2.8 cm
Solution:
(d) Given, BC = 3 cm and ∠C=60°
We know that, the construction of a triangle is possible, if sum of two sides is greater than the third side of the triangle i.e.,
AB+ BC > AC
=> BC > AC – AB
=> 3 > AC – AB
So, if AC – AB = 2.8 cm, then construction of ΔABC with given conditions is possible.

Exercise 11.2: Very Short Answer Type Questions

Write whether True or False and justify your answer
Question 1:
An angle of 52.5° can be constructed.
Solution:
True
To construct an angle of 52.5° firstly construct an angle of 90°, then construct an angle of 120° and then plot an angle bisector of 120° and 90° to get an angle 105° (90° + 15°). Now, bisect this angle to get an angle of 52.5°.

Question 2:
An angle of 42.5° can be constructed.
Solution:
False
We know that, 42.5° = ½ x 85° and an angle of 85° cannot be constructed with the help of ruler and compass.

Question 3:
ΔABC can be constructed in which AB = 5 cm, ∠A = 45° and BC + AC = 5 cm.
Solution:
False
We know that, a triangle can be constructed, if sum of its two sides is greater than third side.
Here, BC + AC = AB = 5 cm
So, ΔABC cannot be constructed.

Question 4:
ΔABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB =4 cm.
Solution:
True
We know that, a triangle can be constructed if sum of its two sides is greater than third side.
i.e., in ΔABC,  AB + BC > AC
=> BC > AC – AB
=> 6 > 4, which is true, so ΔABC with given conditions can be constructed.

Question 5:
A ΔABC can be constructed in which ∠B =105°, ∠C = 90° and AB +BC + CA = 10 cm.
Solution:
False
Here, ∠B = 105°, ∠C = 90°
and AB + BC + CA = 10cm
We know that, sum of angles of a triangle is 180°.
∠A + ∠B + ∠C = 180°
Here, ∠B + ∠C = 105°+90°
= 195° >180° which is not true.
Thus, ΔABC with given conditions cannot be constructed.

Question 6:
A ΔABC can be constructed in which ∠B = 60°, ∠C =45°, and AB + BC + CA = 12 cm.
Solution:
True
We know that, sum of angles of a triangle is 180°.
∠A + ∠B + ∠C = 180°
Here, ∠B + ∠C = 60°+ 45° = 105°< 180°,
Thus, ΔABC with given conditions can be constructed.

Exercise 11.3: Short Answer Type Questions

Question 1:
Draw an angle of 110° with the help of a protractor and bisect it. Measure , each angle.
Solution:
Draw ∠BXA = 110° with the help of a protractor. Now, we use the following steps for required construction
ncert-exemplar-problems-class-9-maths-constructions-1

  1. Taking X as centre and any radius daw an arc to Intersect the rays XA and XB, say at E and D, respectively.
  2. Taking D and E as centres and with the radius more than ½ DE, draw arcs to intersect
    each other, say at F.
  3. Draw the ray XF.
    Thus, ray XF is the required bisector of the angle B X A. On measuring each angle, we get
    ∠BXC = ∠AXC = 55°.
    [∴ ∠BXC = ∠AXC = ½ ∠BXA = ½ x 110° = 55° ]

Question 2:
Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?
Solution:
To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction.

  1. Draw a line segment AB = 4 cm.
  2. Taking 4 as centre and radius more than ½ AB (i.e., 2 cm) draw an arc say it intersect AB at E.
  3. Taking E as centre and with same radius as above draw an arc which intersect previous arc at F.
  4.  Again, taking F as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at G.
    ncert-exemplar-problems-class-9-maths-constructions-2
  5. Taking G and F are centres, draw arcs which intersect each other at H.
  6. Join AH . Thus, AX is perpendicular to AB at A. Similarly, draw BY ⊥ AB at B.
    Now, we know that if two lines are parallel, then the angle between them will be 0° or
    180°.
    Here, ∠XAB = 90° [∴ XA ⊥ AB]
    and ∠YBA = 90° [ ∴ YB ⊥ AB]
    ∠XAB+ ∠YBA = 90° + 90° = 180°
    So, the lines XA and YS are parallel.
    [since, it sum of interior angle on same side of transversal is 180°, then the two lines are
    parallel]

Question 3:
Draw an angle of 80° with the help of protractor. Then, construct angles of (1) 40° (2) 160° and (3) 120°.
Solution:
First, draw an angle of 80° say ∠QOA = 180° with the help of protractor. Now, use the the following steps to construct angles of
ncert-exemplar-problems-class-9-maths-constructions-3
(1)40° 2() 160° (3) 120°

  1. Taking O as centre and any radius draw an arc which intersect OA at E and OO at F.
  2. Taking E and F as centres and radius more than ½ EF draw arcs which intersect each other at P.
  3. Join OP Thus, ∠POA = 40° [∴ 40° = ½ x 80°]
  4. Now, taking F as centre and radius equal to EF draw an arc which intersect previous arc obtained in step ii at S.
  5. Join OS. Thus, ∠SOA = 160° [∴ 160° = 2 x 80°]
  6. Taking S and F as centre and radius more than ½ SF draw arcs which intersect each other at R.
  7. Join OR. Thus, ∠ROA = ∠ROQ = 40° + 80°= 120°.

Question 4:
Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part.
Thinking Process
Angle opposite to smallest side is smallest.
Solution:
To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps.

  1. Draw a line segment BC of length 4.8 cm.
  2. From B, point A is at a distance of 3.6 cm. So, having B as centre, draw an arc of radius 3.6 cm.
    ncert-exemplar-problems-class-9-maths-constructions-4
  3. From C, point A is at a distance of 3 cm. So, having C as centre, draw an arc of radius 3 cm which intersect previous arc at A.
  4. Join AB and AC. Thus, ΔABC is the required triangle.

Flere, angle B is smallest, as AC is the smallest side. To direct angle B, we use the following steps.

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

Question 5:
Construct a ΔABC in which BC = 5 cm, ∠B = 60° and
Solution:
Given, in ΔABC, BC=5 cm, ∠B=60° and AC + AB=7.5 cm. To construct the triangle ABC use the following steps.

  1. Draw the base BC = 5 cm.
  2. At the point B make an ∠XBC = 60°.
  3. Cut a line segment BD equal to AB + AC = 7.5 cm from the ray BX.
    ncert-exemplar-problems-class-9-maths-constructions-5
  4. Join DC.
  5. Make an ∠DCY = ∠BDC.
  6.  Let CY intersect BX at A.
    Then, ΔABC is the required triangle.

Question 6:
Construct a square of side 3 cm.
Thinking Process
Firstly, draw a line segment of given length. In both ends of segment draw an angle of 90° from these line segment draw a line upto the given length. Further draw an angle of 90° from these line and join the other line, to get the required construction.
Solution:
We know that, each angle of a square is right angle (i.e., 90°).
To construct a square of side 3 cm, use the following steps.

  1. Draw a line segment AS of length 3 cm.
  2. Now, generate an angle of 90° at points A and B of the line segment and plot the parallel lines AX and BY at these points.
  3. Cut AD and SC of length 3 cm from AX and BY, respectively.
  4. Draw an angle of 90° at any one of the point C or D and join both points by a line segment CD of length 3 cm.
    Thus, ABCD is the required square of side, 3 cm.
    ncert-exemplar-problems-class-9-maths-constructions-6

Alternate Method
To construct a square ABCD of side 3 cm, use the following steps

  1. Draw a line segment AB of length 3 cm.
  2. Now, draw an angle XAB = 90° at point A of line segment AB.
  3. Cut a line segment AD = 3 cm from the ray AX and join BD.
  4. Now, from D point C is at a distance of 3 cm. So, having D as centre, draw an arc of radius 3 cm.
    ncert-exemplar-problems-class-9-maths-constructions-7
  5. From B, point C is at a distance of 3 cm. So, having 8 as centre, draw an arc of radius 3 cm which intersect previous arc (obtained in step iv) at C.
  6. Join DC and BC. Thus, ABCD is the required square of side 3 cm.

Question 7:
Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.
Solution:
We know that, each angle of a rectangle is right angle (i.e., 90°) and its opposite sides are equal and parallel.
To construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm, use the 1 following steps

  1. Draw a line segment BC of length 5 cm.
  2. Now, generate an angle of 90° at points B and C of the line segment BC and plot the parallel lines BX and CY at these points.
    ncert-exemplar-problems-class-9-maths-constructions-8
  3. Cut AB and CD of length 3.5 cm from BX and CY, respectively.
  4. Draw an angle 90° at one of the point A or D and join both points by a line segment AD of length 5 cm.
    Thus, ABCD is the required rectangle with
    adjacent sides of length 5 cm and 3.5 cm.

Alternate Method
To construct a rectangle ABCD whose adjacent sides are of lengths 5 cm and 3.5 cm, use the following steps

  1. Draw a line segment SC of length 5 cm.
  2. Now, draw an ∠XBC = 90° at point B of line segment SC.
  3. Cut a line segment AB = 3.5 cm from the ray BX and join AC.
    ncert-exemplar-problems-class-9-maths-constructions-9
  4. Now, from A, point D is at a distance of 5 cm. So, having A as centre draw an arc of radius 5 cm.
  5. From C, point D is at a distance of 3.5 cm. So, having C as centre draw an arc of radius 3.5 cm which intersect previous arc (obtained in step iv) at D.
  6. Join AD and CD.
    Thus, ABCD is the required rectangle with adjacent sides of length 5 cm and 3.5 cm.

Question 8:
Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45°.
Solution:
We know that, in rhombus all sides are equal. To construct a rhombus whose side is of length 3.4 cm and one of its angle is 45°, use the following steps

  1. Draw a line segment AS of length 3.4 cm.
    ncert-exemplar-problems-class-9-maths-constructions-10
  2. Now, generate an angle 45° at both ends A and B of line segment AB and plot the parallel lines AX and BY.
  3. Cut AD and SC of length 3.4 cm from AX and BY, respectively.
  4. Draw an angle of 45° at one of the point D or C and join both points by a line segment DC of length 3.4 cm and parallel to AB.
    Thus, ABCD is the required rhombus whose side is of length 3.4 cm and one of its angle is 45°.
    ncert-exemplar-problems-class-9-maths-constructions-11

Alternate Method
To construct a rhombus whose side is of length 3.4 cm and one of its angle is 45°, use the following steps.

  1. Draw a line segment AB of length 3.4 cm. .
  2. Now, draw an angle XAB = 45° at point A of line segment AB.
  3. Cut a line segment AD = 3.4 cm from the ray AX and join BD.
  4. Now, from D point C is at a distance of 3.4 cm. So, having D as centre draw an arc of radius 3.4 cm.
  5. From B, point C is at distance of 3.4 cm. So, having B as centre draw an arc of radius 3.4 cm which intersect previous arc (obtained in step iv) at C.
  6. Join CD and BC.
    Thus, ABCD is required rhombus whose side is of length 3.4 cm and one of its angle is 45°.

Exercise 11.4: Long Answer Type Questions

Construct each of the following and give justification
Question 1:
A triangle if its perimeter is 10.4 cm and two angles are 45° and 120°.
Solution:
Let ABC be a triangle. Then, given perimeter = 10.4 cm i.e., AB+ BC + CA = 10.4 cm and two angles are 45° and 120°.
say ∠B = 45° and ∠C = 120°
Now, to construct the ΔABC use the following steps.

  1. Draw a line segment say XY and equal to perimeter i.e., AB+ BC + CA = 10.4 cm
    ncert-exemplar-problems-class-9-maths-constructions-12
  2. Make angle ∠LXY = ∠B = 45° and ∠MYX = ∠C = 120°.
  3. Bisect ∠LXY and ∠MYX and let these bisectors intersect at a point A (say).
  4. Draw perpendicular bisectors PQ and RS of AX and AY, respectively.
  5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC. Thus, ΔABC is the required triangle.

Justification
Since, B lies on the perpendicular bisector PQ of AX.
Thus, AB+ BC + CA = XB+ BC + CY=XY
Again, ∠BAX = ∠AXB [∴ in ΔAXB, AB = XB] …(i)
Also, ∠ABC = ∠BAX + ∠AXB [ ∠ABC is an exterior angle of ΔAXB]
= ∠AXB + ∠AXB [from Eq. (i)]
= 2 ∠AXB= ∠LXY [ AX is a bisector of ∠LXB]
Also, ∠CAY = ∠AYC [∴ in A AYC, AC = CY]
∠ACB=∠CAY + ∠AYC [ ∠ACB is an exterior angle of ΔAYC]
= ∠CAY + ∠CAY
= 2 ∠CAY= ∠MYX [∴ AY is a bisector of ∠MYX]
Thus, our construction is justified.

Question 2:
A ΔPQR, given that QR = 3 cm, ∠PQR = 45° and QP – PR =2 cm.
Solution:
Given, inv ΔPQR, QR = 3 cm, ∠PQR = 45°
and QP – PR = 2 cm
Since, C lies on the perpendicular bisector RS of AY.
To construct ΔPQR, use the following steps.

  1. Draw the base QR of length 3 cm.
  2. Make an angle XQR = 45° at point Q of base QR.
  3. Cut the line segment QS =QP- PR = 2 cm from the ray QX.
    ncert-exemplar-problems-class-9-maths-constructions-13
  4. Join SR and draw the perpendicular bisector of SR say AB.
  5. Let bisector AB intersect QX at P. Join PR Thus, ΔPQR is the required triangle.

Justification
Base QR and ∠PQR are drawn as given.
Since, the point P lies on the perpendicular bisector of SR.
PS = PR
Now, QS = PQ – PS
= PQ -PR
Thus, our construction is justified.

Question 3:
A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm.
Solution:
Let given right triangle be ABC.
Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps

  1. Draw the base BC = 3.5 cm
  2. Make an angle XBC = 90° at the point B of base BC.
    ncert-exemplar-problems-class-9-maths-constructions-14
  3. Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.
  4. Join DC and make an ∠DCY equal to ∠BDC.
  5. Let Y intersect BX at A.
    Thus, ΔABC is the required triangle.

Justification
Base BC and ∠B are drawn as given.
In ΔACD, ∠ACD = ∠ADC [by construction]
AD = AC …(i)
[sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)]
=> BD = AB + AC
Thus, our construction is justified.

Question 4:
An equilateral triangle, if its altitude is 3.2 cm.
Solution:
We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60°.
Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC use the following steps.

  1. Draw a line PQ.
  2. Take a point D on PQ and draw a ray DE ⊥ PQ.
  3. Cut the line segment AD of length 3.2 cm from DE.
  4. Make angles equal to 30° at A on both sides of AD say ∠CAD and ∠BAD, where B and C lie on PQ.
  5. Cut the line segment DC from PQ such that DC = BD

Join AC
Thus, A ABC is the required triangle.
ncert-exemplar-problems-class-9-maths-constructions-15
Justification
Here, ∠A = ∠BAD + ∠CAD
= 30°+30° =60°.
Also, AD ⊥ SC
∴ ∠ADS = 90°.
In ΔABD, ∠BAD + ∠DBA = 180° [angle sum property]
30° + 90° + ∠DBA = 180° [∠BAD = 30°, by construction ]
∠DBA = 60°
Similary, ∠DCA = 60°
Thus, ∠A = ∠B=∠C = 60°
Hence, ΔABC is an equilateral triangle.

Question 5:
A rhombus whose diagonals are 4 cm and 6 cm in lengths.
Solution:
We know that, all sides of a rhombus are equal and the diagonals of a rhombus are perpendicular bisectors of one another. So, to construct a rhombus whose diagonals are 4 cm and 6 cm use the following steps.

  1. Draw the diagonal say AC = 4 cm
  2. Taking A and C as centres and radius more than ½ AC draw arcs on both sides of the line segment AC to intersect each other.
  3. Cut both arcs intersect each other at P and Q, then join PQ.
  4. Let PQ intersect AC at the point O. Thus, PQ is perpendicular bisector of AC.
  5. Cut off 3 cm lengths from OP and OQ, then we get points B and D.
  6. Now, join AB, BC, CD, and DA .
    Thus, ABCD is the required rhombus.
    ncert-exemplar-problems-class-9-maths-constructions-16

Justification
Since, D and B lie on perpendicular bisector of AC.
DA = DC and BA = BC …(i)
[since, every point on perpendicular bisector of line segment is equidistant from end points
of line segment]
Now, ∠DOC = 90°
Also, OD = OB = 3 cm
Thus, AC is perpendicular bisector or BD.
CD = CB …(ii)
AB = BC =CD = DA
from Eq (i)and (ii)
ABCD is a rhombus.


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