MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-1

  1. Draw a ray OA and with O as centre and any radius, draw an arc, cutting OA at B.
  2. With B as centre and the same radius draw an arc cutting the arc drawn in step (i) at C.
  3. With C as centre draw another arc with same radius cutting the arc drawn in step (i) at D.
  4. with C as centre and the same radius draw an arc.
  5. With D as centre and the same radius draw an arc, cutting the arc drawn in step (iv) at E.
  6. Draw OE ∴ ∠AOF = 90°

MP Board Solutions

Question 2.
Construct an angle of 45° at the initial point of a given ray justify the construction.
Solution:

  1. Draw ∠AOF = 90° by following the same steps for constructing a 90° angle.
  2. Draw OG, the bisector of ∠AOF.
  3. ∠AOF= 45°

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-2

Question 3.
Construct the angle of the following measurements:
Solution:

  1. 30°
  2. 22 \(\frac{1}{2}\)°
  3. 15°

1. 30°

  1. Draw a ray OA.
  2. With O as centre and any radius draw an arc which intersect OA at B.
  3. With B as centre and same radius draw an arc cutting the arc drawn is step (ii) at C.
  4. Join OC and produce upward
  5. ∠BOC = 60°
  6. Draw the bisector OD of ABOC.
  7. ∠BOD = ∠COD = 30°

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-3
2. 22 \(\frac{1}{2}\)°

  1. Draw a ray OA.
  2. With O as centre and any radius draw an arc which intersect intersect CM at B.
  3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii) at C.
  4. With C as centre and same radius draw an arc cutting the arc drawn in step (ii) at D.
    MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-4
  5. With C and D as centres and same radius draw arcs to intersect at E.
  6. Join OE.
  7. ∠AOG – 90°
  8. Draw the bisector OE of ∠AOE, to get ∠AOF = 45°
  9. Draw the bisector OG of ∠AOF.
  10. ∠AOG = 22\(\frac{1}{2}\)°

3. 15°

  1. Draw a ray CM.
  2. With O as centre and any radius draw an arc which intersect CM at B.
  3. With B as centre and same radius draw an arc intersecting the arc drawn in step (ii), at C.
  4. Draw OD as the bisector of ZAOC.
  5. ∠BOD = 30°
  6. Draw the bisector OE of ∠AOD.
  7. ∠AOE = 15°

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-5

Question 4.
Construct the following angles and verily by measuring them by a protractor.

(i) 75°
(ii) 105°
(iii) 135°

Solution:
MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-6

Question 5.
Construct an equilateral triangle given its side and justify the construction.
Solution:

  1. Draw a ray AY with intial point A.
  2. With centred and radius equal to length of a side of the A draw an arc BY, cutting the ray AX at B.
  3. With centre B and the same radius draw an arc cutting are BY at C.
  4. Join AC and BC to obtain the required A.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-7

Construction of Triangles:

Example 1:
Construction of a Triangle when its base, a base angle and sum of other two sides are given
Solution:
Step of Construction:

  1. Draw the base PQ.
  2. At point P draw an angle, MPQ equal to the given angle.
  3. Cut a line segment PM equal to sum of sides i.e., (PR + RQ) from point P.
  4. Join MQ.
  5. Draw the perpendicular bisector of MQ which intersect PM at R.
  6. Join QR. PQR is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-8
Justification:
Mark points S as shown in Fig.
QS = MS (∴ RS is the perpendicular bisector of MQ)
RS = RS (Common)
∠QSR = ∠MSR (Each \({ 90 }^{ \underline { 0 } }\))
∆RSQ ≅ ∆RSQ (By SAS)
and so PQ = RM (By CPCT)
Now PR = PM – RM = PM – RQ
PM = PR + RQ.

MP Board Solutions

Example 2:
Construct a AABC in which BC = 3.6 cm, AB + AC = 4.8 and A RSM QS – MS RS = RS cm and B = ∠60°.
Solution:
Steps of Construction:

  1. Draw BC = 3.6 cm.
  2. Draw ∠CBX= \({ 60 }^{ \underline { 0 } }\) at B.
  3. From BX, cut off line segment BD = 4.8 cm.
  4. Join DC.
  5. Draw the perpcndicub: bisector ofDC meeting BD at A.
  6. joinAC.
  7. ABC is the required Mangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-9
Justification:
‘A‘ lies on the perpendicular bisector of DC
∴ AD = AC
Now BD = 4.8 cm
⇒ BA + AD = 4.8
BA + AC = 4.8 (∴ AD = AC)

Example 3:
Construction of a Triangle when its Base Angle and the Difference of the other two sides are given.
Solution:
Case I:
Given the base BC, a base angle, say ∠B and AB – AC. Steps of Construction:

  1. Draw the base BC.
  2. At point B draw an angle ∠CBX equal to the given angle.
  3. Cut a line segment BD = AB – AC from point B.
  4. Join DC.
  5. Draw the perpendicular bisector of . DC which intersect BX at A.
  6. Join AC.
  7. ΔABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-10
Justification:
As point A lies on the perpendicular bisector of DC
∴ AD = AC
BD = AB – AD = AB – AC (∴ AD = AC)

Case II:
Given the base BC, a base angle say ∠B and (AC – AB).
Steps of Construction:

  1. Draw the base SC.
  2. At point S, draw an angle, ∠CBX equal to the given angle and extend the arm XB backward.
  3. Cut a line segment BD equal to (AC – AB) from the extended arm.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersect BX at A.
  6. Join AC.
  7. ΔABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-11
Justification:
As the perpendicular bisector of DC passes through A.
∴ AD – AC
BD = AD – AB
∴ BD = AC – AB

Example 4:
Construct a ∆ABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

  1. Draw base BC = 3.4 cm.
  2. Draw ∠CBX = \({ 45 }^{ \underline { 0 } }\).
  3. From BX, cut line segment BD =1.5 cm.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersect BX at A.
  6. Join AC.
  7. ABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-12
Justification:
As A lies on perpendicular bisector of DC.
AD = AC
BD = 1.5 cm
⇒ AB – AD =1.5 cm
AB – AC =1.5 cm

MP Board Solutions

Example 5:
Construct a ∆PQR in which QR = 5.8 cm, PR – PQ = 1.8 cm and ∠Q = \({ 45 }^{ \underline { 0 } }\). Justify your construction.
Solution:
Steps of Construction:

  1. Draw base QR = 5.8 cm.
  2. Draw ∠RQP = \({ 45 }^{ \underline { 0 } }\).
  3. Produce arm XQ backward and cut a line segment QS = 1.8 cm.
  4. Join SR.
  5. bisector of SR which intersect QX at P.
  6. Join PR.
  7. PQR is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-13
Justification:
As P lies on perpendicular bisector of SR
PS = PR
QS = 1.8 cm
⇒ PS – PQ = 1.8 cm
∴ PR – PQ = 1.8 cm

Example 6:
Construct a ∆ABC in which BC = 5.8 cm, ∠C = \({ 60 }^{ \underline { 0 } }\) and AB – AC = 2.5 cm.
Solution:
Steps of Construction:

  1. Draw base BC = 5.8 cm
  2. Draw ∠ACB = \({ 60 }^{ \underline { 0 } }\)
  3. Produce arm CX backward and cut a line segment CD = 2.5 cm.
  4. Join BD.
  5. Draw perpendicular bisector of BD which intersect CX at A.
  6. Join AB.
  7. ABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-14
Justification:
A lies on the perpendicular bisector of BD.
∴ AB =AD
CD = 2.5 cm
⇒ AD – AC = 2.5 cm
⇒ AB – AC = 2.5 cm

Example 7:
Construct a ∆ABC in which AC – AB = 3.5 cm, BC = 6.2 cm and ∠C = \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

  1. Draw BC = 6.2 cm
  2. Draw ∠ACB = \({ 45 }^{ \underline { 0 } }\)
  3. Cut CD = 3.5 cm from CX.
  4. Join BD and draw the perpendicular bisector of BD which intersect CX at A.
  5. Join AB.
  6. ABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-15
Justification:
As A lies on the perpendicular bisector of BD.
∴ AB = AD
CD = 3.5 cm
⇒ AC – AD = 3.5 cm
∴ AC – AB = 3.5 cm

Example 8:
Construction of a Triangle when its Perimeter and Two Base Angles are given.
Solution:
Given the base angles, ∠B and ∠C and perimeter, i.e., AB + BC + AC.
Steps of Construction:

  1. Draw a line segment, DE equal to AB + BC + CA.
  2. Draw ∠EDM and ∠DEN equal to the base angles ∠B and ∠C respectively.
  3. Draw bisectors of ∠MDE and ∠NED which intersect at A.
  4. Draw the perpendicular bisectors of AD and AE which intersect DE at B and C respectively.
  5. Join AE and AC.
  6. ABC is the required triangle.

MP Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.1 img-16
Justification:
As B lies on the perpendicular bisector of AD
∴ AB = DB
In ∆ADB
AB = DB
∠ADB = ∠DAB
(∴ Angles opposite to equal sides of a A are equal)
Similarly, AC = CE
and ∠CAE = ∠CEA
Now DE = DB + BC + CE
= AB + BC + AC
In ∆ABD, ∠ABC = ∠DAB + ∠ADB = ∠ADB + ∠ADB = 2∠ADB
In ∆ACE, ∠ACB = ∠CAE + ∠CEA = ∠CEA + ∠CEA = 2∠CEA

MP Board Solutions

Example 9:
Construct a triangle whose perimeter is 6.4 cm and angles at the base are \({ 60 }^{ \underline { 0 } }\) and \({ 45 }^{ \underline { 0 } }\).
Solution:
Steps of Construction:

  1. Draw line segment DE equal to (AB + BC + CA) = 6.4 cm
  2. Draw ∠EDM = \({ 60 }^{ \underline { 0 } }\) and ∠DEN = \({ 45 }^{ \underline { 0 } }\).
  3. Draw AD and AE as bisectors of ∠MDE and ∠NED which inter sect at A.
  4. Draw the perpendicular bisector of AD and AE which intersect DE at B and C respectively.
  5. Join AB and AC.
  6. ABC is the required triangle.

MP Board Class 9th Maths Solutions