PSEB Solutions for Class 9 Maths Chapter 11 Constructions

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 PSEB Solutions for Class 9 Maths Chapter 11 Constructions

PSEB 9th Class 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.
Answer:

Steps of construction :

  1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
  2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
  3. Taking X and Y as centres and radius more than 1/2XY, draw arcs to intersect at P on one side of line MAB.
  4. Draw ray AC passing through E
    Thus, ∠CAB is the required angle of 90°.

Justification:
Draw PX and PY.
In ∆ PAX and ∆ PAY,
AX = AY (Radii of same arc)
PX = PY (Radii of congruent arcs)
PA = PA (Common)
∴ By SSS rule, ∆ PAX ≅ ∆ PAY
∴ ∠PAX = ∠PAY (CPCT)
But, ∠PAX + ∠PAY = 180° (Linear pair)
∴ ∠PAY = 180°/2 = 90°
∴ ∠CAB = 90°

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

Steps of construction:

  1. Ray AB is given. Produce ray AB on the side of A to get line MAB.
  2. Taking A as centre and some radius, draw an arc of a circle to intersect line MAB at X and Y.
  3. Taking X and Y as centres and radius more than 1/2XY, draw arcs to intersect at P on one side of line MAB.
  4. Draw ray AC passing through E Thus, ∠CAB of 90° is received.
  5. Name the point of intersection of the arc with centre A and ray AC as Z.
  6. Taking Y and Z as centres and radius more than 1/2YZ, draw arcs to intersect each other at Q.
  7. Draw ray AQ.
    Thus, ∠QAB is the required angle of 45°.

Justification:
In example 1, we have already justified that ∠CAB = 90°. So, we do not repeat that part’ here.
Draw QZ and QY.
In ∆ AYQ and ∆ AZQ,
AY = AZ (Radii of same arc)
YQ = ZQ (Radii of congruent arcs)
AQ = AQ (Common)
∴ By SSS rule, ∆ AYQ ≅ ∆ AZQ
∴ ∠QAY = ∠QAZ (CPCT)
But, ∠QAY + ∠QAZ = ∠ZAY = ∠CAB = 90°
∴ ∠QAY = 90°/2 = 45°
∴ ∠QAB = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
Answer:

Steps of construction:

  1. Draw any ray AB. With centre A and any radius, draw an arc to intersect AB at X.
  2. With centre X and the same radius [as in step (1)], draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
  3. Draw ray AT, the bisector of ∠YAB.
    Thus, ∠TAB is the required angle of 30°.

(iii) 15°
Answer:

Steps of construction:

  1. Draw any ray AB. Taking A as centre and any radius, draw an arc of a circle to intersect AB at X.
  2. Taking X as centre and the same radius as before, draw an arc to intersect the previous arc at Y. Draw ray AY. Then, ∠YAB = 60°.
  3. Draw ray AL, the bisector of ∠YAB. Then, ∠LAB = 30°.
  4. Draw ray AM, the bisector of ∠LAB. Then, ∠MAB = 15°.
    Thus, ∠MAB is the required angle of 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75° and (ii) 105°
Answer:

Steps of construction:

  1. Draw any ray AB and produce it on the side of A to get line CAB. Taking A as centre and any radius draw an arc of a circle to intersect line CAB at X and Y.
  2. Taking X and Y as centres and radius more than 1/2XY, draw arcs to intersect each other at point L on one side of line CAB. Draw ray AL. Then, ∠LAB = 90°.
  3. Taking X as centre and radius AX, draw an arc of a circle to Intersect the first arc (arc XY) with centre A at Z.
  4. Draw ray AZ. Then, ∠ZAB = 60°.
  5. Now, draw ray AM, the bisector of ∠LAZ. Then, ∠MAB = 75° and ∠MAC = 105°.
    Thus, ∠MAB and ∠MAC are the required angles of measure 75° and 105° respectively.

Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Answer:
Line segment XY is given. We have to construct an equilateral triangle with each side being equal to XY.

Steps of construction:

  1. Draw any ray BM.
  2. With centre B and radius XY, draw an arc of a circle to intersect BM at C.
  3. Taking B and C as centres and rhdius XY, draw arcs to intersect each other at A on one side of line AC.
  4. Draw AB and AC.
    Thus, ∆ ABC is the required equilateral triangle with each side being equal to XY.

Justification:
The arc drawn with centre B and radius XY intersects ray BM at C. ∴ BC = XY. The arcs drawn with centres B and C and radius XY intersect at A.
∴ AB = XY and AC = XY.
Thus, in ∆ ABC, AB = BC = AC = XY.
Hence, ∆ ABC is an equilateral triangle in which all sides are equal to XY.
Note: If the measure of sides are given numerically, e.g., 4 cm, 5 cm, etc., then we have to use graduated scale instead of straight edge.

Ex 11.2

Question 1.
Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13cm.
Answer:

Steps of construction:

  1. Draw any ray BX. With centre B and radius 7 cm draw an arc to intersect BX at C.
  2. At B, construct ∠YBC with measure 75°.
  3. With centre B and radius 13 cm, draw an arc to intersect BY at M.
  4. Draw line segment MC. Draw the perpendicular bisector of MC to intersect BM at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the required triangle.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Answer:

Steps of construction:

  1. Draw any ray BX and from that obtain the line segment BC of length 8 cm.
  2. At B, draw ray BY such that ∠YBC = 45°.
  3. With centre B and radius 3.5 cm, draw an arc to intersect ray BY at D.
  4. Draw line segment DC. Draw the perpendicular bisector of DC to intersect ray BY at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the required triangle.

Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Answer:

Steps of construction:

  1. Draw any ray QX and from that obtain the line segment QR of length 6 cm.
  2. At Q, construct ray QY such that Z YQR = 60°.
  3. Produce ray QY on the side of Q to obtain ray QZ. Obtain point S on ray QZ such that QS = 2 cm.
  4. Draw line segment RS. Draw the perpendicular bisector of RS to intersect QY at E
  5. Draw line segment PR.
    Then, ∆ PQR is the required triangle.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer:

Steps of construction:

  1. Draw any ray AP and from that obtain the line segment AB of length 11 cm.
  2. Construct ray AL such that ∠LAB = 30°.
  3. Construct ray BM such that ∠MBA = 90°.
  4. Draw the bisectors of ∠LAB and ∠MBA to intersect each other at X.
  5. Draw line segment XB. Draw the perpendicular bisector of XB to intersect AB at Z.
  6. Draw line segment XA. Draw the perpendicular bisector of XA to intersect AB at Y.
  7. Draw line segments XY and XZ.
    Then, ∆ XYZ is the required triangle.

Question 5.
Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Answer:

Steps of construction:

  1. Draw any ray BX and from that obtain the line segment BC of length 12 cm.
  2. Construct ray BY such that ∠YBC = 90°.
  3. Taking B as centre and radius 18 cm, draw an arc to intersect BY at M.
  4. Draw line segment CM. Draw the perpendicular bisector of CM to intersect BM at A.
  5. Draw line segment AC.
    Then, ∆ ABC is the require triangle in which ∠B is a right angle, BC = 12 cm and AB + AC = 18 cm.
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