11-CONSTRUCTIONS


Here we are going through the ncert solutions for class 9 maths chapter 11-constructions . so before going through the ncert solutions make sure to go through the textbook that helps you to understand the solutions of constructions more easily

Chapter 11 – Constructions

(ncert solutions for class 9 maths chapter 11)


EXERCISE 11.1

Question 1:                                            

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

ANSWER:

The below given steps will be followed to construct an angle of 90°.

(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.

(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(v) Join PU, which is the required ray making 90° with the given ray PQ.


Justification of Construction:

We can justify the construction, if we can prove ∠UPQ = 90°.

For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.

∴ ∠UPS =  ∠TPS

Also, ∠UPQ = ∠SPQ + ∠UPS

= 60° + 30°

= 90°

Question 2:                                            (ncert solutions for class 9 maths chapter 11)

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

ANSWER:

The below given steps will be followed to construct an angle of 45°.

(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.

(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(v) Join PU. Let it intersect the arc at point V.

(vi) From R and V, draw arcs with radius more than  RV to intersect each other at W. Join PW.

PW is the required ray making 45° with PQ.

Justification of Construction:                                        (ncert solutions for class 9 maths chapter 11)

We can justify the construction, if we can prove ∠WPQ = 45°.

For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.

∴ ∠UPS =  ∠TPS

Also, ∠UPQ = ∠SPQ + ∠UPS

= 60° + 30°

= 90°

In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.

∴ ∠WPQ =  ∠UPQ 

Question 3:                                            (ncert solutions for class 9 maths chapter 11)

Construct the angles of the following measurements:

(i) 30° (ii)   (iii) 15°

ANSWER:

(i)30°

The below given steps will be followed to construct an angle of 30°.

Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.

Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.

Step III: Taking R and S as centre and with radius more than  RS, draw arcs to intersect each other at T. Join PT which is the required ray making 30° with the given ray PQ.

(ii)

The below given steps will be followed to construct an angle of .

(1) Take the given ray PQ. Draw an arc of some radius, taking point P as its centre, which intersects PQ at R.

(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(5) Join PU. Let it intersect the arc at point V.

(6) From R and V, draw arcs with radius more than  RV to intersect each other at W. Join PW.

(7) Let it intersect the arc at X. Taking X and R as centre and radius more than  RX, draw arcs to intersect each other at Y.

Joint PY which is the required ray making  with the given ray PQ.

(iii) 15°

The below given steps will be followed to construct an angle of 15°.

Step I: Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.

Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.

Step III: Taking R and S as centre and with radius more than  RS, draw arcs to intersect each other at T. Join PT.

Step IV: Let it intersect the arc at U. Taking U and R as centre and with radius more than  RU, draw an arc to intersect each other at V. Join PV which is the required ray making 15° with the given ray PQ.

Question 4:                                        (ncert solutions for class 9 maths chapter 11)

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

(i) 75° (ii) 105° (iii) 135°

ANSWER:

(i) 75°

The below given steps will be followed to construct an angle of 75°.

(1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.

(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(5) Join PU. Let it intersect the arc at V. Taking S and V as centre, draw arcs with radius more than  SV. Let those intersect each other at W. Join PW which is the required ray making 75° with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 75º.

(ii) 105°

The below given steps will be followed to construct an angle of 105°.

(1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.

(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(5) Join PU. Let it intersect the arc at V. Taking T and V as centre, draw arcs with radius more than  TV. Let these arcs intersect each other at W. Join PW which is the required ray making 105° with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 105º.

(iii) 135°

The below given steps will be followed to construct an angle of 135°.

(1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of some radius taking point P as its centre, which intersects PQ at R and W.

(2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).

(4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(5) Join PU. Let it intersect the arc at V. Taking V and W as centre and with radius more than  VW, draw arcs to intersect each other at X. Join PX, which is the required ray making 135°with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 135º.

Question 5:                                (ncert solutions for class 9 maths chapter 11)

Construct an equilateral triangle, given its side and justify the construction

ANSWER:

Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60º.

The below given steps will be followed to draw an equilateral triangle of 5 cm side.

Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P.

Step II: Taking P as centre, draw an arc to intersect the previous arc at E. Join AE.

Step III: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 5 cm.

Justification of Construction:

We can justify the construction by showing ABC as an equilateral triangle i.e., AB = BC = AC = 5 cm and ∠A = ∠B = ∠C = 60°.

In ΔABC, we have AC = AB = 5 cm and ∠A = 60°.

Since AC = AB,

∠B = ∠C (Angles opposite to equal sides of a triangle)

In ΔABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

⇒ 60° + ∠C + ∠C = 180°

⇒ 60° + 2 ∠C = 180°

⇒ 2 ∠C = 180° − 60° = 120°

⇒ ∠C = 60°

∴ ∠B = ∠C = 60°

We have, ∠A = ∠B = ∠C = 60° … (1)

⇒ ∠A = ∠B and ∠A = ∠C

⇒ BC = AC and BC = AB (Sides opposite to equal angles of a triangle)

⇒ AB = BC = AC = 5 cm … (2)

From equations (1) and (2), ΔABC is an equilateral triangle.

MISCELLANEOUS EXERCISE

                          (ncert solutions for class 9 maths chapter 11)

Question 1:

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

ANSWER:

The below given steps will be followed to construct the required triangle.

Step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 75°, say ∠XBC.

Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.

Step III: Join DC and make an angle DCY equal to ∠BDC.

Step IV: Let CY intersect BX at A. ΔABC is the required triangle.

Question 2:

Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB − AC = 3.5 cm.

ANSWER:

The below given steps will be followed to draw the required triangle.

Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 45°, say ∠XBC.

Step II: Cut the line segment BD = 3.5 cm (equal to AB − AC) on ray BX.

Step III: Join DC and draw the perpendicular bisector PQ of DC.

Step IV: Let it intersect BX at point A. Join AC. ΔABC is the required triangle.

Question 3:                                                        

Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR − PQ = 2 cm

ANSWER:

The below given steps will be followed to construct the required triangle.

Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠XQR.

Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and PR − PQ = 2 cm). Join SR.

Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.

ΔPQR is the required triangle.

Question 4:                                    

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

ANSWER:

The below given steps will be followed to construct the required triangle.

Step I: Draw a line segment AB of 11 cm.

(As XY + YZ + ZX = 11 cm)

Step II: Construct an angle, ∠PAB, of 30° at point A and an angle, ∠QBA, of 90° at point B.

Step III: Bisect ∠PAB and ∠QBA. Let these bisectors intersect each other at point X.

Step IV: Draw perpendicular bisector ST of AX and UV of BX.

Step V: Let ST intersect AB at Y and UV intersect AB at Z.

Join XY, XZ.

ΔXYZ 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:

The below given steps will be followed to construct the required triangle.

Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90° with AB.

Step II: Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX.

Step III: Join DB and make an angle DBY equal to ADB.

Step IV: Let BY intersect AX at C. Join AC, BC.

ΔABC is the required triangle.

 


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