Class 9 - Maths - Constructions

Exercise 11.1

Question 1:

Construct an angle of 900 at the initial point of a given ray and justify the construction.

Steps of Construction:

(i) Let us take a ray AB with initial point A.

(ii) Taking A as centre and some radius, draw an arc of a circle, which intersects AB at C.

(iii) With C as centre and the same radius as before, draw an arc, intersecting the previous  arc

at E.

(iv) With E as centre and the same radius, as before, draw an arc, which intersects the arc

drawn in step (ii) at F.

(v) With E as centre and some radius, draw an arc.

(vi) With F as centre and the same radius as before, draw another arc, intersecting the

previous arc at G.

(vii) Draw the ray AG.

Then BAG is the required angle of 900.

Justification:  Join AE, CE, EF, FG and  GE

AC = CE = AE                                               [By construction]

ΔACE is an equilateral triangle

Angle CAE = 60°      .……..………(i)

Similarly, ΔAEF = 60°  ………... (ii)

From (i) and (ii), FE || AC   ….(iii)            [Alternate angles are equal]

Also,  FG = EG                                             [By construction]

=> G lies on the perpendicular bisector of EF in

=> angle GIE = 900   ... (iv)

So, angle GAB = angle GIE = 900              [Corresponding angles]

GF = GE                                                        [Arcs of equal radii]

Question 2:

Construct an angle of 450 at the initial point of a given ray and justify the construction.

The below given steps will be followed to construct an angle of 450.

(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

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

(vi) From R and V, draw arcs with  radius more than RV/2 to intersect each other at W.

Join PW.

PW is the required ray making 450 with PQ.

Justification of Construction:

We can justify the construction, if we can prove ∠WPQ = 450.

For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 600.

In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.

So, ∠UPS = ∠TPS/2 = 600/2 = 300

Also, ∠UPQ = ∠SPQ + ∠UPS

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

So, ∠WPQ = ∠UPQ/2 = 900/2 = 450

Question 3:

Construct the angles of the following measurements:

(i) 300                      (ii)  22.50                (iii) 150

(i) 300

The below given steps will be followed to construct an angle of 300.

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/2, draw arcs to intersect each

other at T. Join PT which is the required ray making 30° with the  given ray PQ.

(ii)22.50

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,

(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.

(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/2 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/2, draw arcs

to intersect each other at Y.

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

(iii) 150

The below given steps will be followed to construct an angle of 150.

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.

Step III: Taking R and S as centre and with radius more than RS/2, 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/2,

draw an arc to intersect each other at V.

Join PV which is the required ray making 150 with the given ray PQ.

Question 4:

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

(i) 750                         (ii) 1050                    (iii) 1350

(i) 750

The below given steps will be followed to construct an angle of 750.

(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.

(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 the angle so

formed can be measured with the help of a protractor. It comes to be 750.

(ii) 1050

The below given steps will be followed to construct an angle of 1050.

(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.

(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/2. Let these arcs intersect each other at W.

Join PW which is the required ray making 1050 with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 1050.

(iii) 1350

The below given steps will be followed to construct an angle of 1350.

(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.

(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/2, draw arcs to intersect each other at X.

Join PX, which is the required ray making 1350 with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 1350.

Question 5:

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

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 600.

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.

∆ABC 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 = 600.

In ∆ABC, we have AC = AB = 5 cm and ∠A = 600.

Since AC = AB, ∠B = ∠C                            [Angles opposite to equal sides of a triangle]

In ∆ABC,

∠A + ∠B + ∠C = 180°                                 [Angle sum property of a triangle]

∠ 600 + ∠C + ∠C = 1800

∠ 600 + 2 ∠C = 1800

2 ∠C = 1800 − 600 = 1200

∠C = 1200/2

∠C = 600

∠B = ∠C = 600

We have, ∠A = ∠B = ∠C = 600 ………….....1

So, ∠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, ∆ABC is an equilateral triangle.

Exercise 11.2

Question 1:

Construct a triangle ABC in which BC = 7 cm, ∠B = 750 and AB + AC = 13 cm.

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. ∆ABC is the required triangle.

Question 2:

Construct a triangle ABC in which BC = 8 cm, ∠B = 450 and AB − AC = 3.5 cm.

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 450, say

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

Step IV: Let it intersect BX at point A. Join AC.

∆ABC is the required triangle.

Question 3:

Construct a triangle PQR in which QR = 6 cm, ∠Q = 600 and PR − PQ = 2 cm

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 600, 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.

∆PQR is the required triangle.

Question 4:

Construct a triangle XYZ in which ∠Y = 300, ∠Z = 90° and XY + YZ + ZX = 11 cm.

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 intersects AB at Y and UV intersects AB at Z. Join XY, XZ.

∆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.

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 900 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.

∆ABC is the required triangle.