Class 9 - Maths - Constructions

**Exercise 11.1**

**Question 1:**

Construct an angle of 90^{0} at the initial point of a given ray and justify the construction.

Answer:

**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 90^{0}.

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 = 90^{0} ... (iv)

So, angle GAB = angle GIE = 90^{0} [Corresponding angles]

GF = GE [Arcs of equal radii]

**Question 2:**

Construct an angle of 45^{0} 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^{0}.

(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 45^{0} with PQ.

**Justification of Construction:**

We can justify the construction, if we can prove ∠WPQ = 45^{0}.

For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60^{0}.

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

So, ∠UPS = ∠TPS/2 = 60^{0}/2 = 30^{0}

Also, ∠UPQ = ∠SPQ + ∠UPS

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

So, ∠WPQ = ∠UPQ/2 = 90^{0}/2 = 45^{0 }

**Question 3:**

Construct the angles of the following measurements:

(i) 30^{0} (ii) 22.5^{0} (iii) 15^{0}

Answer:

(i) 30^{0}

The below given steps will be followed to construct an angle of 30^{0}.

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)22.5^{0}

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) 15^{0}

The below given steps will be followed to construct an angle of 15^{0}.

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 15^{0} with the given ray PQ.

**Question 4:**

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

(i) 75^{0} (ii) 105^{0} (iii) 135^{0}

Answer:

(i) 75^{0}

The below given steps will be followed to construct an angle of 75^{0}.

(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 75^{0}.

(ii) 105^{0}

The below given steps will be followed to construct an angle of 105^{0}.

(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 105^{0} with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 105^{0}.

(iii) 135^{0}

The below given steps will be followed to construct an angle of 135^{0}.

(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 135^{0} with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 135^{0}.

**Question 5:**

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 60^{0}.

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 = 60^{0}.

In ∆ABC, we have AC = AB = 5 cm and ∠A = 60^{0}.

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]

∠ 60^{0} + ∠C + ∠C = 180^{0}

∠ 60^{0} + 2 ∠C = 180^{0}

2 ∠C = 180^{0} − 60^{0} = 120^{0}

∠C = 120^{0}/2

∠C = 60^{0}

∠B = ∠C = 60^{0}

We have, ∠A = ∠B = ∠C = 60^{0} ………….....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 = 75^{0} 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. ∆ABC is the required triangle.

**Question 2:**

Construct a triangle ABC in which BC = 8 cm, ∠B = 45^{0} 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^{0}, 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 = 60^{0} 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^{0}, 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 = 30^{0}, ∠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 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.

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^{0} 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.

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