# NCERT Solutions for Class 9 Maths Chapter 11 - Constructions

Learn to draw triangles and bisectors of line segments with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 11 Constructions. Get step-by-step instructions to construct angles, bisectors and line segments. Revise the chapter solutions to understand how to give an appropriate justification for your construction.

Practise drawing arcs, triangles etc. according to the given data using the CBSE Class 9 Maths textbook solutions created by our subject experts. You can access these solutions any time at our online learning portal to revisit the topic of constructions and thus do well in your Maths exam.

## Chapter 11 - Constructions Exercise Ex. 11.1

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

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

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

(v) Join PU, which is the required ray making 90

^{o}with given ray PQ.

**Justification of Construction:**

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

^{o}.

For this let us join PS and PT

^{o}. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.

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

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

(vi) Now from R and V draw arcs with other at W with radius more than RV to intersect each other. PW is the required ray making 45

^{o}with PQ.

**Justification of Construction:**

To justify the construction, we have to prove WPQ = 45

^{o}. Join PS and PT

^{o}. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.

= 60

^{o}+ 30

^{o}

= 90

^{o}

In step (vi) of this construction, we constructed PW as the bisector of UPQ

^{o}

^{o}

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

^{o}with the given ray PQ.

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

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

(vii) Let it intersects 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 22 with the given ray PQ.

^{0}

Step I: Draw the given ray PQ. Now 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: Now 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 is intersects the arc at U. Now taking U and R as centre and with radius more than RU draw arc to intersect each other at V. Join PV which is the required ray making 15

^{o}with given ray PQ.

^{o}

The steps of construction are as follows:

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

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

^{o}with the given ray PQ.

^{o}.

(B) 105

^{o}

The steps of construction are as follows:

(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect 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, drawn an arc intersecting the arc at T (see figure)

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

(v) Join PU. Let it intersects the arc at V. Now 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

^{o}with the given ray PQ.

^{o}.

(C) 135

^{o}

The steps of construction are as follows:

(i) Take the given ray PQ. Extend PQ on opposite side of Q. Draw a semicircle of some radius taking point P as its centre, which intersect PQ at R and W.

(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, drawn an arc intersecting the arc at T (see figure)

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

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

^{o }with the given line PQ.

^{o}.

Also, each angle of an equilateral triangle is 60.

The steps of construction are as follows:

**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:**Now 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.

To justify the construction, we have to prove that ABC is an equilateral triangle i.e.AB = BC = AC = 5 cm and A = B = C = 60

^{o}.

Now, in ABC, we have AC = AB = 5 cm and A = 60

^{o}

Since, AC = AB, we have

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

Now, in ABC

A + B + C = 180

^{o}(angle sum property of a triangle)

60

^{o}+ C + C = 180

^{o}

60

^{o}+ 2 C = 180

^{o}

2 C = 180

^{o}- 60

^{o}= 120

^{o}

C = 60

^{o}

^{o}... (1)

## Chapter 11 - Constructions Exercise Ex. 11.2

**Step I:**Draw a line segment BC of 7 cm. At point B draw an angle of 75

^{o}sayXBC.

**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 intersects BX at A. ABC is the required triangle.

**Step I:**Draw the line segment BC = 8 cm and at point B make an angle of 45

^{o}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. ABC is the required triangle.

**Step I:**Draw line segment QR of 6 cm. At point Q draw an angle of 60

^{o}say XQR.

**Step II:**Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). 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.

^{o}at point A and an angle QBA of 90

^{o}at point B.

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

^{o}with AB.

**Step II:**Cut a line segment AD of 18 cm. (As sum of other two side is 18) from ray AX.

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

**Step IV:**Let BY intersects AX at C. Join AC, BC.ABC is the required triangle.

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