# NCERT Solution for Class 9 Mathematics Chapter 11 - Constructions

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All our solutions for Chapter 11 - Constructions are prepared considering the latest CBSE syllabus, and they are amended from time to time. Our free NCERT Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

## NCERT Solution for Class 9 Mathematics Chapter 11 - Constructions Page/Excercise 11.1

Following are the steps of construction:

(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, 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
We have SPQ = TPS = 60^{o}. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.

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 intersect arc at point V.

(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
We have SPQ = TPS = 60^{o}. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
UPS = TPS=
Now, UPQ = SPQ + UPS

= 60^{o} + 30^{o}

= 90^{o}

In step (vi) of this construction, we constructed PW as the bisector of UPQ
WPQ = UPQ = = 45^{o}

(i) 30^{o}
The steps of construction are as follows:

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 which is the required ray making 30^{o} with the given ray PQ.
(ii) 22
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 intersect arc at point V.
(vi) Now from R and V draw arcs with radius more than RV to intersect each other at W. Join PW.

(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.
(iii) 15^{0}
The steps of construction are as follows:

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.

(A) 75^{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 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^{o} with the given ray PQ.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 75^{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.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 105^{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.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 135^{o}.

We know that all sides of an equilateral triangle are equal. So, all sides of this equilateral triangle will be 5 cm.

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.
Justification of Construction:

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}
Now, we have A = B = C = 60^{o} ... (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)
Equations (1) and (2) show that the ABC is an equilateral triangle.

## NCERT Solution for Class 9 Mathematics Chapter 11 - Constructions Page/Excercise 11.2

The steps of construction for the required triangles are as follows:
**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.

The steps of construction for the required triangles are as follows:**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.

The steps of construction for the required triangles are as follows:**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.

The steps of construction for the required triangles are as follows:
Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle PAB of 30^{o} at point A and an angle QBA of 90^{o} 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.

The steps of construction for the required triangles are as follows:**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.

## NCERT Mathematics - IX Class 9 Chapter Solutions

- Chapter 1 - Number Systems
- Chapter 2 - Polynomials
- Chapter 3 - Coordinate Geometry
- Chapter 4 - Linear Equations in Two Variables
- Chapter 5 - Introduction to Euclid's Geometry
- Chapter 6 - Lines And Angles
- Chapter 7 - Triangles
- Chapter 8 - Quadrilaterals
- Chapter 9 - Areas of Parallelograms and Triangles
- Chapter 10 - Circles
- Chapter 11 - Constructions
- Chapter 12 - Heron's Formula
- Chapter 13 - Surface Areas and Volumes
- Chapter 14 - Statistics
- Chapter 15 - Probability

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