# NCERT Solutions for Class 9 Maths Unit 11

## Constructions Class 9

### Unit 11 Constructions Exercise 11.1, 11.2 Solutions

Introduction to Constructions
Constructions: The drawing of various shapes using only a pair of compasses and straightedge or ruler. No measurement of lengths or angles is allowed.
The word construction in geometry has a very specific meaning: the drawing of geometric items such as lines and circles using only compasses and straightedge or ruler. Very importantly, you are not allowed to measure angles with a protractor, or measure lengths with a ruler.

Why we learn about constructions
Picture of Euclid The Greeks formulated much of what we think of as geometry over 2000 years ago. In particular, the mathematician Euclid documented it in his book titled "Elements", which is still regarded as an authoritative geometry reference. In that work, he uses these construction techniques extensively, and so they have become a part of the geometry field of study. They also provide insight into geometric concepts and give us tools to draw things when direct measurement is not appropriate.

Compasses
Pair of compasses Compasses are a drawing instrument used for drawing circles and arcs. It has two legs, one with a point and the other with a pencil or lead. You can adjust the distance between the point and the pencil and that setting will remain until you change it.
This kind of compass has nothing to do with the kind used find the north direction when you are lost. A compass used to find the north direction is usually referred to in the singular - a compass. The kind we are talking about here is usually referred to in the plural - compasses. This plural reference is similar to the way we talk about scissors - with an 's' on the end.

### Exercise 11.1 : Solutions of Questions on Page Number : 191

Q1 :

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

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°

Q2 :

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

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:

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

Q3 :

Construct the angles of the following measurements:

(i) 30° (ii) (iii) 15°

Q4 :

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

(i) 75° (ii) 105° (iii) 135°

Q5 :

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

### Exercise 11.2 : Solutions of Questions on Page Number : 195

Q1 :

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

Q2 :

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

Q3 :

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

Q4 :

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

Q5 :

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Maths : CBSE NCERT Exercise Solutions for Class 9th for Constructions ( Exercise 11.1, 11.2 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

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 Exercise 11.1 Question 1 Question 2 Question 3 Question 4 Question 5
 Exercise 11.2 Question 1 Question 2 Question 3 Question 4 Question 5