﻿ Class 9 NCERT Math Solution
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TOPICS
Exercise - 11.1

Question-1 :-  Construct an angle of 900 at the initial point of a given ray and justify the construction.

Solution :-
```  Steps of Constructions :
(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.
∠UPS = 1/2 x ∠TPS  = 60 x 1/2 = 30°
Also, ∠UPQ = ∠SPQ + ∠UPS
= 60° + 30°
= 90°
```

Question-2 :-  Construct an angle of 450 at the initial point of a given ray and justify the construction.

Solution :-
```  Steps of Constructions :
(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/2 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 = 1/2 x ∠TPS = 1/2 x 60° = 30°.
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 = 1/2 x ∠UPQ = 1/2 x 90 = 45°.
```

Question-3 :-  Construct the angles of the following measurements:
(i) 30°  (ii) 22 by 1/2° (iii) 15°

Solution :-
```(i)30°

Steps of Constructions :

(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.
(II) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
(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 by 1/2°

Steps of Constructions :

(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 1/2RV to intersect each other at W.
Join PW.
(VII) Let it intersect the arc at X. Taking X and R as centre and radius more than 1/2RX, draw arcs to intersect each other at Y.
Joint PY which is the required ray making 22 by 1/2° with the given ray PQ. ```
```(iii) 15°

Steps of Constructions :

(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.
(II) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
(III) Taking R and S as centre and with radius more than RS, draw arcs to intersect each other at T. Join PT.
(IV) Let it intersect the arc at U. Taking U and R as centre and with radius more 1/2 ray making 15° with the given ray PQ. ```

Question-4 :-  Construct the following angles and verify by measuring them by a protractor:
(i) 75°  (ii) 105°  (iii) 135°

Solution :-
```(i) 75°

Steps of constructions :

(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 (see figure).
(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 radius more than 1/2 SV.
Let those intersect each other at W.
Join PW which is the required ray making 75° with the given ray PQ. ```
```(ii) 105°

Steps of constructions :

(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 (see figure).
(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 1/2 TV. Let these arcs intersect each other at W.
Join PW which is the required ray making 105° with the given ray PQ. ```
```(iii) 135°

Steps of constructions :

(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 (see figure).
(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 1/2 VW, draw arcs to intersect each other at X.
Join PX, which is the required ray making 135° with the given line PQ. ```

Question-5 :-  Construct an equilateral triangle, given its side and justify the construction.

Solution :-
```  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 60º.

Steps of Construction :

(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.
(II) Taking P as centre, draw an arc to intersect the previous arc at E. Join AE.
(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°.
In ∆ABC, we have AC = AB = 5 cm and  ∠A = 60°.
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° + ∠C + ∠C = 180°
60° + 2 ∠C = 180°
2 ∠C = 180° − 60° = 120°
∠C = 60°
∠B = ∠C = 60°
We have, ∠A = ∠B = ∠C = 60° ... (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)
From equations (1) and (2),
∆ABC is an equilateral triangle.
```
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