## EXERCISE 11.1

#### Page No 191:

#### Question 1:

Construct an angle of 90° 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 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°

#### Question 2:

Construct an angle of 45° 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°.

(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

#### Question 3:

Construct the angles of the following measurements:

(i) 30° (ii) (iii) 15°

#### Answer:

(i)30°

The below given steps will be followed to construct an angle of 30°.

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, draw arcs to intersect each other at T. Join PT which is the required ray making 30° with the given ray PQ.

(ii)

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, 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 point V.

(6) From R and V, draw arcs with radius more than RV 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, draw arcs to intersect each other at Y.

Joint PY which is the required ray making with the given ray PQ.

(iii) 15°

The below given steps will be followed to construct an angle of 15°.

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, 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, draw an arc to intersect each other at V. Join PV which is the required 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°

#### Answer:

(i) 75°

The below given steps will be followed to construct an angle of 75°.

(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 SV. Let those intersect each other at W. Join PW which is the required ray making 75° with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 75º.

(ii) 105°

The below given steps will be followed to construct an angle of 105°.

(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 TV. Let these arcs intersect each other at W. Join PW which is the required ray making 105° with the given ray PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 105º.

(iii) 135°

The below given steps will be followed to construct an angle of 135°.

(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.

(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, draw arcs to intersect each other at X. Join PX, which is the required ray making 135°with the given line PQ.

The angle so formed can be measured with the help of a protractor. It comes to be 135º.

#### 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 60º.

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. ΔABC 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 ΔABC, we have AC = AB = 5 cm and ∠A = 60°.

Since AC = AB,

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

In ΔABC,

∠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), ΔABC is an equilateral triangle.