Learn Exercise 11.1 with Free Lessons & Tips

The steps required to construct an angle of 90° is as follows.

(i)Consider the given ray PQ. Draw an arc of some radius making point P as its centre, which intersects PQ at R.

(ii) Now, taking R as centre with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Mark S as centre 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°

(i)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)

(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°

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.

(i) 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.

(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.

(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.

(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 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.

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 steps mentioned below 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.