TOPICS

Exercise - 11.1

Constructions

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

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.

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°

(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°

(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.

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.

CLASSES