Page 1 Cbse-spot.blogspot.com 1 Class IX Chapter 11 – Constructions Maths Exercise 11.1 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. Page 2 Cbse-spot.blogspot.com 1 Class IX Chapter 11 – Constructions Maths Exercise 11.1 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. Cbse-spot.blogspot.com 2 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: ? Page 3 Cbse-spot.blogspot.com 1 Class IX Chapter 11 – Constructions Maths Exercise 11.1 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. Cbse-spot.blogspot.com 2 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: ? Cbse-spot.blogspot.com 3 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. Page 4 Cbse-spot.blogspot.com 1 Class IX Chapter 11 – Constructions Maths Exercise 11.1 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. Cbse-spot.blogspot.com 2 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: ? Cbse-spot.blogspot.com 3 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. Cbse-spot.blogspot.com 4 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. ? ? Page 5 Cbse-spot.blogspot.com 1 Class IX Chapter 11 – Constructions Maths Exercise 11.1 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. Cbse-spot.blogspot.com 2 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: ? Cbse-spot.blogspot.com 3 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. Cbse-spot.blogspot.com 4 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. ? ? Cbse-spot.blogspot.com 5 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 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 given ray PQ. ( ii )Read More

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