Chapter 12 Constructions Ex 12.1
Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction
Justification
(i) Join BC.
Then, OC=OB = BC (By construction)
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60°
∴ ∠EOA = 60°
(ii) Join CD.
Then, OD=OC=CD (By construction)
∴ ∆DOC is an equilateral triangle.
∴ ∠DOC = 60°
∴ ∠FOE = 60°
Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction
 Taking O as centre and some radius, draw an arc of a circle which intersects OA, say at a point B.
 Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
Justification
(i) Join BC. (By construction)
Then, OC = OB = BC
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60°
∴ ∠EOA = 60°
Question 3.
Solution:
(i) Steps of construction
(ii) Steps of construction
(iii) Steps of construction
 Taking O as centre and some radius, draw an arc of a circle which intersects OA say at a point B.
 Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C..
 Draw the ray OE passing through C. Then, ∠EOA = 60°.
Question 4.
Construct the following angles and verify by measuring them by a protractor
(i) 75°
(ii) 105°
(iii) 135°
Solution:
(i) Steps of construction
 On measuring the ∠IOA by protractor, we find that ∠IOA = 15°
Thus, the construction is verified.
(ii) Steps of construction
 Next, taking H and D as centres and with the radius more than 12 HD, draw arcs to intersect each other, say at l.
 Draw the ray Ol. This ray Ol is the bisector of the ∠FOG, i.e.,

Thus, ∠lOA = ∠IOG + ∠GOA = 15° + 90° = 105°. On measuring the ∠lOA by protractor, we find that ∠FOA = 105°.
Thus, the construction is verified.
(iii) Steps of construction
 Produce AO to A’ to form ray OA’.
 Taking O as centre and some radius, draw an arc of a circle which intersects OA at a point B and OA’ at a point B’.
 Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.
 Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
 Draw the ray OE passing through C, then ∠EOA = 60°.
 Draw the ray OF passing through D, then ∠FOE = 60°.
 Next, taking C and D as centres and with the radius more than 12 CD, draw arcs to intersect each other, say at G.
 Draw the ray OGintersecting the arc drawn in step 1 at H. This ray OG is the bisector of the ∠FOE i,e.,
 Next, taking B’ and H as centres and with the radius more than 12 B’H, drawn arcs to intersect each other, say at l.
 Draw the ray Ol. This ray Ol is the bisector of the ∠B’OG i.e.,

On measuring the ∠IOA by protractor, we find that ∠lOA = 135°.
Thus, the construction is verified.
Question 5.
Construct an equilateral triangle, given its side and justify the construction.
Solution:
Steps of construction
 Take a ray AX with initial point A From AX, cut off AB = 4 cm.
 Taking A as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point B.
 Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.
 Draw the ray AE passing through C.
 Next, taking B as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point A

Taking A as centre and with the same radius as in step 5, draw an arc intersecting the previously drawn arc, say at a point C.
Draw the ray BF passing through C.
Then, ∆ ABC is the required triangle with gives side 4 cm.
Justification
AB = BC (By construction)
AB = AC (By construction)
∴ AB = BC = CA
∴ ∆ ABC is an equilateral triangle.
∴ The construction is justified.