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Question-1

Construct an angle of 90°  at the initial point of a given ray and justify the construction.

Solution:

Given: A ray OA.

Required: To construct an angle of 90° at O and justify the construction.

Steps of Construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.


3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C.
Then EOA = 60°.
Draw the ray OF passing through D. Then FOE = 60°.


6. Next, taking C and D as centres and with the radius more than CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e., FOG = EOG = FOE = (60°) = 30°.

Thus, GOA = GOE + EOA = 30° + 60° = 90°.

Justification:

(i) Join BC.
Then, OC = OB = BC (By construction)
Δ COB is an equilateral triangle.
COB = 60°.
EOA = 60°.


(ii) Join CD.
Then, OD = OC = CD (By construction) Δ DOC is an equilateral triangle. DOC = 60°.
∴ ∠ FOE = 60°.


(iii) Join CG and DG.
In Δ ODG and Δ OCG,
OD = OC              
| Radii of the same arc

DG = CG                |  Arcs of equal radii

OG = OG                       |Common Δ ODG  Δ OCG          |SSS Rule DOG = COG         |CPCT

FOG = EOG = FOE = (60°) = 30°
Thus, ∠ GOA =    GOE +   EOA = 30° + 60° = 90°.

Question-2

Construct an angle of 45° at the initial point of a given ray and justify the construction.

Solution:
Given: A ray OA.
Required: To construct an angle of 45° at O and justify the construction.
Steps of Construction:

1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.


4. Draw the ray OE passing through C. Then EOA = 60°.

5. Draw the ray OF passing through D. Then FOE = 60°.

6. Next, taking C and D as centres and with radius more than CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e., FOG = EOG =
FOE = (60°) = 30°.

Thus, GOA = GOE + EOA = 30° + 60° = 90°.

8. Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I respectively.

9. Next, taking H and I as centres and with the radius more than HI, draw arcs to
intersect each other, say at J.

10. Draw the ray OJ. This ray OJ is the required bisector of the angle GOA.

Thus, GOJ = AOJ = GOA = (90°) = 45°.
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)

D DOC is an equilateral triangle. DOC = 60°. FOE = 60°.

(iii) Join CG and DG.
In Δ ODG and Δ OCG,
OD = OC                                  | Radii of the same arc
DG = CG                                  | Arcs of equal radii
OG = OG                                  | Common Δ ODG = Δ OCG                      | SSS Rule DOG = COG                     | CPCT

FOG = EOG = FOE = (60°) = 30°
Thus, GOA = GOE + EOA = 30° + 60° = 90°.

(iv) Join HJ and IJ.
In Δ OIJ and Δ OHJ,
OI = OH                                 | Radii of the same arc
IJ = HJ                                   | Arcs of equal radii
OJ = OJ                                  | Common
∴ Δ OIJ = Δ OHJ                       | SSS Rule ∴ ∠ IOJ = HOJ                      | CPCT AOJ = GOJ = GOA = (90°) = 45°.

Question-3

Construct the angles of the following measurements:
(i) 30° (ii) 22° (iii) 15°

Solution:
(i) 30°
Given: A ray OA.
Required: To construct an angle of 30° at O.

Steps of Construction:

1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Draw the ray OE passing through C. Then EOA = 60°.

4. Taking B and C as centres and with the radius more than BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD. This ray OD is the bisector of the angle EOA, i.e., EOD = AOD = EOA = (60°) = 30°.

(ii) 22°
Given: A ray OA.

Required: To construct an angle of 22 ° at O.
Steps of Construction:

1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.


3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then EOA = 60°.

5. Draw the ray OF passing through D. Then FOE = 60°.

6. Next, taking C and D as centres and with radius more than CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the angle FOE, i.e., FOG = EOG = FOE = (60°) = 30°.
Thus, ZGOA = GOE + EOA = 30° + 60° = 90°.


8. Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OG, say at H and I respectively.

9. Next, taking H and I as centres and with the radius more than HI, draw arcs to
intersect each other, say at J.


10. Draw the ray OJ. This ray OJ is the bisector of the angle GOA.
i.e., GOJ = AOJ = GOA = (90°) = 45°.


11. Now, taking O as centre and any radius, draw an arc to intersect the rays OA and OJ, say at K and L respectively.

12. Next, taking K and L as centres and with the radius more than KL, draw arcs to
intersect each other, say at M.


13. Draw the ray OM. This ray OM is the bisector of the angle AOJ,
i.e., JOM = AOM = AOJ = (45°) = 22° 
(iii) 15°
Given: A ray OA.
Required: To construct an angle of 15° at O.
Steps of construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.

2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Draw the ray OE passing through C.
Then EOA = 60°.


4. Now, taking B and C as centres and with the radius more than BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the angle EOA,
     i.e., EOD = AOD = EOA = (60°) = 30°.

6. Now, taking B and F as centres and with the radius more than BF, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the angle AOD, i.e., DOG = AOG = AOD =(30°) = 15°.

Question-4

Construct the following angles and verify by measuring them by a protractor:
(i) 
75° (ii) 105° (iii)135°

Solution:
(i) 75°
Given: A ray OA.
Required: To construct an angle of 75° at O.
Steps of Construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.


2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Join the ray OE passing through C. Then EOA = 60°.

5. Draw the ray OF passing through D. Then FOE - 60°.

6. Next, taking C and D as centres and with the radius more than CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG intersecting the arc of step 1 at H. This ray OG is the bisector of the
angle FOE, i.e., FOG = EOG = FOE = (60°) = 30°.


8. Next, taking C and H as centres and with the radius more than CH, draw arcs to intersect each other, say at I.

9. Draw the ray OI. This ray OI is the bisector of the angle GOE, i.e., GOI = EOI = GOE =  (30°) = 15°.
Thus, IOA = IOE + EOA = 15° + 60° = 75°.
On measuring the IOA by protractor, we find that IOA = 75°.
Thus the construction is verified.

(ii) 105°

Given: A ray OA.
Required: To construct an angle of 105° at O.
Steps of Construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.


2. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point C.

3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then EOA = 60°.

5. Draw the ray OF passing through D. Then FOE = 60°.

6. Next, taking C and D as centres and with the radius more than CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector of the angle FOE, i.e., FOG = EOG =        FOE = (60°) = 30°.
Thus, GOA = GOE + EOA = 30° + 60° = 90°.


8. Next, taking H and D as centres and with the radius more than HD, draw arcs to intersect each other, say at I.

9. Draw the ray OI. This ray OI is the bisector of the angle FOG, i.e., FOI = GOI = FOG = (30°) = 15°.
Thus, IOA = IOG + GOA = 15° + 90° = 105°. On measuring the IOA by protractor, we find that FOA = 105°.
Thus the construction is verified.
(iii) 135°
Given:
A ray OA.
Required: To construct an angle of 135° at O.
Steps of Construction:
1. Produce AO to A' to form ray OA'.


2. 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'.

3. Taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at a point C.

4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.

5. Draw the ray OE passing through C. Then EOA = 60°.

6. Draw the ray OF passing through D. Then FOE = 60°.

7. Next, taking C and D as centres and with the radius more than CD, draw arcs to intersect each other, say at G.

8. Draw the ray OG intersecting the arc drawn in step 1 at H. This ray OG is the bisector
of the angle FOE, i.e., FOG = EOG = FOE = (60°) = 30°.

Thus, GOA = GOE + EOA = 30° + 60° = 90°.
B'OH = 90°.


9. Next, taking B' and H as centres and with the radius more than B'H, draw arcs to intersect each other, say at I.

10. Draw the ray OI. This ray OI is the bisector of the angle B'OG, i.e., B'OI = GOI = B'OG = (90°) = 45°.
Thus, IOA = IOG + GOA
                          = 45° + 90° = 135°.
On measuring the IOA by protractor, we find that IOA = 135°.

Thus the construction is verified.

Question-5

Construct an equilateral triangle, given its side and justify the Construction.

Solution:
Given: Side (say 6 cm) of an equilateral triangle.
Required: To construct the equilateral triangle and justify the construction.
Steps of Construction:

1. Take a ray AX with initial point A.

2. Taking A as centre and radius (= 6 cm), draw an arc of a circle, which intersects AX,
say at a point B.


3. Taking B as centre and with the same radius as before, draw an arc intersecting the
previously drawn arc, say at a point C.


4. Join AC and BC.

Then Δ ABC is the required triangle with side 6 cm.

Justification:
AB = BC                           | By construction
AB = AC                           | By construction AB = BC = CA Δ ABC is an equilateral triangle.

The construction is justified.

Question-6

Construct a triangle ABC in which BC = 7Cm, B =  75° and AB + AC = 13cm.

Solution:
In Δ ABC, BC = 7 cm, B =  75° and AB + AC = 13 cm
Required: To construct the triangle ABC.
Steps of Construction:
1. Draw the base BC = 7 cm.

2. At the point B make an angle XBC = 75°.

3. Cut a line segment BD equal to AB + AC (= 13 cm) from the ray BX.

4. Join DC.
 


5. Make an DCY = BDC.

6. Let CY intersect BX at A.
Then ABC is the required triangle.

Question-7

Construct a triangle ABC in which BC = 8 cm, B = 45° and AB - AC = 3.5 cm.

Solution:
Given: In ABC, BC = 8 cm, B = 45° and AB - AC = 3.5 cm.
Required: To construct the triangle ABC.

Steps of Construction:

1. Draw the base BC = 8 cm.

                                                                                              
2. At the point B make an angle CBX = 45°.

3. Cut the line segment BD equal to 3.5 cm from the ray BX.

4. Join DC.

5. Draw the perpendicular bisector, say PQ of DC.

6. Let it intersect BX at a point A.

7. Join AC.
Then ABC is the required triangle.

Question-8

Construct a triangle PQR in which QR = 6 cm.  Q = 60° and PR - PQ = 2 cm.

Solution:
Given: In Δ PQR, QR = 6 cm, Q = 60° and PR - PQ = 2 cm.
Required: To construct the Δ PQR. In Δ PQR, QR = 6 cm, Q = 60° and PR - PQ = 2 cm.
Steps of Construction:

1. Draw the base QR = 6 cm.

2. At the point Q make an XQR = 60°.


3. Cut line segment QS = PR - PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.

4. Join SR.

5. Draw the perpendicular bisector LM of SR.

6. Let LM intersect QX at P.

7. Join PR.
Then, PQR is the required triangle.

Question-9

Construct a triangle XYZ in which Y= 30°, Z = 90° and XY + YZ + ZX = 11 cm.

Solution:
Given: In triangle XYZ, Y= 30°, Z = 90° and XY + YZ + ZX = 11 cm.
Required: To construct the ΔXYZ.
Steps of Construction:
1. Draw a line segment BC = XY + YZ + ZX (= 11 cm).


2. Make LBC = Y (= 30°) and MCB = Z (= 90°).

3. Bisect LBC and MCB. Let these bisectors meet at a point X.

                                                                                                                             
4. Draw perpendicular bisectors DE of XB and FG of XC. 

5. Let DE intersect BC at Y and FC intersect BC at Z.

6. Join XY and XZ.
Then XYZ is the required triangle.

Question-10

Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Solution:
Given: In right ΔABC, base BC = 12 cm, B = 90° and AB + AC = 18 cm.
Required: To construct the right triangle ABC.
                                                  
 Steps of Construction:
1. Draw the base BC = 12 cm.


2. At the point B, make an XBC = 90°.

3. Cut a line segment BD = AB + AC(= 18 cm) from the ray BX.

4. Join DC

5. Draw the perpendicular bisector PQ of CD to intersect BD at a point A.

6. Join AC.
Then, ABC is the required right triangle.




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