1. Construct an angle of \(90^\circ\) at the initial point of a given ray 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, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (1/2)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., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

Justification:

i)Join BC.

Then, OC = OB = BC ...(By construction)

Therefore, \(\angle{COB}\) is an equilateral triangle.

Thus, \(\angle{COB}\) = \(60^\circ\)

Also, \(\angle{EOA}\) = \(60^\circ\)

ii)Join CD.

Then, OD = OC = CD ...(By construction)

Therefore, \(\angle{DOC}\) is an equilateral triangle.

Thus, \(\angle{DOC}\) = \(60^\circ\)

Also, \(\angle{FOE}\) = \(60^\circ\)

iii)Now, join CG and DG.

In \(\triangle{ODG}\) and \(\triangle{OCG}\), we have,

OD = OC ...((Radii of the same arc)

DC = CG ...(Arcs of equal radii)

and OG = OG ...(Given)

By SSS criterion, we get,

\(\triangle{ODG}\) \(\displaystyle \cong\) \(\triangle{OCG}\)

Hence, \(\angle{DOG}\) = \(\angle{COG}\) ...(By CPCT)

So, \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

2. Construct an angle of \(45^\circ\) at the initial point of a given ray 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, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (1/2)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., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

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 (1/2)HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the \(\angle{GOA}\).

Thus, \(\angle{GOJ}\) = \(\angle{AOJ}\) = (1/2)\(\angle{GOA}\) = (1/2)\(90^\circ\) = \(45^\circ\)

Justification:

i)Join BC.

Then, OC = OB = BC ...(By construction)

Therefore, \(\angle{COB}\) is an equilateral triangle.

Thus, \(\angle{COB}\) = \(60^\circ\)

Also, \(\angle{EOA}\) = \(60^\circ\)

ii)Join CD.

Then, OD = OC = CD ...(By construction)

Therefore, \(\angle{DOC}\) is an equilateral triangle.

Thus, \(\angle{DOC}\) = \(60^\circ\)

Also, \(\angle{FOE}\) = \(60^\circ\)

iii)Now, join CG and DG.

In \(\triangle{ODG}\) and \(\triangle{OCG}\), we have,

OD = OC ...((Radii of the same arc)

DC = CG ...(Arcs of equal radii)

and OG = OG ...(Given)

By SSS criterion, we get,

\(\triangle{ODG}\) \(\displaystyle \cong\) \(\triangle{OCG}\)

Hence, \(\angle{DOG}\) = \(\angle{COG}\) ...(By CPCT)

So, \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

iv) Now, join HJ and IJ.

In \(\triangle{OIJ}\) and \(\triangle{OHJ}\), we have,

OI = OH ...((Radii of the same arc)

IJ = HJ ...(Arcs of equal radii)

and OJ = OJ ...(Given)

By SSS criterion, we get,

\(\triangle{OIJ}\) \(\displaystyle \cong\) \(\triangle{OHJ}\)

Hence, \(\angle{IOJ}\) = \(\angle{HOJ}\) ...(By CPCT)

So, \(\angle{AOJ}\) = \(\angle{GOJ}\) = (1/2)\(\angle{GOA}\) = (1/2)\(90^\circ\) = \(45^\circ\).

3. Construct the angles of the following measurements:

I)\(30^\circ\)

II) \(22(1/2)^\circ\)

III)\(15^\circ\)

I)\(30^\circ\)

II) \(22(1/2)^\circ\)

III)\(15^\circ\)

I) 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, \(\angle{EOA}\) = \(60^\circ\).

4. Taking B and C as centres and with the radius more than (1/2)BC, draw arcs to intersect each other, say at D.

5. Draw the ray OD, this ray OD is the bisector of the \(\angle{EOA}\) = \(60^\circ\).

i.e., \(\angle{EOD}\) = \(\angle{AOD}\) = (1/2)\(\angle{EOA}\) = (1/2)\(60^\circ\) = \(30^\circ\)

II) 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, drawn an arc intersecting the arc drawn in step 1, say at D.

4. Draw the ray OE passing through C. Then, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (1/2)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., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

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 (1/2)HI, draw arcs to intersect each other, say at J.

10. Draw the raw OJ. This ray OJ is the required bisector of the \(\angle{GOA}\).

Thus, \(\angle{GOJ}\) = \(\angle{AOJ}\) = (1/2)\(\angle{GOA}\) = (1/2)\(90^\circ\) = \(45^\circ\).

11. Now, taking O as centre and any radius, drawn 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 (1/2)KL, draw arcs to intersect each other, say at H.

13. Draw the ray OM. This ray OM is the bisector of the \(\angle{AOJ}\)

i.e., \(\angle{JOM}\) = \(\angle{AOM}\) = (1/2) \(\angle{AOJ}\) = (1/2) \(45^\circ\) = \(22(1/2)^\circ\)

III) 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 intersecting the previously drawn arc, say at a point c.

3. Draw the ray OE passing through C. Then, \(\angle{EOA}\) = \(60^\circ\)

4. Next, taking B and C as centres and with the radius more then (1/2)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 the \(\angle{EOA}\).

i.e., \(\angle{EOD}\) = \(\angle{AOD}\) = (1/2)\(\angle{EOA}\) = (1/2)\(60^\circ\) = \(30^\circ\)

6. Now, taking B and F as centers and with the radius more then (1/2) 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., \(\angle{DOG}\) = \(\angle{AOG}\) = (1/2)\(\angle{AOD}\) = (1/2)\(30^\circ\) = \(15^\circ\)

4. Construct the following angles and verify by measuring then by a protractor:

I)\(75^\circ\)

II) \(105^\circ\)

III)\(135^\circ\)

I)\(75^\circ\)

II) \(105^\circ\)

III)\(135^\circ\)

I) 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, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (1/2)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., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

8. Next, taking C and H as centres and with the radius more than (1/2)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., \(\angle{GOI}\) = \(\angle{EOI}\) = (1/2)\(\angle{GOE}\) = (1/2)\(30^\circ\) = \(15^\circ\)

Thus, \(\angle{IOA}\) = \(\angle{IOE}\) + \(\angle{EOA}\)

i.e., \(\angle{IOA}\) = \(15^\circ\) + \(60^\circ\) = \(75^\circ\)

On measuring the IOA by protractor, we find that \(\angle{IOA}\) = \(75^\circ\).

Thus, the construction is verified.

II) 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, \(\angle{EOA}\) = \(60^\circ\)

5. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

6. Next, taking C and D as centres and with the radius more then (1/2)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., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

8. Next, taking H and D as centres and with the radius more then (1/2)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., \(\angle{FOI}\) = \(\angle{GOI}\) = (1/2)\(\angle{FOG}\) = (1/2)\(30^\circ\) = \(15^\circ\)

Thus, \(\angle{IOA}\) = \(\angle{IOG}\) + \(\angle{GOA}\)

i.e., \(\angle{IOA}\) = \(15^\circ\) + \(90^\circ\) = \(105^\circ\)

On measuring the IOA by protractor, we find that \(\angle{FOA}\) = \(105^\circ\).

Thus, the construction is verified.

III) 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 O.

5. Draw the ray OE passing through C. Then, \(\angle{EOA}\) = \(60^\circ\)

6. Draw the ray OF passing through D. Then, \(\angle{FOE}\) = \(60^\circ\).

7. Next, taking C and D as centres and with the radius more then (1/2)CD, draw arcs to intersect each other, say at G.

8. Draw the ray OG intersecting the arc of step 1 at H.. This ray OG is the bisector of the \(\angle{FOE}\)

i.e., \(\angle{FOG}\) = \(\angle{EOG}\) = (1/2)\(\angle{FOE}\) = (1/2)\(60^\circ\) = \(30^\circ\)

Thus, \(\angle{GOA}\) = \(\angle{GOE}\) + \(\angle{EOA}\)

i.e., \(\angle{GOA}\) = \(60^\circ\) + \(30^\circ\) = \(90^\circ\).

i.e., \(\angle{B'OH}\) = \(90^\circ\)

9. Next, taking B' and H as centres and with the radius more then (1/2)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., \(\angle{B'OI}\) = \(\angle{GOI}\) = (1/2)\(\angle{B'OG}\) = (1/2)\(90^\circ\) = \(45^\circ\)

Thus, \(\angle{IOA}\) = \(\angle{IOG}\) + \(\angle{GOA}\)

i.e., \(\angle{IOA}\) = \(45^\circ\) + \(90^\circ\) = \(135^\circ\)

On measuring the IOA by protractor, we find that \(\angle{IOA}\) = \(135^\circ\).

Thus, the construction is verified.

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

Steps of construction:

1. Take a ray AX with initial point A. From AX, cut off AB = 4cm.

2. Taking A as centre and radius (= 4 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. Draw the ray AE passing through C.

5. Next, taking B as centre and radius (= 4 cm), draw an arc of a circle, which intersects AX, say at a point A.

6. 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.

7. 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)

Thus, AB = BC = CA

Therefore, \(\triangle{ABC}\) is an equilateral triangle.

Hence, The construction is justified.

1. Construct a \(\triangle{ABC}\) in which BC = 7 cm, \(\angle{B}\) = \(75^\circ\) and AB + AC = 13cm.

Given: In \(\triangle{ABC}\) in which BC = 7 cm, \(\angle{B}\) = \(75^\circ\) and AB + AC = 13cm.

Steps of construction:

1. Draw the base BC = 7 cm.

2. At the point B make an \(\angle{XBC}\) = \(75^\circ\).

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

4. Join DC.

5. Make \(\angle{DCY}\) = \(\angle{BDC}\).

6. Let CY intersect BX at A.

Hence, ABC is the required triangle.

2. Construct a \(\triangle{ABC}\) in which BC = 8 cm, \(\angle{B}\) = \(45^\circ\) and AB - AC = 35cm.

Given: In \(\triangle{ABC}\) in which BC = 8 cm, \(\angle{B}\) = \(45^\circ\) and AB - AC = 35cm.

Steps of construction:

1. Draw the base BC = 8cm

2. At the point B make an \(\angle{XBC}\) = \(45^\circ\).

3. Cut the line segment BD equal to AB - AC = 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.

Hence, ABC is the required triangle.

3. Construct a \(\triangle{ABC}\) in which QR = 6 cm, \(\angle{Q}\) = \(60^\circ\) and PR - PQ = 2cm.

Given: In \(\triangle{ABC}\) in which QR = 6 cm, \(\angle{Q}\) = \(60^\circ\) and PR - PQ = 2cm.

Steps of construction:

1. Draw the base QR = 6 cm.

2. At the point Q make an \(\angle{XQR}\) = \(60^\circ\).

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

6. Let LM intersect QX at P.

7. Join PR.

Hence, ABC is the required triangle.

4. Construct a \(\triangle{XYZ}\) in which \(\angle{Y}\) = \(30^\circ\), \(\angle{Z}\) = \(90^\circ\) and XY + YZ + ZX = 11cm.

Given: In \(\triangle{XYZ}\) in which \(\angle{Y}\) = \(30^\circ\), \(\angle{Z}\) = \(90^\circ\) and XY + YZ + ZX = 11cm.

Steps of construction:

1. Draw a line segment BC = XY + YZ + ZX = 11cm.

2. Make \(\angle{LBC}\) = \(\angle{Y}\) = 30° and \(\angle{MCB}\) = \(\angle{Z}\) = 90°.

3. Bisect \(\angle{LBC}\) and \(\angle{MCB}\). 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, \(\triangle{XYZ}\) is the required triangle.

5. Construct right triangles whose base is 12 cm and sum its hypotenuse and other
side is 18 cm.

Given: In \(\triangle{ABC}\), base BC = 12 cm, \(\angle{B}\) = \(90^\circ\) and
AB + BC = 18 cm.

Steps of construction:

1. Draw the base BC =12cm.

2. At the point B, make an \(\angle{XBC}\) = \(90^\circ\).

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, \(\triangle{ABC}\) is the required right triangle.