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

SOLUTION:**Steps of Construction :**

**Step I :** Draw a ray .

**Step II :** Taking O as centre and suitable radius, draw a semicircle, which cuts OA at B.

**Step III :** Keeping the radius same, divide the semicircle into three equal parts such that

**Step IV :** Draw .

**Step V :** Draw , the bisector of ∠COD.

Thus, ∠AOF = 90°

**Justification :**

O is the centre of the semicircle and it is divided into 3 equal parts.

Therefore,

⇒ ∠BOC = ∠COD = ∠DOE [Equal chords
subtend equal angles at the centre]

∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC + ∠BOC + ∠BOC = 180°

⇒ 3∠BOC = 180° ⇒ ∠BOC = 60°

Similarly, ∠COD = 60° and ∠DOE = 60°

OF is the bisector of ∠COD

Now, ∠BOC + ∠COF = 60° + 30°

⇒ ∠BOF = 90° or ∠AOF = 90°

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

SOLUTION:**Steps of Construction :**

**Step I : **Draw a ray .

**Step II :** Taking O as centre and with a suitable radius, draw a semi

circle such that it intersects
at B.

**Step III :** Taking B as centre and keeping the same radius, cut the semicircle at C. Now, taking C as centre and keeping the same radius, cut the semicircle at D and similarly, cut at E, such that

Join

**Step IV :** Draw , the angle bisector of ∠BOC.

**Step V :** Draw , the angle bisector of ∠FOC.

Thus, ∠BOG = 45° or ∠AOG = 45°

**Justification :**

⇒ ∠BOF + ∠FOG = 45° [∵ ∠COF = ∠BOF]

⇒ ∠BOG = 45°

Construct the angles of the following measurements.

(i) 30°

(ii) 22½°

(iii) 15°

(i) Angle of 30°

**Steps of Construction :**

**Step I :** Draw a ray .

**Step II :** With O as centre and having a suitable radius, draw an arc cutting
at B.

**Step III :** With centre at B and the same radius as above, draw an arc to cut the previous arc at C.

**Step IV :** Join which gives ∠BOC = 60°.

**Step V :** Draw , bisector of ∠BOC, such

(ii) 22½°

**Steps of Construction :**

**Step I :** Draw a ray

**Step II :** Construct ∠AOB = 90°

**Step III :** Draw , the bisector of ∠AOB, such that

**Step IV :** Now, draw ,
the bisector of ∠AOC, such that

(iii) Angle of 15°

**Steps of Construction :**

**Step I :** Draw a ray .

**Step II :** Construct ∠AOB = 60°

**Step III :** Draw , the bisector

of ∠AOB, such that

Construct the following angles and verify by measuring them by a protractor.

(i) 75°

(ii) 105°

(iii) 135°

**(i) Steps of Construction :**

**Step I :** Draw a ray .

**Step II :** With O as centre and having a suitable radius, draw an arc which cuts
at B.

**Step III :** With centre B and keeping the radius same, mark a point C on the previous arc.

**Step IV :** With centre C and having the same radius, mark another point D on the arc of step II.

**(ii) Steps of Construction :**

**Step I :** Draw a ray .

**Step II :** With centre O and having a suitable radius, draw an arc which cuts
at B.

**Step III :** With centre B and keeping the same radius, mark a point C on the arc of step II.

**Step IV :** With centre C and having the same radius, mark another point

D on the arc drawn in step II.

**Step V :** Draw , the bisector of

**Step VI :** Draw , the bisector of

Thus, ∠AOQ = 90° + 15° = 105°

**(iii) Steps of Construction :**

**Step I :** Draw a ray

**Step II : **With centre O and having a suitable radius, draw an arc which cuts OP at A.

**Step III :** Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that

**Step IV :** Draw ,
the bisector of

**Step V :** Draw ,
the bisector of

Thus, ∠POM = 135°

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

SOLUTION:Let us construct an equilateral triangle, each of whose side = 3 cm(say).

**Steps of Construction :**

**Step I :**Draw a ray .

**Step II :**Taking O as centre and radius equal to 3 cm, draw an arc to cut
at B such that OB = 3 cm

**Step III :** Taking B as centre and radius equal to OB, draw an arc, to intersect the previous arc at C.

**Step IV : **Join OC and OB.

Thus, ∆OBC is the required equilateral triangle.

**Justification :**

[Chords corresponding to equal arcs are equal]

OC = OB = BC

∴ ∆OBC is an equilateral triangle.

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

SOLUTION:**Step I **Draw a ray .

**Step II :** Along , cut off a line segment

BC = 7 cm.

**Step III :** At B, construct ∠CBY = 75°

**Step IV : **From , cut off BD = 13 cm

(= AB + AC)

**Step V :** Join DC.

**Step VI : **Draw perpendicular bisector of CD which meets BD at A.

**Step VII :**Join AC.

Thus, ∆ABC is the required triangle.

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

SOLUTION:**Steps of Construction:**

**Step I :** Draw a ray .

**Step II :** Along , cut off a line segment
BC = 8 cm.

**Step III :** Construct ∠CBY = 45°

**Step IV :** From BY, cut off BD = 3.5 cm.

**Step V :** Join DC.

**Step VI :** Draw PQ, perpendicular bisector of DC, which intersects
at A.

**Step VII :** Join AC.

Thus, ABC is the required triangle.

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

SOLUTION:**Steps of Construction:**

**Step I :** Draw a ray .

**Step II :** Along , cut a line segment
QR = 6 cm.

**Step III :** Construct a line YQY ′ such that
∠RQY = 60°.

**Step IV :** Cut off QS = 2 cm on QY ′.

**Step V :** Join SR.

**Step VI :** Draw MN, perpendicular bisector of SR, which intersects QY at P.

**Step VII :** Join PR.

Thus, ∆PQR is the required triangle.

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

SOLUTION:**Steps of Construction:**

**Step I :** Draw a line segment AB = 11 cm = (XY + YZ + ZX)

**Step II : **Construct ∠BAP = 30°

**Step III :** Construct ∠ABQ = 90°

**Step IV :** Draw AR, the bisector of ∠BAP.

**Step V :** Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X.

**Step VI :** Draw perpendicular bisector of AX, which intersects AB at Y.

**Step VII :** Draw perpendicular bisector of XB, which intersects AB at Z.

**Step VIII :** Join XY and XZ.

Thus, ∆XYZ is the required triangle.

Construct a right triangle whose base is

12 cm and sum of its hypotenuse and other side is 18 cm.

**Steps of Construction:**

**Step I :** Draw BC = 12 cm.

**Step II :** Construct ∠CBY = 90°

.
**Step III :** Along , cut a line segment

BX = 18 cm.

**Step IV :** Join CX.

**Step V :** Draw PQ, perpendicular bisector of CX, which meets BX at A.

**Step VI :** Join AC.

Thus, ∆ABC is the required triangle.