Constructions

Given, to construct an angle of 90° from the initial point of a ray and then we should justify if the constructed angle is 90°.

Steps of Construction:

i. Draw a ray AB of any length.

ii. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

iii. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

iv. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

v. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

vi. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

Justification:

Firstly, join AE, AD, DE, DC, EF and FD.

From the construction, we can say that

AD = DC = AC

Thus, as all the sides are equal ΔADC is an equilateral triangle.

[If all the sides and angles in a triangle are equal then the triangle is an equilateral triangle]

Now, in an equilateral triangle each angle is equal to 60°.

Hence, ∠ DAC = 60° ----- (1)

Similarly, if we consider ΔADF,

AD = DE = EA (by construction)

So, ΔADF is equilateral and ∠ DAE = 60° ----- (2)

From (1) and (2) we can say that, ∠ DAE and ∠ DAC are alternative angles for the line AD, thus making ED parallel to AC.

Now, ED || AC ------- (3)

Also, we know that DF = EF ----- (4)

(As F is formed by the arcs with same radius)

From (4) we can say that, point F lies on the perpendicular bisector of DE.

Thus ∠ FID = 90° ------- (5)

(where I is the point where FA bisects DE.)

Consider the lines AF, AC and DE. Here AF is the transverse to the parallel lines AC and DE.

∠ FID = ∠ FAC ------- (6)

(since the angles are corresponding angles made by the transverse AF with the parallel lines DE and AC)

Thus,

∠ FID = ∠ FAC = 90°

[from (5) and (6)]

Hence, the constructed angle at A is 90°.

Given, to construct an angle of 45° from the initial point of a ray and then we should justify if the constructed angle is 45°.

Steps of Construction:

i. Draw a ray AB of any length.

ii. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

iii. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

iv. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

v. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

vi. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

vii. Now, consider G as the point where the first arc (from step ii) meets the line AF and point C where the first arc (from step ii) meets the line AB.

With G and C as centers and with any radius more than half the length of GC, draw two arcs which intersect at point H.

viii. Join points A and H, thus the line forms an angle 45° with the ray AB.

Justification:

Firstly, join GH and CH.

From the construction, we can say that

GH= CH ----- (1)

[as they are arcs of equal radii]

Also,

AG = CG ------ (2)

[as they are the radii of the same arc in step (ii) of the construction ]

For the ΔAGH & ΔACH, side AH is common. ---- (3)

From (1), (2) and (3), we can say that ΔAGH & ΔACH are similar, i.e.

------ (4)

(From SSS property of triangles)

Now, from (4), as the triangles are congruent,

∠HAG =∠HAC ------- (5)

[Since the angles are the corresponding parts of congruent triangles]

But, from step (vi) ∠ FAB = 90°. ----- (6)

Also from the figure, we can clearly say that

∠ HAC + ∠ HAG = ∠ FAB

Now from (5) and (6),

∠ HAC + ∠ HAC = 90°

2 ∠ HAC = 90°

∠ HAC = 45°

Hence, the constructed ∠ HAC is 45°.

(i) To construct 30°.

Given to construct an angle of 30°

Steps for construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. With D and C are centers and with same radius (which is more than half of the length of CD), draw two arcs intersecting at E.

e. Draw the ray AE, which forms an angle of 30° at point A.

Hence the ∠ EAB = 30°.

(ii) To construct

Given to construct an angle of

Steps for construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

e. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

f. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

g. Now, consider G as the point where the first arc (from step ii) meets the line AF and point C where the first arc (from step ii) meets the line AB.

With G and C as centers and with any radius more than half the length of GC, draw two arcs which intersect at point H.

h. Join points A and H, thus the line forms an angle 45° with the ray AB.

i. The point where the first arc (from step b) intersects the ray AH is the point I. With I and C as centers, with any radius more than half of the length of IC, draw two arcs meeting at point J.

j. Now join points J and A, then the angle formed by the ray AJ with the ray AB is

Hence

(iii) To construct 15°.

Given to construct an angle of 15°

Steps for construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. With D and C are centers and with same radius (which is more than half of the length of CD), draw two arcs intersecting at E.

e. Draw the ray AE, which forms an angle of 30° at point A.

f. The ray AE intersects the first arc (from step b) at point F. With F and C as centers draw two arcs with any radius more than half of the length of FC, which intersect at G.

g. Now join points A and G. The angle formed between ray AB and AG is 15°.

Hence the ∠ GAB = 15°.

(i) Given, to construct an angle of 75° from the initial point of a ray.

Steps of Construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

e. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

f. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

g. Now the point G will be the point of intersection of the ray AF and the first arc (from step b). With points G & D as centers, with any radius more than half the length of GD, draw two arcs such that they meet at point H.

h. By joining point H and A, we get the ray AH which forms 75° with ray AB.

Hence ∠ HAB = 75°

(ii) Given, to construct an angle of 105° from the initial point of a ray.

Steps of Construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

e. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

f. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

g. Now the point G will be the point of intersection of the ray AF and the first arc (from step b). With points G & E as centers, with any radius more than half the length of GE, draw two arcs such that they meet at point H.

h. By joining point H and A, we get the ray AH which forms 105° with ray AB.

Hence ∠ HAB = 105°

(iii) Given, to construct an angle of 135° from the initial point of a ray.

Steps of Construction:

a. Draw a ray AB of any length.

b. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

c. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

d. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.

e. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.

f. Join points A and F. The angle formed by FAB is 90°. i.e.

∠ FAB = 90°.

g. Now extend AB to the left till point X. With A as centre and with some convenient radius draw an arc which intersects ray AF at G and AX at H.

h. With H and G as centers and with any radius more than half of the length of GH, draw two arcs intersecting each other at I.

i. Join I and A. The angle formed by ray IA and AB is 135°.

Hence ∠ IAB = 135°

Given to construct an equilateral triangle and justify the construction.

Let us construct an equilateral triangle of 3 cms length on each side.

Steps of construction:

i. Draw a ray AX.

ii. With 3 cms as radius, draw an arc with centre as A. The place where the arc meets the ray AZ is point B.

iii. Now, with B as centre, draw another arc with the same 3 cms radius intersecting the previous arc at C.

iv. Now join AC and BC to get the required construction where ΔABC is an equilateral triangle.

Justification:

As per the above construction,

AB = BC ---- (1)

[as they are the radii of the same arc]

Also

AB = AC ---- (2)

[as they are the radii of the same arc]

From (1) & (2), we have AB = BC= AC

Given base BC = 7 cm

∠ B = 75°

And AB + BC = 13 cms.

Steps of construction:

i. Draw a base line BC of 7 cms.

ii. Construct ∠ B = 75°.

a. With B as centre and with any radius, draw another arc cutting the line BC.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.

c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.

d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

∠ GBC = 90°.

f. Now the point H will be the point of intersection of the ray BG and the first arc (from step b). With points H & E as centers, with any radius more than half the length of HE, draw two arcs such that they meet at point I.

g. By joining point I and B, we get the ray BI which forms 75° with ray BD.

iii. With B as centre draw an arc with length 13 cms ( = AB + BC given), such that it intersects ray BI at J.

iv. Join CJ and we draw a perpendicular bisector for CJ.

a. By drawing arcs on both sides of the line CJ, with C and J as centers and with same lengths. These arcs intersect at K and L on either side of line CJ.

v. The perpendicular bisector for CJ will intersect the ray BJ at point A. Join AC.

Thus the formed triangle ABC is the required triangle.

Given base BC = 8 cm

∠ B = 45°

And AB - AC = 3.5 cms.

Steps of construction:

i. Draw a base line BC of 8 cms.

ii. Construct ∠ B = 45°.

a. With B as centre and with any radius, draw another arc cutting the line BC.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.

c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.

d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

∠ GBC = 90°.

f. Now the point H will be the point of intersection of the ray BG and the first arc (from step b). With points H & E as centers, with any radius more than half the length of HE, draw two arcs such that they meet at point I. By joining points I and B, we get the ray BI which forms 45° with ray BD.

As, AB – AC = 3.5

AB > AC

So the ray BI will be above BC

iii. With B as centre draw an arc with length 3.5 cms ( = AB - AC given), such that it intersects ray BI at J.

iv. Join CJ and we draw a perpendicular bisector for CJ.

a. By drawing arcs on both sides of the line CJ, with C and J as centers and with same lengths. These arcs intersect at K and L on either side of line CJ.

v. The perpendicular bisector for CJ will intersect the ray BJ at point A. Join AC.

Thus the formed triangle ABC is the required triangle.

Given base QR = 6 cm

∠ Q = 60°

And PR – PQ = 2 cms.

Steps of construction:

i. Draw a base line QR of 6 cms.

ii. Construct ∠ Q = 60°.

a. With Q as centre and with any radius, draw another arc cutting the line QR at A.

b. With A as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point B.

c. Now join the ray QB which forms an angle of 60° with the line QR.

As, PR – PQ = 2

PR > PQ (negative)

So the ray QB will be below QR

iii. Extend the ray QB below QR. With Q as centre draw an arc with length 2 cms ( = PR- PQ given), such that it intersects ray QB at C below QR.

iv. Join CR and draw a perpendicular bisector for CR.

a. By drawing arcs on both sides of the line CR, with C and R as centers and with same lengths which is more than half length of CR. These arcs intersect at D and E on either side of line CR.

v. The perpendicular bisector for CR will intersect the ray QB at point P. Join PR.

Thus the formed triangle PQR is the required triangle.

Given ∠ Y = 30°

∠ Z = 90°

XY + YZ + ZX = 11 cm

Steps of construction:

i. Draw a line segment AB of length 11 cm (= XY + YZ + ZX as given).

ii. At point A, draw angle 30°.

a. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.

b. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.

c. With D and C are centers and with same radius (which is more than half of the length of CD), draw two arcs intersecting at E.

d. Draw the ray AE, which forms an angle of 30° at point A.

Hence the ∠ EAB = 30°.

iii. At point B, draw an angle of 90°.

a. With B as center draw an arc with any radius which cuts the line AB at G.

b. With G as centre and with the same radius, draw another arc which cuts the first arc ( in step a ) at point H.

c. With H as centre and with same radius, draw an arc which cuts the first arc ( in step a. ) at point I.

d. Now with H and I as centers, and with radius more than half the length of IH, draw two arcs that intersect at point J.

e. Join JB. The angle thus formed is 90° i.e. ∠JBA = 90°

iv. Draw an angle bisector QR for ∠ A = 30° and OP for ∠ B = 90°. The place where both these angle bisectors intersect is point Y. Join AY and BY.

v. Now draw perpendicular bisector for AY which intersects the line AB at point X.

vi. Similarly draw perpendicular bisector for BY which intersects the line AB at point Z.

vii. Now by joining XY and YZ the required triangle is constructed.

Given base BC = 12 cm

∠ B = 90°

And AB (other side)+ AC (hypotenuse)= 18 cms.

Steps of construction:

i. Draw a base line BC of 12 cms.

ii. Construct ∠ B = 90°.

a. With B as centre and with any radius, draw another arc cutting the line BC.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.

c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.

d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

∠ GBC = 90°.

iii. With B as centre draw an arc with length 18 cms ( = AB + AC given), such that it intersects ray BG at H.

iv. Join CH and we draw a perpendicular bisector for CH.

a. By drawing arcs on both sides of the line CH, with C and H as centers and with same lengths. These arcs intersect at I and J on either side of line CH.

v. The perpendicular bisector for CH will intersect the ray BH at point A. Join AC.

Thus the formed triangle ABC is the required triangle with right angle formed at B.