Question

CONSTRUCT AN ANGLE

OF 90° at the intial point of

a given ray and justify the

construction.


XD XD XD​

Answer 1

$\Large{\underbrace{\sf{\red{Given:}}}}$

• A ray OA.

$\Large{\underbrace{\sf{\orange{Required:}}}}$

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

$\Large{\underbrace{\sf{\purple{Steps\:of\:Construction:}}}}$

1. Taking O as centre and some radius.Draw an arc of a circle, which intesects 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 $\displaystyle{\sf{ \dfrac{1}{2}CD}}$, draw arcs to intersect each other, say G.
7. Draw the ray OG. This ray OG is the bisector of the angle $\displaystyle{\sf{ \angle FOE, i.e., \angle FOG = \angle EOG}}$ $\displaystyle{\sf{ \dfrac{1}{2} \angle FOE = \dfrac{1}{2}(60)° = 30°.}}$

Thus, $\displaystyle{\sf{ \angle GOA = \angle GOE + \angle EOA = 30° + 60° = 90°.}}$

$\Large{\underbrace{\sf{\pink{Justification:}}}}$

(i) Join BC.

Then, OC = OB = BC (By construction)

$\displaystyle{\sf{ \therefore \triangle COB\:is\:an\:equilateral\:triangle.}}$

$\displaystyle{\sf{ \therefore \angle COB = 60°}}$

$\displaystyle{\sf{ \therefore \angle EOA = 60°}}$

(ii) Join CD.

Then, OD = OC = CD (By construction)

$\displaystyle{\sf{ \therefore \triangle DOC\:is\:an\:equilateral\:triangle.}}$

$\displaystyle{\sf{ \therefore \angle DOC = 60°}}$

$\displaystyle{\sf{ \therefore \angle FOE = 60°}}$

(iii) Join CG and DG.

In ∆ ODG and ∆OCG,

OD = OC | Radii of the same arc

DG = CG | arcs of equal radii

OG = OG | Common

$\displaystyle{\sf{ \therefore \triangle ODG \cong \triangle OCG}}$ | SSS rule

$\displaystyle{\sf{ \therefore \angle DOG = \angle COG}}$ | CPCT

$\displaystyle{\sf{ \therefore \angle FOG = \angle EOG = \dfrac{1}{2} \angle FOG}}$

:$\implies{\sf{ 30°= \dfrac{1}{2}(60°)}}$

Thus, $\displaystyle{\sf{ \angle GOA = \angle GOE + \angle EOA}}$

$\large{\boxed{\sf{\red{= 30° + 60° = 90°.}}}}$

$\rule{400}4$

Answer 2

Answer:

Steps of construction

Draw a line segment OA.

Taking O as center and any radius, draw an arc cutting OA at B.

Now, taking B as center and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

With C as center and the same radius, draw an arc cutting the arc at D.

With C and D as center and radius more than

2

1

CD, draw two arc intersecting at P.

Join OP.

Thus, ∠AOP=90

o

Justification

Join OC and BC

Thus,

OB=BC=OC [Radius of equal arcs]

∴△OCB is an equilateral triangle

∴∠BOC=60

o

Join OD,OC and CD

Thus,

OD=OC=DC [Radius of equal arcs]

∴△DOC is an equilateral triangle

∴∠DOC=60

o

Join PD and PC

Now,

In △ODP and △OCP

OD=OC [Radius of same arcs]

DP=CP [Arc of same radii]

OP=OP [Common]

∴△ODP≅△OCP [SSS congruency]

∴∠DOP=∠COP [CPCT]

So, we can say that

∠DOP=∠COP=

2

1

∠DOC

∠DOP=∠COP=

2

1

×60=30

o

Now,

∠AOP=∠BOC+∠COP

∠AOP=60+30

∠AOP=90

o

Hence justified.