Question

construct an angle of 90°at the untial point of a given ray and justify the constrution.
​

Answer 1

Answer:

kkfauplcdilkcfolvzgkehdkflyef hdvbxmwldjfgbwleodhejeleoeydhelelridbdlelwieegdhdhfie

Answer 2

Answer:

Step-by-step explanation:

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.

solution

expand

Was this answer helpful?