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Construct an angle of 90° at the initial point of a given ray and justify the construction |
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Answer»
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
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. thank you (*_*)
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