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Construct an angle of 90 0 at the initial point of a given ray and justify the construction.

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The steps required to construct an angle of 90° is as follows. (i)Consider the given ray PQ. Draw an arc of some radius making point P as its centre, which intersects PQ at R. (ii) Now, taking R as centre with the same radius as before, draw an arc intersecting the previously drawn arc at S. (iii)...
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The steps required to construct an angle of 90° is as follows.

(i)Consider the given ray PQ. Draw an arc of some radius making point P as its centre, which intersects PQ at R.

(ii) Now, taking R as centre with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Mark S as centre with the same radius as before, draw an arc intersecting the arc at T (see figure).

(iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.

(v) Join PU, which is the required ray making 90° with the given ray PQ.



Justification of Construction:

We can justify the construction, if we can prove ∠UPQ = 90°.

For this, join PS and PT.

We have, ∠SPQ = ∠TPS = 60°. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of ∠TPS.

∴ ∠UPS = ∠TPS

Also, ∠UPQ = ∠SPQ + ∠UPS

= 60° + 30°

= 90°

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Maths tutor with 4 years experience

Following are the steps of construction:(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R. (ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.(iii) Taking S as centre and with...
read more
Following are the steps of construction:

(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R.
 
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.

(iii) Taking S as centre and with the same radius as before, drawn an arc intersecting the arc at T (see figure)

(iv) Taking S and T as centre, draw arc of same radius to intersect each other at U.

(v) Join PU, which is the required ray making 90o with given ray PQ.



We have SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
 
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