Question

Question 1
Construct an angle of ${90}^{\circ }$ at the initial point of a given ray and justify the construction.

Answer 1

Steps to construct an angle of ${90}^{\circ }$

(i) Let us take a straight ray PQ with initial point P and draw an arc of some radius taking point P as its centre, which intersects PQ at R.

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

(iii) Taking S as the centre and with the same radius as before, draw an arc intersecting the initial arc at T(as shown in figure).

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

(v) Join PU, which is the required ray that makes an angle of ${90}^{\circ }$ with the given ray PQ.



Justification
We can justify the construction if we can prove $\angle UPQ={90}^{\circ }$
For this, join PS and PT.


We have, $\angle SPQ=\angle TPS={60}^{\circ }$
In (iii) and (iv) steps of this construction, PU was drawn as the bisector of $\angle TPS$.
$\therefore \angle UPS=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$
Also, $\angle UPQ=\angle SPQ+\angle UPS$
$={60}^{\circ }+{30}^{\circ }$
$={90}^{\circ }$