Q. 13.9( 264 Votes )

Construct an angl

Answer :

Given, to construct an angle of 90° from the initial point of a ray and then we should justify if the constructed angle is 90°.


Steps of Construction:


i. Draw a ray AB of any length.



ii. Now as point A is the initial point as the centre, draw an arc any radius such that, the arc meets the ray AB at point C.



iii. With C as centre and with the same radius as before, draw another arc cutting the previous one at point D.



iv. Now with D as centre and with the same radius, draw an arc cutting the first arc (drawn in step ii) at point E.



v. With E and D as centers, and with a radius more than half the length of DE, draw two arcs intersecting at point F.



vi. Join points A and F. The angle formed by FAB is 90°. i.e.


FAB = 90°.



Justification:


Firstly, join AE, AD, DE, DC, EF and FD.



From the construction, we can say that


AD = DC = AC


Thus, as all the sides are equal ΔADC is an equilateral triangle.


[If all the sides and angles in a triangle are equal then the triangle is an equilateral triangle]


Now, in an equilateral triangle each angle is equal to 60°.


Hence, DAC = 60° ----- (1)


Similarly, if we consider ΔADF,


AD = DE = EA (by construction)


So, ΔADF is equilateral and DAE = 60° ----- (2)


From (1) and (2) we can say that, DAE and DAC are alternative angles for the line AD, thus making ED parallel to AC.


Now, ED || AC ------- (3)


Also, we know that DF = EF ----- (4)


(As F is formed by the arcs with same radius)


From (4) we can say that, point F lies on the perpendicular bisector of DE.


Thus FID = 90° ------- (5)


(where I is the point where FA bisects DE.)


Consider the lines AF, AC and DE. Here AF is the transverse to the parallel lines AC and DE.


FID = FAC ------- (6)


(since the angles are corresponding angles made by the transverse AF with the parallel lines DE and AC)


Thus,


FID = FAC = 90°


[from (5) and (6)]


Hence, the constructed angle at A is 90°.

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