Get Answers to all your Questions

header-bg qa

Q1.    Construct an angle of 90o at the initial point of a given ray and justify the construction.

Answers (1)



The steps of construction to follow:

Step 1: Draw a ray OP.

Then, take O as the centre and any radius draw an arc cutting OP at Q.

Step 2: Now, taking Q as the centre and with the same radius as before draw an arc cutting the previous arc at R. Repeat the process with R to cut the previous arc at S.


Step 3: Take R and S as centre draw the arc of radius more than the half of RS and draw two arcs intersecting at A. Then, join OA.


Hence, \angle POA = 90^{\circ }.


We need to justify, \angle POA = 90^{\circ }

So, join OR and OS and RQ. we obtain

construction 4







By construction OQ = OS = QR.

So, \triangle ROQ is an equilateral triangle. Similarly \triangle SOR is an equilateral triangle.

So, \angle SOR = 60^{\circ}

Now, \angle ROQ = 60^{\circ} that means \angle ROP = 60^{\circ}.

Then, join AS and AR:

constructino 5

Now, in triangles OSA and ORA:

SR = SR  (common) 

AS = AR  (Radii of same arcs)

OS = OR  (radii of the same arcs)

So, \angle SOA = \angle ROA = \frac{1}{2}\angle SOR

Therefore, \angle ROA = 30^{\circ}

and \angle POA = \angle ROA+\angle POR = 30^{\circ} +60^{\circ} =90^{\circ}

Hence, justified.

Posted by

Divya Prakash Singh

View full answer