Q. 204.6( 29 Votes )

Construct a

Answer :

Steps of construction:

Step 1: A ray OP is drawn.
Taking O as a centre and any radius draw an arc cutting OP at Q.




Step 2: Taking Q as a centre and any radius draw an arc cutting previous arc at R.
Repeat the process with R to cut the previous arc at S.




Step3: Taking R and S as centre draw the arc of radius more tha half of RS
and draw two arcs intersecting at A.
Join OA

Hence
POA = 90°
Justification:



We need to prove
POA = 90°
Join OR and OS and RQ.

By construction
OQ = OS = QR
So, ΔROQ is an equilateral triangle.
Similarly ΔSOR is an equilateral triangle.
So,
SOR = 60°
As
ROQ = 60°
It means
ROP = 60°
Join AS and AR.


Now in ΔOSA and ΔORA,
SR = SR (common)
AS = AR (radii of same arcs)
OS = OR (radii of same arcs)
ΔOSA
ΔORA
SOA = ROA (CPCT)
So,

SOA = ROA = 1/2 SOR
= 1/2 (60°)
= 30°
Now,
POA = ROA + POR
= 30° + 60°
= 90°


hence Justified


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses