Question

if p q r and s are the points on the circle. o is the centre of the circle, pq=qr=rs, angle ops=15 find opr​

Answer 1

Answer:

Step-by-step explanation:

These are three points on circle and it very unlikely that all three points are collinear that means, when they are joined they create a isosceles triangle,

if we drop a side bisector from P to line segment joining Q and R it will pass through the center of the circle as it is an isosceles triangle.

Let the center be at O and the foot of perpendicular/side bisector of QR be D

then OP=r(radius of circle)........Which we need to find out???

OD=x(say)

angle(PDQ)=90

PQ=13 and QD=5

hence from Pythagoras theorem, PD=12

Now PD=r+x

=>x=12-r---------------(1)

Triangle OQD also form a right angled triangle

Hence,

QD^2+OD^2=OQ^2

=>5^2+x^2=r^2

=>25+(12-r)^2=r^2

=>r=169/24

=>r=7.04167cms

Thanks!

Answer 2

Answer:

This is a triangle made out of three points on a circle, and it is highly rare that all three of them are collinear, which means that when they are brought together, they form an isosceles triangle.

Because it is an isosceles triangle, if we drop a side bisector from P to the line segment connecting Q and R, it will pass through the circle's center.

Let O serve as the center and D as the foot of the perpendicular or side bisector of QR.

so OP=r(circle radius), which we need to determine:

OD=x(say)

angle(PDQ)=90

PQ = 13; QD = 5.

PD=12 is the result of Pythagoras's Theorem.

Now, PD=r+x.

=>x=12-r———————-(1)

Triangle OQD can also be formed into a right triangle.

Hence,

QD^2+OD^2=OQ^2

=>5^2+x^2=r^2

=>25+(12-r)^2=r^2

=>r=169/24

=>r=7.04167cms

#SPJ3

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