What is the perimeter of triangle APQ in this geometry problem?

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Homework Help Overview

The problem involves a circle with a radius of 9 cm centered at point O, with segment OA measuring 15 cm. The tangents AB, CA, and PQ touch the circle at points B, C, and R, respectively. The objective is to find the perimeter of triangle APQ.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss two methods for calculating the perimeter, with one method assuming the perimeter is invariant with respect to point R's location. There is also mention of using similar triangles to derive segment lengths.

Discussion Status

Some participants express uncertainty about the validity of their assumptions regarding the perimeter's invariance and the relationships between segments. There is a recognition of the similarity of triangles, but not all participants agree on the conclusions drawn. Guidance is offered in the form of confirming relationships between segments.

Contextual Notes

Participants question the assumptions made about the tangents and the positioning of point R, indicating a lack of explicit restrictions in the problem statement. There is also a note of confusion regarding the relationships between segments BP, PR, RQ, and CQ.

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Homework Statement


upload_2017-6-22_15-44-26.png

there is a circle with radius 9cm centred at o
segment oa is 15cm
ab, ca and pq are tangents to circle at points b,c and r respectively
find perimeter of triangle apq

Homework Equations

The Attempt at a Solution


i have solved the problem and got the right answer with two methods but i feel none of the methods is proper per say
method 1:
since the question is not imposing any restriction on the location of point r i assumed the perimeter is invariant with respect to r and hence let the segment pq fall on segment ca essentially the perimeter of two triangles is now just the two times the length of ac
ac is 12 cm by pythagoras hence the perimeter of triangle is 24 cm
a really crude method really don't think it is correct
method 2:
once with the assumption that the perimeter is invariant with respect to r
i shifted r right up to the point where da intersects the circle
then the triangle rpa is similar to triangle boa then rp is 9/2 and pa is 7.5 cm
then perimeter is once again 15 + 9 = 24 cm
however i feel my assumption is wrong and the question can be solved without the assumption so someone please help? thanks
 
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vishnu 73 said:

Homework Statement


View attachment 205891
there is a circle with radius 9cm centred at o
segment oa is 15cm
ab, ca and pq are tangents to circle at points b,c and r respectively
find perimeter of triangle apq

Homework Equations

The Attempt at a Solution


i have solved the problem and got the right answer with two methods but i feel none of the methods is proper per say
method 1:
since the question is not imposing any restriction on the location of point r i assumed the perimeter is invariant with respect to r and hence let the segment pq fall on segment ca essentially the perimeter of two triangles is now just the two times the length of ac
ac is 12 cm by pythagoras hence the perimeter of triangle is 24 cm
a really crude method really don't think it is correct
method 2:
once with the assumption that the perimeter is invariant with respect to r
i shifted r right up to the point where da intersects the circle
then the triangle rpa is similar to triangle boa then rp is 9/2 and pa is 7.5 cm
then perimeter is once again 15 + 9 = 24 cm
however i feel my assumption is wrong and the question can be solved without the assumption so someone please help? thanks
upload_2017-6-22_14-37-57.png


How are the segments BP and PR related? The same with RQ and CQ?
 
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BP = BR and RQ = CQ
and i also note that triangle PRO is similar to ORQ
ok that is smart thanks
perimeter = PA + PQ + AQ = BA - BP + CA - CQ + PQ = BA + CA- PQ + PQ = 24 thanks
 
vishnu 73 said:
BP = BR and RQ = CQ
and i also note that triangle PRO is similar to ORQ
Are they??
vishnu 73 said:
ok that is smart thanks
perimeter = PA + PQ + AQ = BA - BP + CA - CQ + PQ = BA + CA- PQ + PQ = 24 thanks
It was simple, was not it? :smile:
 
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wait no they are not sorry my bad

yup it was simple
 

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