geomerty sum

THAT IS AN EXCELLENT DIAGRAM CHRIS!   I am very impressed !

I was being a bit lazy but it is always better when a good pic is added.

Hi Manas, it is nice to meet you.  :)

I got 7  

This is an OUTLINE of my logic.

I drew up the diagram. Let X be the centre of the cirlce.

Draw in each radii and draw in t he intervals between the centre ifthe circle and the vertices

The angle between a radius and a tangent is 90 degrees 

So you have now split your triangle up into 6 right angled triangles

cr=cp=8   tangents subtended from a point intersect the circle at equal distances.

br=bQ=6

ap=aq = ?

tan(< rbX)=tan(<Xbq)=4/6     so       < rbX = 33.69 degrees (2dp)  

tan(<rcX) = tan(<pcX) = 4/8      so      those angles each =26.56  degrees (2dp)

Now < pax= 29.75 degrees         Working involves angle sum of a triangle

so

tan 29.75 = 4/ap

ap=aq=6.99 = near enough to 7.

It may be exactly 7.  You would have to do the working without any rounding to find out :))

NOTE:  I have not checked my working it is quite possible that some of the numbers are incorrect.  

You need to check it!

If something doesn't make sense then ask about it.  :)

Following Melody's logic that, from a point outside a circle, two tangents drawn to the circle will have equal lengths.....look at the following diagram of the problem.....

Angle OCR will ≈  tan^-1(1/2) ≈ 26.565051177078°

And we can imagine kite PORC  with angle PCR = 2* 26.565051177078°  ≈ 53.130102354156°

And the tangent of this angle will   = 4/3

So.......the line with the equation  y = 4/3x  will intersect this circle at  (4.8, 6.4)

Angle OBR ≈ tan^-1(2/3) ≈ 33.69006752598°

And, like before, we can imagine kite QORB with angle QBR = 2 * 33.69006752598° ≈ 67.38013505196°

And the tangent of this angle = 12/5

So....the line with the equation   y = (-12/5)(x - 14)  will intersect the circle at  about ( 152/13 , 72/13)

And these two tangent lines will intersect at (9, 12)  = "A"

So ....  AQ = sqrt [(9 - 152/13)^2 + ( 12 - 72/13)^2 ]  = 7

And  since AP  is a tangent to the circle drawn from "A,"  it will have the same length as AQ   = 7

Looks like you were correct, Melody.....

melody thx for ans. can you try this with herons formula

Manas has asked me to explain my answer more fully. (by private message)

I drew up the diagram. Let X be the centre of the cirlce.  (I fogot to put that on the diagram) 

Draw in each radii and draw in t he intervals between the centre ifthe circle and the vertices

The angle between a radius and a tangent is 90 degrees 

So you have now split your triangle up into 6 right angled triangles

cr=cp=8   tangents subtended from a point intersect the circle at equal distances.

br=bQ=6

ap=aq = ?

\(tan\beta = 4/6\\ so\\ \beta = tan^{-1}(4/6) = 33.6900675\; degrees \)

\(tan\delta=4/8\\ \delta=tan^{-1}(4/8)=26.5650511\; degrees\)

\(2\alpha+2\beta+2\delta=180\;\;angle\;sum\;of\;a\;triangle\\ \alpha+\beta+\delta=90\\ \alpha=90-\beta-\delta\\ \alpha=90-33.6900675-26.5650511\\ \alpha=29.74488\\ tan\;\alpha=\frac{4}{distance\;Aq}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\;Aq=\frac{4}{tan\;29.74488}\\ distance\; Aq\approx6.99888\\ Aq=Ap\approx 7\;units \)

If you still do not understand manas you can say so but you need to ask specific questions. 

If you do not understand any of it then perhaps it is just way over your head. :)

Oh, you cannot use Heron's Rule.

that is for finding the area when you have the sides.

OR

somtimes you may be able to use it fo find a side when given an area 

BUT this question has nothing to do with area :))