anonymous
  • anonymous
Hi folks! I'm a bit stuck on this construction question: Construct the triangle 'pqr' such that: |pq| = 5cm, |qr| = 12cm, |pr| = 13cm. Prove (i) |
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What is the center of the circle? :O
anonymous
  • anonymous
The incircle, or the circumcircle?
anonymous
  • anonymous
The circumcircle.

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anonymous
  • anonymous
It's halfway along the line |pr|
anonymous
  • anonymous
Is there anything given about the inner circle? I am assuming that it touches all three sides of the triangle.
anonymous
  • anonymous
It does, and we also know that at those points, the radius, and triangle side make a 90 degree angle.
anonymous
  • anonymous
The radius of the \(circum\)circle is 6.5, but part (iv) asks about the radius of the \(in\)circle.
experimentX
  • experimentX
sorry .. for that
anonymous
  • anonymous
|dw:1332842959991:dw|
anonymous
  • anonymous
|dw:1332843250907:dw|
anonymous
  • anonymous
Co-ordinate of Incenter should be (r,r), where r is the radius of the incircle.
anonymous
  • anonymous
Apply the formula for incenter, \((\frac{12*5}{5+12+13},\frac{5*12}{5+12+13})=( 2,2)\)
anonymous
  • anonymous
I am thinking of another way, equations of sides of triangle are x=0,y=0 and 12y+5x+60=0. Now the three sides are tangent to the incircle. We can assume the point of contact (tangent-circle) to be \((x_1,y_1)\), (r,0) and (0,r).
anonymous
  • anonymous
The equation \(x - x_1 = \frac{y_1 - r}{x_1 - r} ( y - y_1)\) must be perpendicular to the equation 12y+5x+60=0. \[\implies \frac{x_1 - r}{y_1 - r} = \frac{5}{12}\]
anonymous
  • anonymous
I just realised that I can add the area of 3 triangles (assuming 'o' is the centre of the incircle): \[A_1=\frac{1}{2}\times 5 \times r\]\[A_2 = \frac{1}{2} \times 12 \times r\]\[A_3 = \frac{1}{2} \times 13 \times r\]\[A_1+A_2+A_3=30\]\[\Leftrightarrow r\left(\frac{5}{2}+\frac{12}{2}+\frac{13}{2}\right)=30\]\[\Leftrightarrow 15r=30\]\[\Leftrightarrow r=2\]
anonymous
  • anonymous
I've not seen that formula for the incentre before, but it also works!
anonymous
  • anonymous
Yeah (y) a better way
anonymous
  • anonymous
Thanks Ishy!
experimentX
  • experimentX
draw perpendicular line from center of circle to sides and equate area 1/2*5*12 = x^2+(12-x)*x+(5-x)*x

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