Question

A circle is inscribed in a triangle ABC, having sides 8cm, 10cm and 12cm. Find AD, BE and CF (these 3 are altitudes of triangle ABC).

Answer 1

Hint: To find this first we will draw a figure of a circle inscribed in a triangle ABC where the sides of triangle can touch the edge of the circle by making points D, E, & F. From the figure we can easily understand that AD, BE & CF are tangents of a circle. Then with the help of the given sides of a triangle we will find the length of tangent made by external point of circle by substituting method

Complete step-by-step answer:

From the figure we get to know that the external points of the circle of circle A, B, & C are the tangents and $AB=12cm,BC=8cm,AC=10cm$ .
We know that the tangents drawn from the external points of the circle are equal.
Therefore, $AD=AF,CF=CE,BE=BD.$
Now let us consider –
$\begin{align}
  & AD=AF=x \\
 & BD=BE=y \\
 & CE=CF=z \\
\end{align}$
It is given that $AB=12cm,BC=8cm,AC=10cm$.
Where,
$\Rightarrow AB=AD+BD$
$x+y=12cm$ ……………….. (1)
$\Rightarrow BC=BE+CE$
$y+z=8cm$ …………………… (2)
$\Rightarrow AC=AF+CF$
$x+z=10cm$ ………………… (3)
By adding equation (1), (2) and (3), we get –
$\begin{align}
  & \left( x+y \right)+\left( y+z \right)+\left( x+z \right)=12cm+8cm+10cm \\
 & 2x+2y+2z=30cm \\
\end{align}$
By taking 2 as common we get –
$2\left( x+y+z \right)=30cm$
By dividing both sides by 2, we get –
$x+y+z=15cm$ ………………………….. (4)
Now, we will substitute equation (1) in equation (4). So, we get –
$\begin{align}
  & \left( x+y \right)+z=15 \\
 & 12+z=15 \\
\end{align}$
By subtracting 12 from both the sides, we get –
$\begin{align}
  & z=15-12 \\
 & z=3 \\
\end{align}$
Now, we will substitute the value of ‘z’ in equation (3).
$\begin{align}
  & x+z=10 \\
 & x+3=10 \\
\end{align}$
By subtracting 3 from both sides, we get –
$\begin{align}
  & x=10-3 \\
 & x=7 \\
\end{align}$
By substituting the value of ‘z’ in equation (2), we get –
$\begin{align}
  & y+z=8 \\
 & y+3=8 \\
\end{align}$
By subtracting 3 from both sides, we get –
$\begin{align}
  & y=8-3 \\
 & y=5 \\
\end{align}$
Therefore,
$\begin{align}
  & AD=x=7cm \\
 & BE=y=5cm \\
 & CF=z=3cm. \\
\end{align}$

Note: Generally students make mistakes while solving this problem by taking x, y & z as the sides of the triangle then divide it by 2 to get an answer which is completely wrong. Students should know that the external points of the circle which is forming a triangle are the tangents of a circle. The lengths of two tangents from an external point of a circle are equal.