In the given figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.
From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, we have
BQ = BP
Let us denote BP and BQ by x
AP = AR
Let us denote AP and AR by y
RC = QC
Let us denote RC and RQ by z
We have been given that is a right triangle and BC = 6 cm and AB = 8 cm. let us find out AC using Pythagoras theorem. We have,
$A C^{2}=A B^{2}+B C^{2}$
$A C^{2}=6^{2}+8^{2}$
$A C^{2}=36+64$
$A C^{2}=100$
$A C=\sqrt{100}$
$A C=10$
Consider the perimeter of the given triangle. We have,
AB + BC + AC = 8 + 6 + 10
AB + BC + AC = 24
Looking at the figure, we can rewrite it as,
AP + PB + BQ + QC + AR + RC = 24
Let us replace the sides with the respective x, y and z which we have decided to use.
$y+x+x+z+y+z=24$
$2 x+2 y+2 z=24$
$2(x+y+z)=24$
$x+y+z=12$.....................(1)
Now, consider the side AC of the triangle.
AC = 10
Looking at the figure we can say,
AR + RC = 10
y + z = 10 …… (2)
Now let us subtract equation (2) from equation (1). We have,
x + y + z = 12
y + z = 10
After subtracting we get,
x = 2
That is,
BQ = 2, and
BP = 2
Now consider the quadrilateral BPOQ. We have,
BP = BQ (since length of two tangents drawn to a circle from the same external point are equal)
Also,
PO = OQ (radii of the same circle)
It is given that .
From the property of tangents, we know that the tangent will be at right angle to the radius of the circle at the point of contact. Therefore,
$\angle O P B=90^{\circ}$
$\angle O Q B=90^{\circ}$
We know that sum of all angles of a quadrilateral will be equal to . Therefore,
$\angle P B Q+\angle O P B+\angle O Q B+\angle P O Q=360^{\circ}$
$90^{\circ}+90^{\circ}+90^{\circ}+\angle P O Q=360^{\circ}$
$270^{\circ}+\angle P O Q=360^{\circ}$
$\angle P O Q=90^{\circ}$
Since all the angles of the quadrilateral are equal to and the adjacent sides also equal, this quadrilateral is a square. Therefore, all sides will be equal. We have found out that,
BP = 2 cm
Therefore, the radii
PO = 2 cm
Thus the radius of the incircle of the triangle is 2 cm.
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