Question
A circle of diameter 8 inches is inscribed in a triangle ABC where \(\angle ABC = {90^^\circ }\). If BC = 10 inches then the area of the triangle in square inches is
A circle of diameter 8 inches is inscribed in a triangle ABC where \(\angle ABC = {90^^\circ }\). If BC = 10 inches then the area of the triangle in square inches is
Option: 120
Solution:
We know that Inradius=\(\frac{{\left( {Perpendicular + Base - Hypotenuse} \right)}}{2}\)
\(4 = \frac{{\left( {p + 10 - h} \right)}}{2}\)
h-p = 2 or h= p+2.
Now, \({p^2} + 100\; = \;{h^2}\)
\({p^2} + 100\; = \;{\left( {p + 2} \right)^2}\)
\({p^2} + 100\; = \;{p^2} + 4p + 4\)
4p = 96
p=24.
Hence, Area = \(\frac{1}{2} \times \;10 \times \;24\; = \;120\).
Solution:
We know that Inradius=\(\frac{{\left( {Perpendicular + Base - Hypotenuse} \right)}}{2}\)
\(4 = \frac{{\left( {p + 10 - h} \right)}}{2}\)
h-p = 2 or h= p+2.
Now, \({p^2} + 100\; = \;{h^2}\)
\({p^2} + 100\; = \;{\left( {p + 2} \right)^2}\)
\({p^2} + 100\; = \;{p^2} + 4p + 4\)
4p = 96
p=24.
Hence, Area = \(\frac{1}{2} \times \;10 \times \;24\; = \;120\).
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