| If the lengths of the sides of a triangle are in the ratio 4 : 5 : 6 and the in radius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is [SSC (CGL) 2013] |
A) 10 cm
B) 8 cm
C) 7.5 cm
D) 6 cm
Correct Answer: B
Solution :
| We know that radius of incircle, |
| \[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}\] |
| Let the sides of triangle be 4x, 5x and 6x. |
| \[\therefore \]\[s=\frac{4x+5x+6x}{2}=7.5x\] |
| \[\Rightarrow \]\[3=\sqrt{\frac{(7.5x-4x)(7.5x-5x)(7.5x-6x)}{7.5x}}\] |
| On solving, we get \[x=2.27\] |
| \[\therefore \]Sides are\[(2.27\times 4)=9.08\] |
| \[(2.27\times 5)=11.35\] |
| and \[(2.27\times 6)=13.62\] |
| \[\therefore \] \[s=\frac{9.80+11.35+13.62}{2}=17.385\] |
| Now, area of isosceles triangle, |
| \[\Delta =\sqrt{s\,\,(s-a)(s-b)(s-c)}\] |
| \[=\sqrt{17.385\,\,(8.305)(6.035)(3.765)}=57.27\] |
| \[\therefore \]Area of triangle \[=\frac{1}{2}\times \]Base \[\times \]Altitude |
| \[\Rightarrow \] \[57.27=\frac{1}{2}\times 13.62\times h\] |
| \[\Rightarrow \] \[h=8.40\approx 8\,\,cm\] |
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