Question: 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
Options:
A) 10 cm
B) 8 cm
C) 7.5 cm
D) 6 cm
Show Answer
Answer:
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 8cm $
BETA
