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In DeltaABC, the incircle touches the si...

In `DeltaABC`, the incircle touches the sides BC, CA and AB, respectively, at D, E,and F. If the radius of the incircle is 4 units and BD, CE, and AF are consecutive integers, then

A

Sides are also consecutive integers

B

Perimeter of the triangle is 42 units

C

Area of triangle is 84 sq. units

D

Diameter of circumcircle is 65 units

Text Solution

Verified by Experts

The correct Answer is:
21

Let `BD = x, CE = x + 1 and AF = x+2`. Then
`CD = CE = x + 1 = s -a`
`AE = AF = x + 2 = s -b`
`BF = BD = x = s -c`
On adding, we get
`3s - (a + b + c) = 3x + 3`
`:. S = 3x + 3`
Now `r = (Delta)/(s) = (1)/(s) sqrt(s(s-a) (s-b) (s-c))`
or `4 = sqrt(((x+2) x(x+1))/(3x + 3))`
or `16 = (x(x+2))/(3)`
or `x^(2) + 2x = 48`
or `(x +8) (x-6) = 0`
or `x = 6`
Hence, `s = 21`
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