Problem Solving

In the figure above, fi the area of the triangle on the right is twice the area of the triangle on the left, the in terms of s, S=

o ((sqrt2)/2)s

o ((sqrt3)/2)s

o (sqrt2)s

o (sqrt3)s

o 2s

I've attached a screen shot of the question. Not sure how they got the answer. Many Thanks in advance!

Area of triangle given its sides a, b and c = sqrt [ s(s-a)(s-b)(s-c) ]

where s = (a+b+c) / 2

Now, consider a,b,c to be the sides of the smaller triangle and A,B,C to
be the sides of the larger triangle.

we know A/a = B/b = C/c (because they are proportional). Lets call this x

s1 = (a+b+c)/2
s2 = (A+B+C)/2

s2/s1 = x = A/a = B/b = C/c

We have

2 * sqrt [ s1(s1-a)(s1-b)(s1-c) ] = sqrt [ s2(s2-A)(s2-B)(s2-C) ]

2 * sqrt [ s1(s1-a)(s1-b)(s1-c) ] = sqrt [ x*s1 * x*(s1-a) *x*(s1-b) *x*(s1-c) ]

4 * s1(s1-a)(s1-b)(s1-c) = x^4 * s1(s1-a)(s1-b)(s1-c)

So, x = sqrt(2)

So, if smaller triangle has side s, the bigger one has S = sqrt(2) * s

Hence C

Jayhawk2001 Thanks a bunch!!