See the picture below.

You are given AB, AC and BC.
DE is parallel to BC. You are also given the area ratio between
ADE and BDEC. You have to find the value of
AD.

Input starts with an integer T (≤ 25), denoting the number of test cases.

Each case begins with four real numbers denoting AB, AC, BC and the ratio of
ADE and BDEC(ADE / BDEC). You can safely assume that the given triangle is a valid triangle with positive area.

For each case of input you have to print the case number and AD. Errors less than
10^-6 will be ignored.

4

100 100 100 2

10 12 14 1

7 8 9 10

8.134 9.098 7.123 5.10

Case 1: 81.6496580

Case 2: 7.07106781

Case 3: 6.6742381247

Case 4: 7.437454786

#include<stdio.h>

#include<string.h>

#include<math.h>

int main()

{

	int t;

	int K=1;

	scanf("%d",&t);

	while(t--)

	{

		double a,b,c,k;

		scanf("%lf%lf%lf%lf",&a,&b,&c,&k);

		double ad=a*sqrt(k/(1+k));

		printf("Case %d: %.6lf\n",K++,ad);

	}

	return 0;

}