Description
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 ofAD.
Input
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.
Output
For each case of input you have to print the case number and AD. Errors less than 10-6 will be ignored.
Sample Input
4
100 100 100 2
10 12 14 1
7 8 9 10
8.134 9.098 7.123 5.10
Sample Output
Case 1: 81.6496580
Case 2: 7.07106781
Case 3: 6.6742381247
Case 4: 7.437454786
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int main() {
int T,p=0;
scanf("%d",&T);
while(T--) {
double a,b,c,d;
scanf("%lf%lf%lf%lf",&a,&b,&c,&d);
double cnt=sqrt(d/(d+1));
printf("Case %d: %.6lf\n",++p,cnt*a);
}
return 0;
}

本文介绍了一个简单的几何问题:已知三角形ABC的两边及平行于BC的直线DE,且给出ADE与BDEC区域面积比,求解AD的长度。通过相似三角形原理,利用输入数据计算出AD的值。
uDebug
280

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