Given the height of an isosceles triangle and two integers , . The task is to find the height from top of the triangle such that if we make a cut at height h from top parallel to base then the triangle must be divided into two parts with the ratio of their areas equal to n:m.
Input : H = 4, n = 1, m = 1 Output : 2.82843 Input : H = 4, n = 1, m = 0 Output : 4
First of all before proceeding let us mention some of the basic properties of an isosceles triangle.
Let ▲ABC is an isosceles triangle with AB = AC and BC being unequal side and base of the triangle. If D is mid-point of BC, then AD is our height H. Also, if we draw a parallel line to BC which cuts AB and AC at points E and F respectively and G being the midpoint of EF then ▲AEG is similar to ▲ABD, ▲AFG is similar to ▲ACD, ▲AEF is similar to ▲ABC, and by using properties of similar triangles we can conclude the following points:
AE/AB = AG/AD = EG/BD = EF/BC = AF/AC —–(i)
As per problem’s requirement, we have to find out the height h, such that the ratio of the area of ▲AEF to the area of trapezium EFCB = n:m.
Let, h is the height of cut from the top of the triangle.
Now, area of ▲AEF = 0.5 * AG * EF and area of trapezium EFCB = 0.5 * GD * (EF+BC)
also, ratio of both is n:m.
So, we can say that ratio of area of ▲AEF to area of ▲ABC is equal to n :(n+m).
=> area of ▲AEF / area of ▲ABC = n / (n+m)
=> (0.5 * AG * EF) / (0.5 * AD * BC) = n / (n+m)
=> (AG/AD) * (EF/BC) = n / (n+m)
=> (EF/BC) * (EF/BC) = n / (n+m)
=> h2 /H2 = n / (n+m)
=> h = H*sqrt(n/(n+m))
Below is the implementation of the above approach:
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