If a triangle is divided into four pieces with areas 8 , 5 , 10 and x then find x.

if the question is a triangle is divided into four regions by two straight lines. the area of the three regions are given.

what is the area of the fourth region?

let 1/2 AB= u, let the area of the triangle AEF=5, area(ΔDFB)=8, area(ΔAFB)=10, and area(CEFD)=x

here [by the crossed ladder theorem(thm is given after the solution)]

area of the triangle AEB is 5+10=15

since area of the triangle is 1/2*base*height

area(tri AEB)=1/2AB*e=eu

therefore

similarly

substitute these values in (1) we have:

eliminating u from both sides of the equation:

**crossed ladder theorem:**

let AE=x, BF=y, GA=a and HB=b

traingle EAB and triangle CDB will be similar triangle. therefore

traingle FBA and triangle CDA will be similar triangle. therefore

taking the reciprocal of equation (1) and (2) and adding

similarly

from (3) ,(4),(5) and (6) we have

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