If a triangle is divided into four pieces with areas 8 , 5 , 10 and x then find x - Maths - Areas of Parallelograms and Triangles

Question

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

Ajanta Trivedi · Accepted Answer

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(Î”DFB)=8, area(Î”AFB)=10, and area(CEFD)=x here $\frac{1}{c} + \frac{1}{f} = \frac{1}{e} + \frac{1}{d} (1)$ [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 $e u = 15 \Rightarrow \frac{1}{e} = \frac{u}{15}$ similarly $\frac{1}{f} = \frac{u}{10}, \frac{1}{d} = \frac{u}{10 + 8} = \frac{u}{18} \frac{1}{c} = \frac{u}{x + 8 + 10 + 5} = \frac{u}{x + 23}$ substitute these values in (1) we have: $\frac{u}{x + 23} + \frac{u}{10} = \frac{u}{15} + \frac{u}{18}$ eliminating u from both sides of the equation: $\frac{1}{x + 23} + \frac{1}{10} = \frac{1}{15} + \frac{1}{18} \frac{10 + x + 23}{10 * (x + 23)} = \frac{12 + 10}{180} \frac{x + 33}{x + 23} = \frac{22}{18} = \frac{11}{9} 9 (x + 33) = 11 (x + 23)$ $9 x + 297 = 11 x + 253 2 x = 297 - 253 2 x = 44 x = 22$ crossed ladder theorem:

let AE=x, BF=y, GA=a and HB=b traingle EAB and triangle CDB will be similar triangle therefore $\frac{x}{A B} = \frac{c}{B D} (1)$ traingle FBA and triangle CDA will be similar triangle therefore $\frac{y}{A B} = \frac{c}{A D} (2)$ taking the reciprocal of equation (1) and (2) and adding $\frac{A B}{x} + \frac{A B}{y} = \frac{B D}{c} + \frac{A D}{c} \Rightarrow A B (\frac{1}{x} + \frac{1}{y}) = \frac{1}{c} (B D + A D) = \frac{A B}{c} \Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{c} (3)$ similarly $\frac{1}{f} = \frac{1}{b} + \frac{1}{a} (4) \frac{1}{d} = \frac{1}{x} + \frac{1}{b} (5) \frac{1}{e} = \frac{1}{y} + \frac{1}{a} (6)$ from (3) ,(4),(5) and (6) we have $\frac{1}{c} + \frac{1}{f} = \frac{1}{x} + \frac{1}{y} + \frac{1}{a} + \frac{1}{b} \frac{1}{d} + \frac{1}{e} = \frac{1}{x} + \frac{1}{b} + \frac{1}{y} + \frac{1}{a} \Rightarrow \frac{1}{c} + \frac{1}{f} = \frac{1}{d} + \frac{1}{e}$

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