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Solution

 

draw DG line so that DG||BC

draw CM line so that CM||AD

here, AB,BC,CD,DA are given.

DG=BC(known)

CM = AD (Known)

BG = CD (Known)

AG  = AB – BG = AB –CD (Known)

let,

CO = DH = H (Unknown)

FN = EK = h (Unknown)

In ΔADG

S=(AG+DG+AD)/2

area of ΔADG =(S * (S – AD) + (S – AG) * (S – DG) =A (known)……… (1)

area of ΔADG=(1/2)* AG * DH =(1/2)* AG * H ……………………..(2)

From equation (1) & (2) we get ,

(1/2)* AG * H=A

=>H=(2*A)/AG

Area of trapezium ABCD= 2*Area of trapezium ABFE=2*Area of trapezium EFCD

Area of trapezium ABCD=Area(known)

Area of trapezium ABFE= (AB + EF) * EK = (AB + EF) * h

Area of trapezium EFCD=(EF + CD) (DH – Ek) = (EF + CD) (H – h)

Area = 2 * (AB + EF) * h ……………….(3)

Area = 2 * (EF + CD) (H – h) ……………(4)

From (3) =>

h=Area/(2*(AB+EF))……………..(5)

putting the value of in equation (4)

Area = 2 * (EF + CD) * {H – Area/(2*(AB+EF))  }

=> Area (AB + EF) = (EF + CD) * {2H * (AB + EF) – Area}

=>   Area * AB + Area * EF = EF * 2H * AB + 2H * EF2 + AB * 2H*CD + 2H * CD * EF – Area * EF – Area * CD

=>     2H * EF2 + (2H * AB + 2H * CD – 2 Area) EF + (AB * 2H * CD – Area * CD – Area * AB) = 0

=>   aEF2 + bEF + c = 0

where,

a       = 2H

b       = 2H * AB + 2H * CD – 2 Area = 2H (AB + CD) – 2 Area

= 2Area – 2 Area = 0

c        = AB * 2H * CD – Area * CD – Area * AB

= AB*CD*2H – Area(AB+CD)

= AB*CD*2H – (AB+CD)*H(AB+CD) [GLv‡b  Area = (AB+CD)*H ]

= 2AB*CD*H –  (AB2 +CD2)*H – 2AB*CD*H

=  – (AB2+CD2)*H

so, aEF2 + bEF + c = aEF2 + c =0 [hence b=0]

=>EF2 = –c/a

Here the negative value of EF will not be accepted.

putting the value of EF in equation (5) we can calculate the value of h.

In equiangular triangles ΔAEK & ΔADH we get,

AE/AD=EK/DH

=>AE=(EK/DH)*AD

=>AE=(h/H)*AD

Again , in equiangular triangles ΔCOB & ΔBFN :

BF/BC=FN/CO

=>BF=(FN/CO)*BC

=>BF=(h/H)*BC

Source Code

 


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