Question

Two poles of height a metres and b metres are p metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is gives by ab/ a+ b metres

Answer 1

Let AB & CD  be two poles of heights 'a' m and 'b' m .Let the point of intersection of lines AD & BC be E and the its height be 'h' m. Let the perpendicular drawn from E be EF. Let DF = 'x' m and BF = 'y' m.

BD = 'p' m    [ GIVEN]
Then x + y= p

In ∆ DEF and ∆DAB.
∠DBA = ∠DFE    [each equal to 90°]
∠D = ∠D               [common]
∆DEF ~ ∆DAB      [by AA Similarity ]
DF/DB = EF /AB
x/p = h/a
x = ph/a………………………. (1)

In ∆ BFE and ∆BDC.
∠BFE = ∠BDC          [each equal to 90°]
∠B = ∠B                     [common]
∆ BFE ~ ∆BDC           [by AA Similarity ]
BF/BD = EF /CD
y/p = h/b
y= ph/b……………………... (2)


On Adding eq (1) and (2)
x + y = ph/a + ph/b
p = ph(1/a +1/b)        [x +y = p]
p/p = h(1/a +1/b)
1 = h[(a+b)/ab]
ab = h(a+b)

h = ab/(a+b) m

HOPE THIS WILL HELP YOU....