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Question
A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
Solution
Let the height of the tree AD be “h”.
In ∆ACD and ∆BCF,
∠A = ∠B = 90°
∠C is common
∆ACD ~ ∆BCF by AA similarity
`"AD"/"BF" = "AC"/"BC"`
`"h"/x = 24/2` = 6
h = 6x ...(1)
In ∆ACE and ∆ABF,
∠C = ∠B = 90°
∠A is common
∴ ∆ACE ~ ∆ABF
`"CE"/"BF" = "AC"/"AB"`
`2/x = 24/20`
24x = 20 × 2
x = `(20 xx 2)/24 = (5 xx 2)/6 = 10/6`
x = `5/3`
Substitute the value of x in (1)
h = `6 xx 5/3` = 10 m
∴ Height of the tree is 10 m
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