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A tree is broken at a height of 8m from the ground and it's top touches the ground at a distance of 15 m from the base of tree. Find the original height of the tree.
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Let height of the tree = BD
= 8 + x ,
If it was broken at C , and
touching the ground at A .
AB = x ,
BC = 8 m
Now ,
In ∆ABC , <B = 90°
By Phythogarian theorem ,
AC² = AB² + BC²
= 15² + 8²
= 225 + 64
= 289
AC= √289
AC = x = 17 m
Therefore ,
Height of the tree = BD
= BC + CD
= BC + x
= 8 + 17
= 25 m
••••
= 8 + x ,
If it was broken at C , and
touching the ground at A .
AB = x ,
BC = 8 m
Now ,
In ∆ABC , <B = 90°
By Phythogarian theorem ,
AC² = AB² + BC²
= 15² + 8²
= 225 + 64
= 289
AC= √289
AC = x = 17 m
Therefore ,
Height of the tree = BD
= BC + CD
= BC + x
= 8 + 17
= 25 m
••••
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