Question

tree is broken at a height of 6 m from the ground and its top touches the ground at a distance of 10 m from the base of the tree. Find the original height of the tree.





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Answer 1

Answer:

using pythagoras theorem

hypotenuse square= p^2 + b^2

h^2 = (6)^2+ 10^2

h^2 =36+100

h^2=136

$h = \sqrt{136}$

$h = \sqrt{2 \times 2 \times 2 \times 17}$

$h = 2 \sqrt{34}$

h=11.66 metres

original height of the tree=6+11.66=17.66metres

Answer 2

Given:

• A tree is broken at a height of 6 m from the ground
• Its top touches the ground at a distance of 10 m from the base of the tree.

To Find:

• Original height of tree =?

Solution:

Let,

• ∆ABC be the right traingle, right angles at B .

Since tree is vertical 8

• ∠B = 90°
• AB = 6 m
• BC = 10 m

Using, Phytgeores Theorem,

$\\ \tt:\implies\: (AC)^2 = (AB)^2 + (BC)^2$

$\tt:\implies\: (AC)^ 2 = (10)^ 2+ (6)^2$

$\tt:\implies\: (AC)^2 = 100 + 36$

$\tt:\implies\: (AC)^2 = 136$

$\tt:\implies\: AC = 2 \sqrt{4} m$

$\tt:\implies\: AC =17.66 m$

• Original height of tree = AB + AC

• Original height of tree = 6 +11.66 = 17.66 m

Hence,

Original height of tree is 17.66 m