Question

A tree is broken at a height of 5 m from the groun
and its top touches the ground at a distance of
12 m from the base of the tree. Find the origina
height of the tree.​

Answer 1

Let A’CB be the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then ΔABC is a right angled triangle, at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem,

In ΔABC

(AC)²=(AB)²+(BC)²

(AC)²=(12)²+(5)²

(AC)²=144+25

(AC)²=169

AC = √169

AC= 13 m

Hence, the total height of the tree(A’B) = A’C + CB = 13 + 5 = 18 m.

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Hope this will help you...

Answer 2

Given :-

Height of the tree broken down = 5 m

Distance it reaches the ground from the base of the tree = 12 m

To Find :-

The original height of the tree.​

Solution :-

Let ABC is the triangle and B is the point where tree is broken at the height 5 m from the ground.

Tree top touches the ground at a distance of AC = 12 m from the base of the tree,

Figure:

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{B}}\put(7.7,1){\large\sf{A}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.6,0.7){\sf{\large{12 m}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.55,1.8){\sf 5 m}\put(9.2,2.2){\large\sf ?}\end{picture}$

According to the figure,

The figure we came to conclude that right angle triangle is formed at A.

From the rule of Pythagoras theorem,

$\sf BC^{2} = AB^{2} + AC^{2}$

Substituting the values of AB² and AC²

$\sf BC^{2} = 5^{2} + 12^{2}$

$\sf BC^{2} = 25 + 144$

$\sf BC^{2} = 169$

$\sf BC=\sqrt{169}$

$\sf BC = 13 \ m$

According to the given info,

The original height of the tree = AB + BC

Substituting their values,

$\sf =5+13$

$\sf =18 \ m$

Therefore, the original height of the tree is 18 m