AB is a vertical trunk of a huge tree with A being the point where the base of the trunk touches the ground. Due to a cyclone, the trunk has been broken at C which is at a height of 12 metres, broken part is partially attached to the vertical portion of the trunk at C. If the end of the broken part B touches the ground at D which is at a distance of 5 metres from A, then the original height of the trunk is:

The correct option is 25 m.

Let us break down the problem step-by-step to understand why this answer is correct.

Step 1: Visualize the scenario
Initially, the tree trunk is vertical and represented by the straight line segment AB, where A is the base of the trunk on the ground and B is the top of the tree.
During the cyclone, the trunk breaks at point C. The lower part of the trunk, AC, remains vertical. We are given that the height of this part is 12 metres, so:
AC = 12 m.

Step 2: Understand the broken part
The broken upper part of the tree, BC, bends down but remains attached at C. Its top end B touches the ground at a point D.
Therefore, the length of the broken part BC is equal to the length of the slanted segment CD lying on the ground:
CD = BC.

Step 3: Set up the right-angled triangle
Since the trunk AC is vertical, it makes a right angle with the horizontal ground AD. This forms a right-angled triangle CAD, where:
- The angle at A is a right angle (90 degrees).
- AC is the perpendicular side (height = 12 m).
- AD is the base on the ground, and we are given that the distance from the base of the tree A to the point D where the top touches the ground is 5 metres (AD = 5 m).
- CD is the hypotenuse of the triangle.

Step 4: Apply the Pythagorean theorem
According to the Pythagorean theorem, in the right-angled triangle CAD, the square of the hypotenuse is equal to the sum of the squares of the other two sides:

${\mathrm{CD}}^2={\mathrm{AC}}^2+{\mathrm{AD}}^2$

Substitute the given values into the equation:

${\mathrm{CD}}^2={12}^2+5^2$

${\mathrm{CD}}^2=144+25$

${\mathrm{CD}}^2=169$

Taking the square root on both sides:

$\mathrm{CD}=\sqrt{169}=13\text{m}$

Since CD = BC, the length of the broken part of the tree is 13 m.

Step 5: Calculate the original height of the trunk
The original height of the tree trunk is the sum of the vertical standing part (AC) and the broken part (BC):
Original height = AC + BC

$\text{Original height}=12+13=25\text{m}$

Thus, the original height of the tree trunk is 25 metres.