Question

A tree broke at a point but did not seperate . Its top touched the ground at w distance of 6 dm from its base .if the point where it broke was at a height of 2.5 dm from the ground , what was the total height of tree before it broke

Answer 1

Hey dear!!

I solved a similar question like this in class today!

Anyways here's your answer.

The figure:
Form a right-angled triangle ABC, right-angled at B. Mark A at the top vertex. Extend line AB. Mark the extended line as D. Name the base vertex as C.
So we get to know that line AB is the height of the unbroken tree, So it is of 2.5dm. 
The broken part of the tree is 6dm away from its base.That is the length of line BC.
And line BD is the original length of the tree.
AD is the length of the broken tree.


Time for the solution!

By pythagoras theorem,

AD²= AB²+BC²
AD²= (2.5)²+ (6)²
AD²= 6.25+36
AD²= 42.25
AD²= (6.25)²
AD²= 6.25 dm

Therefore, the original length of the tree is AD+AB. That is equal to 6.25+2.5= 8.75dm.

Therefore, the original length of the tree is 8.75dm.

Hope this helps you!

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Answer 2

in ∆ ABC,

${ab}^{2} + {bc}^{2} = {ac}^{2} \\ {2.5}^{2} + {6}^{2} = ac {}^{2} \\ 6.25 + 36 = {ac}^{2} \\ 42.25 = {ac}^{2} \\ ac = \sqrt{42.25} = 6.5$

length of tree = ab+ac = 2.5+6.5 = 9 dm

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