Question

9. A tree broke at a point but did not separate. Its top touched the ground at a
distance of 6 dm from its base. If the point where it broke be at a height 2.5 dm
from the ground, what was the total height of the tree before it broke?
​

Answer 1

${ \sf Height \: Of \: Tree \: before \: it \: broke \: = \red{9cm }}$

Step by step explanation

Given

• Height of Tree = 2.5 cm
• Base of tree = 6 cm

To Find

• total height of the tree before it broke .

$\\\sf \: AB² +BC²= AC² \: \: \: \: \: \: \: \: \: \green{ By \: Pythagoras \: theorem}$

$\sf \: 2.5² +6²= AC²$

$\sf \: 6.25+36= AC²$

$\sf \:AC = \sqrt{ 42.25}$

$\sf \:AC = 6.25 \: cm$

$~~~~ \underline { {\boxed{ \sf \: The \: Height \: Of \: Tree \: before \: it \: broke \: = \: \pink{6.25+2.5 }= \red{9cm } }}}$

Figure Of The Solution

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf 2.5cm}\put(2.8,.3){\large\bf 6cm}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf }\end{picture}$