Question

a tree broken at the hight of 4m the ground its top touches the ground at a distance of 3 m from the tree. find the original hight of a tree​

Answer 1

Let A'CB represents 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, right angled at B.

⠀⠀⠀⠀⠀⠀⠀

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(8,1){\line(1,0){4}}\put(12,1){\line(0,2){3.8}}\qbezier(8,1)(11.2,2.9)(12,3.5)\put(10,0.6){\sf{\large{3 m}}}\put(12.2,1.9){\sf{\large{4 m}}}\put(12,.7){\sf\large B}\put(12.1,3.5){\sf\large C}\put(7.7,.7){\sf\large A}\put(12.1,4.8){\sf\large A'}\end{picture}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\frak{Here} \begin{cases} & \sf{AB = \bf{3\;m}} \\ & \sf{BC = \bf{4\;m}} \end{cases}$

$\bigstar\;{\underline{\sf{Using\; Pythagoras\; Theorem\;In\; \Delta\;ABC\;:}}}$

$\star\;{\boxed{\sf{\purple{H^2 = B^2 + P^2}}}}$

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

$:\implies\sf (AC)^2 = (3)^2 + (4)^2$

$:\implies\sf (AC)^2 = 9 + 16$

$:\implies\sf (AC)^2 = 25$

$:\implies\sf \sqrt{AC^2} = \sqrt{25}$

$:\implies{\boxed{\frak{\pink{AC = 5\;m}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;The\;total\;height\;of\;tree\;is\;AC + BC = \bf{9\;m}.}}}$

Answer 2

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \thicklines \put(2, 2){\line(0, 1){6}}\put(2, 2){\line(1,0){4}} \put(6,2){\line( -1,1){4}} \put(2,1.5){\sf B} \put(2,6){\sf A}\put(6,1.5){\sf C}\put(1.8,7.8){\line(1,0){0.4}}\put(2.1,7.9){\sf C'} \put(3.4,1.7){\sf 3m}\put(1.3,4){\sf 4m} \end{picture}$

Given:

• Height of Tree when tree broken = 4m
• Distance between foot of the Tree and the top of the Tree = 3m

Find:

• Orginal Height of the Tree

Solution:

In $\triangle$ ABC

we, know that

$\boxed{\sf H^2 = P^2 + B^2}$

where,

• AB, P = 4m
• BC, B = 3m

So,

$\dashrightarrow\sf H^2 = AB^2 + BC^2$

$\dashrightarrow\sf H^2 = {4}^{2} + {3}^{2}$

$\dashrightarrow\sf H^2 = 16+ 9$

$\dashrightarrow\sf H^2 = 25$

$\dashrightarrow\sf H = \sqrt{25}$

$\dashrightarrow\sf H = 5m$

$\therefore\sf AC = 5m$

$\rule{300}{2}$

Total Height of the Tree $\red\star$

➟ AC' = AB + AC

where,

• AB = 4m
• AC = 5m

So,

➟ AC' = 4 + 5

➟ AC' = 9m

$\rule{300}{2}$

Hence, The Total Height of the Tree = 9m