Question

960°
Q7 A tree is broken at a height of 5 m from the ground and its top touches the
ground at a distance of 12 m from the base of the tree. Find the original height of the
tee.​

Answer 1

Step-by-step explanation:

$\sf \underline{Given} :$

• A tree is broken at a height of 5 m from the ground.

• its top touches the ground at a distance of 12 m from the base of the tree.

$\sf \underline{To \: Find} :$

• Find the original height of the tree.

$\sf \underline{Solution \: } :$

Diagram :

$\setlength{\unitlength}{20mm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.9,0.8){\large\sf{B}}\put(10.4,0.8){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.5,1.9){\sf{\large{5 m}}}\put(9,0.7){\sf{\large{12 m}}}\put(9.8,2.2){\sf{\large{ ?}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\multiput(8.1,2.8)(.33,-.25){8}{\line(2,1){.3}}\qbezier(11,1)(10.5,1.4)(8.35,3)\end{picture}$

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.

Let Ab = 12m , BC = 5m

Using Pythagoras theory :

$\boxed{ \sf \: {ab}^{2} \: + {bc}^{2} + {ac}^{2} }$

Substitute all values :

$\sf \leadsto \: {ac}^{2} = {12}^{2} + {5}^{2} \\ \\ \sf \leadsto \: {ac}^{2} = 144 + 25 \\ \\ \sf \leadsto \: {ac}^{2} = 169 \\ \\ \sf \leadsto \: ac = \sqrt{169} \\ \\ \sf \leadsto \: ac = 13$

Hence the total height of the tree :

$\sf \leadsto \: ac \: + bc \\ \\ \sf \leadsto \: 13 + 5 \\ \\ \ \sf \leadsto \underline{18 \: m}$

Answer 2

Answer:

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.

AB=12m and BC=5m

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=13m

Hence, the total height of the tree=AC+CB=13+5=18m.