Question

A tree is broken at a height of 8 m from the ground and its top touches the ground at a distance of 15m from the base of the tree.Find the original height of the tree.

Answer 1

Tree is perpendicular to the ground =8
distance from ground to base of tree= 15cm

on applying pythagoras theorem, we can find hupotenuse

H^2= P^2+B^2
H^2= (15)^2+ (8)^2
H^2= 225+64
H^2= 289
H= plus minus 17

height of tree= 17+8=25

Hope this will help u... :-)

Answer 2

Given :

A tree is broken at a height of 8 m from the ground and its top touches the ground at a distance of 15m from the base of the tree .

To Find :

Original Height of the tree .

Solution :

$\longmapsto\tt{AB=8\:m}$

$\longmapsto\tt{BC=15\:m}$

Using Pythagoras Theorem :

$\longmapsto\tt\bf{{(AC)}^{2}={(AB)}^{2}+{(BC)}^{2}}$

$\longmapsto\tt{{(AC)}^{2}={(8)}^{2}+{(15)}^{2}}$

$\longmapsto\tt{{(AC)}^{2}=64+225}$

$\longmapsto\tt{{(AC)}^{2}=289}$

$\longmapsto\tt{AC=\sqrt{289}}$

$\longmapsto\tt\bf{AC=17\:m}$

Now ,

Original Height of Tree :

$\longmapsto\tt{8+17}$

$\longmapsto\tt\bf{25\:m}$

So , The Original Height of the tree is 25 m .✅