Question

7, A pole broke at a point but did not separate. Its top touched the ground at a distance
of 5 m from its base. If the point where it broke is at a height of 12 m from the ground,
what was the total height of the pole before it broke?​

Answer 1

Answer:

$\huge \bold \pink{Given: }$

• A pole broke at a point but did not separate. Its top touched the ground at a distanceof 5 m from its base. If the point where it broke is at a height of 12 m from the ground

$\huge \bold \pink{To \: Find: }$

• total height of the pole before it broke?

$\huge \bold \pink{Solution: }$

Here BD is the original pole.

Let the height of the pole,BD be 'h' m

The top of the pole touches the ground A then,

AB=5m,CD=AC=(h-12)m

Applying Pythagoras theorem in right ∆ ABC

$\sf \: AC {}^{2} =AB {}^{2} +BC { }^{2} \\ \sf \: = > (h - 12 {)}^{2} = {5}^{2} + {12}^{2} \\ \sf \: = > (h - 12 {)}^{2} = 169 = {13}^{2} \\ \sf \: = > h - 12 = 13 \\ \sf \: = > h = 12 + 13 \\ \ \pink{= > \: h = 25}$

Thus, the total height of the pole before it broke is 25m

Answer 2

Answer:

Let the height of the pole,BD be 'h' m

The top of the pole touches the ground A then,

AB=5m,CD=AC=(h-12)m

Applying Pythagoras theorem in right ∆ ABC

\begin{gathered}\sf \: AC {}^{2} =AB {}^{2} +BC { }^{2} \\ \sf \: = > (h - 12 {)}^{2} = {5}^{2} + {12}^{2} \\ \sf \: = > (h - 12 {)}^{2} = 169 = {13}^{2} \\ \sf \: = > h - 12 = 13 \\ \sf \: = > h = 12 + 13 \\ \ \pink{= > \: h = 25}\end{gathered}

AC

2

=AB

2

+BC

2

=>(h−12)

2

=5

2

+12

2

=>(h−12)

2

=169=13

2

=>h−12=13

=>h=12+13

=>h=25