Question

answer with a good explanation​

Answer 1

Question:-

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 of 2.5 dm from the ground, what was the total height of tree before it broke?

Given:-

• Its top touched the ground at a distance of 6 dm.

• If the point where it broke be at a height of 2.5 dm from the ground.

To Find:-

• what was the total height of tree before it broke?

Solution:-

• Let, BD be the tree of height h m broken at the point C and

• Let, CD take the position CA as shown in the figure.

From the right - angled triangle ABC,

[(h – 2.5)m]²  = (6m)² + (2.5m)²

[Pythagoras Theorem]

= 36m² + 6.25m² 

= 42.25m².

=> (h – 2.5)m = 6.5m.

=> h = 6.5 + 2.5 = 9m.

Answer:-

$\sf \large \red{Hence, \: the \: original \: height \: of \: the tree was 9m.} \: \\ \sf \large \red{the \: tree \: was \: 9m.}$

Hope you have satisfied. ⚘

Answer 2

Answer:

Question:-

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 of 2.5 dm from the ground, what was the total height of tree before it broke?

Given:-

Its top touched the ground at a distance of 6 dm.

If the point where it broke be at a height of 2.5 dm from the ground.

To Find:-

what was the total height of tree before it broke?

Solution:-

Let, BD be the tree of height h m broken at the point C and

Let, CD take the position CA as shown in the figure.

From the right - angled triangle ABC,

[(h – 2.5)m]² = (6m)² + (2.5m)²

[Pythagoras Theorem]

= 36m² + 6.25m²

= 42.25m².

=> (h – 2.5)m = 6.5m.

=> h = 6.5 + 2.5 = 9m.

Answer:-

\begin{gathered} \sf \large \red{Hence, \: the \: original \: height \: of \: the tree was 9m.} \: \\ \sf \large \red{the \: tree \: was \: 9m.}\end{gathered}

Hence,theoriginalheightofthetreewas9m.

thetreewas9m.

Hope you have satisfied. ⚘