Question

A 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. Find the height of the wall where the top of the ladder touches it?



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Answer 1

Step-by-step explanation:

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Answer 2

⇢Answer:

$\sf \underline{ \therefore \: height \: of \: wall = 6m}$

⇢Given:

• Length of ladder = 6.5m

• Length of foot of wall = 2.5m

⇢To find:

• Height of the wall

⇢ATQ

• Given the length of ladder and foot of wall so Now we have to use Pythagoras theorem to find the height

⇢Solution:

As we know that Pythagoras theorem formula:

From the figure ( Refer to Attachment)

• AB is foot of wall (2.5)m

• BC is height (h)

• AC is ladder (6.5)m

$\: \: \: \: \: \sf \implies{AB² + BC² = AC²}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \therefore \: BC = h}$

$\: \: \: \: \: \: \sf \implies{(2.5) {}^{2} + {h}^{2} = (6.5) {}^{2} }$

$\: \: \: \: \: \sf \implies{ {h}^{2} = (6.5) {}^{2} - (2.5) {}^{2} }$

$\: \: \: \: \: \sf \implies{ {h}^{2} = 36}$

$\: \: \: \sf \implies{h = \sqrt{36}}$

$\: \: \: \: \sf \implies{h = 6}$

$\sf \underline{ \therefore \: height \: of \: wall = 6m}$