Question

A 15m long ladder reached a window 12m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.​

Answer 1

Answer:

Step-by-step explanation:

Let,

PR be the ladder

PQ bethe wall

Given

PR = 15 m

PQ = 12 m

QR = a = ?

Here, ZQ = 90° as the wall is straight.

In A PQR

PR² = PQ² + QR²

(15)² = (12)² + a²

225=144+ a²

225-144-a²

81=q²

(9)² =a²

Cancelling squares

9=a

a = 9m

foot of the ladder from the wall is 9m.

Answer 2

♣ Given :-

• A 15m long ladder reached a window 12m high from the ground on placing it against a wall at a distance a.

♣ To Find :-

• The distance of the foot of the ladder from the wall. (Base)

♣ Solution :-

• We can apply Pythagoras theorem to find distance of the foot of the ladder from the wall.

$\: \: \: \: \: \large\sf\underline{✯ \: Calculating \: the \: Base :-}</p><p>$

$\: \: \: \: \: \large\sf↦ \: \: (Hypotenuse)² = (Perpendicular)² + (Base)²$

$\: \: \: \: \: \large\sf↦ \: \: (15)² = (12)² + (Base)²$

$\: \: \: \: \: \large\sf↦ \: \: (15)² - (12)² = (Base)²$

$\: \: \: \: \: \large\sf↦ \: \: \sqrt{(15)² - (12)² = Base²}$

$\: \: \: \: \: \large\sf↦ \: \: \sqrt{225 - 144 = Base }$

$\: \: \: \: \: \large\sf↦ \: \: \sqrt{81 = Base }$

$\: \: \: \: \: \large\sf↦ \: \: Base = 9m$

$\large\fbox{Hence, the distance of the foot of the ladder from the wall is 9m.}$

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