A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.

Activity: As shown in figure suppose

PR is the length of ladder = 10 m

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ^{2} + `square` = PR^{2} .....(I)

Here, PR = 10, PQ = `square`

From equation (I)

8^{2} + QR^{2} = 10^{2}

QR^{2} = 10^{2} – 8^{2}

QR^{2} = 100 – 64

QR^{2} = `square`

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.

#### Solution

As shown in figure suppose PR is the length of ladder = 10 m,

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ^{2} + **QR ^{2}** = PR

^{2}.....(I)

Here, PR = 10, PQ = **8**

From equation (I)

8^{2} + QR^{2} = 10^{2}

QR^{2} = 10^{2} – 8^{2}

QR^{2} = 100 – 64

QR^{2} = **36**

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.