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,
∴ PQ2 + `square` = PR2 .....(I)
Here, PR = 10, PQ = `square`
From equation (I)
82 + QR2 = 102
QR2 = 102 – 82
QR2 = 100 – 64
QR2 = `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,
∴ PQ2 + QR2 = PR2 .....(I)
Here, PR = 10, PQ = 8
From equation (I)
82 + QR2 = 102
QR2 = 102 – 82
QR2 = 100 – 64
QR2 = 36
QR = 6
∴ The distance of foot of the ladder from the base of wall is 6 m.