Question

Aa ladder 13 metre long reaches which is 5 metre above the ground on one side of the street Keeping It foot at the same point the ladder is turned to the other side of the street to reach a window at a height of 12 metre find the width of the street Please send me answer ​

Answer 1

Answer:

use pythogras theorum to calculate bases of both

Answer 2

${\underline{\large {\sf{✠ \: Diagram \: ✠}}}}$

${ \underline{\large{ \sf{✠ \: Solution \: ✠}}}}$

⇒Let PQ be the street and R be the foot of the ladder.

⇒Let S an T be the windows at the height of 5m and 12m respectively from the ground.

⇒Then, RS and RT are the two positions of the ladder.

• ⇛Clearly, PS = 5m, QT = 12m, RT = RS = 13m.

From right ∆SPR, we have

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: P {R}^{2} + P {S}^{2} = R {S}^{2} \\ \\ \implies \: \: \: \: \ \sf \: P{R}^{2} = R {S}^{2} - P {S}^{2} \\ \\ \implies \: \: \: \: \sf \: P {R}^{2} = ({13}^{2} - {5}^{2}) {m}^{2} \\ \\ \: \: \: \implies \: \: \: \: \: \: \sf \: P {R}^{2} = (169 - 25) {m}^{2} \\ \\ \implies \: \: \: \: \: \: \sf \: P {R}^{2} = {144} {m}^{2} \: \: \: \: \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \sf P {R}^{2} = ( {12) }^{2} {m}^{2} \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \sf \: P {R} = 12m \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

From right ∆RQT,

$\\ \: \sf \: RQ²+ QT²= RT² \\ \\ \implies \: \: \: \: \: \: \sf \: RQ² = RT² - QT² \: \: \: \: \\ \\ \implies \: \: \: \: \: \sf \: RQ² = ( {13}^{2} - {12}^{2} ) {m}^{2} \\ \\ \implies \: \: \: \: \: \: \sf \: RQ² = (169 - 144) {m}^{2} \\ \\ \implies \: \: \: \: \: \: \: \sf \: RQ² = 25 {m}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \: \: \: \: \: \: \sf \: RQ² = ( {5})^{2} {m}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \: \: \: \: \: \: \: \: \: \sf \: RQ = 5m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\\ \\ {\large{ \ { \boxed{ \underline {\pink{ \rm{ \therefore{width \: of \: the \: street= PQ=PR+RQ=(12+5)m=17m}}}}}}}}$