Question

A 10m long ladder reaches a window 8m high from the ground on placing it against a wall at a distance x.Find the distance 'x'.​

Answer 1

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Let AC represent the ladder and AB represent the distance of window from the ground and BC is the required distance.

Here, ∆ ABC is a right angled triangle where ABC = 90°

AC = 10 m

AB = 8 m

BC = x m

Using Pythagoras theorem,

(AC)² = (AB)² + (BC)²

(10)² = (8)² + (x)²

x² = (10)²- (8)²

x² = 100 - 64

x² = 36

x= 6 m

Ans --> The distance 'x' is 6m.

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Answer 2

$\rule{400}4$

☆ ANSWER:-

▪︎ Given:-

• Length of ladder = 10m.

• Height of window from ground = 8m.

▪︎ To Find:-

• The value of x.

$\rule{400}2$

▪︎ Now Here,

• We can apply pythagoras theorem as the angle made by ground and the wall is 90°.

$\sf \mapsto(AC) {}^{2} = (AB) {}^{2} + (BC) {}^{2} \\ \\ \sf \mapsto(10m) {}^{2} = (8m) {}^{2} + (x) {}^{2} \\ \\ \sf \mapsto100m {}^{2} = 64m {}^{2} + {x}^{2} \\ \\ \sf \mapsto {x}^{2} = 100m {}^{2} - 64m {}^{2} \\ \\ \sf \mapsto {x}^{2} = 36m {}^{2} \\ \\ \sf \mapsto \: x = \sqrt{36m {}^{2} } \\ \\ \sf \mapsto \: x = \sqrt{6m \times 6m} \\ \\ \sf \large\red{\fbox{\mapsto \: x = 6m}}$

$\rule{400}4$

${\small{\tt{\green{\boxed{\bold{\therefore{The\:Distance\:x\:=\:6\:meters.}}}}}}}$

$\rule{400}4$