Question

a ladder 6m long leaned against a wall. the ladder reaches the wall to a height of 4.8m.find the distnace between the wall and the foot of the ladder

Answer 1

By using Pythagoras method

h²+b²=p²
(4.8)²+b²=6²
b²=6²-(4.8)²
b²=36-23.04
b²=12.96
b=√12.96
b=3.6



so,the distance between wall and ladder=3.6m

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Distance\:between\:wall\:and\:foot\:of\:ladder=3.6\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about the top of the ladder 6 long when placed against a wall reaches a height of 4.8 metres.

• We have to find how far is the foot of the ladder from the wall?

$\green{\underline \bold{Given :}} \\ : \implies \text{Length\:of\:ladder=6\:m} \\ \\ : \implies \text{Height\:of\:wall=4.8\:m}\\\\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Distance\:between\:wall\:and\:foot\:of\:wall= ?}$

• Accroding to given question :

$\bold{ By \: pythagoras \: theorem} \\ : \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ : \implies {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \\ : \implies {6}^{2} = {4.8}^{2} + BC^{2} \\ \\ : \implies 36 = 23.04 + {BC}^{2} \\ \\ : \implies 36 - 23.04 = {BC}^{2} \\ \\ : \implies {BC}^{2} = 12.96 \\ \\ : \implies BC= \sqrt{12.96} \\ \\ \green{: \implies \text{BC = 3.6 \: m}}$