Math, asked by YashneilSahabji, 7 months ago

A 13m LONG LADDER AB IS ADJUSTED AGAINST A WALL SO THAT ITS TOP PART TOUCHES THE WALL AT A HEIGHT AC OF 12m. FIND THE VALUE OF THE DISTANCE BC OF THE FOOT OF THE LADDER FROM THE WALL ON THE GROUND.

Answers

Answered by Anonymous
1

ANSWER

Let BC be the ladder which forms a right angled triangle with the wall.

The distance of the foot of the ladder and the wall, makes AB the base and the distance of the other end of the ladder from the ground, Ac makes the height of the wall.

By Pythagoras theorem, we have 13

2

=5

2

+AC

2

=>AC

2

=169−25=144

=>AC=12 m

So, distance of the other end of the ladder from the ground=12 m.

solution

hope it helps you

thanks

Answered by av1266108
1

Answer:

let \: the \: ladder \: be \:  \\ ab = 13m \:  \\  \\ let \: the \: vertex \: height \: height \:  \\ reach \: be \:  = 12m \:  \\  \\ let \: the \: distance \: between \: the \: ladder \:  \\ and \: the \: wall \: be \:  = x \\  \\ by \: pythagoras \: theorem \\  {ab}^{2}  =  {ac}^{2}   +  {bc}^{2}  \\  \\  {13}^{2}   =   {12}^{2}   +   {x}^{2}  \\  \\ 169 - 144 \\  \\  {x}^{2}  = 25 \\  \\ x = 5 \\  \\

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