Math, asked by ayu5789, 5 months ago

A ladder 10m long reaches a window 8m above the ground. Find the distance of the foot of the bladder from base of the wall.​

Answers

Answered by Anonymous
28

Given :

  • A ladder 10m long reaches a window 8m above the ground.

To Find :

  • Distance from foot of the ladder to the base of the wall = ?

Solution :

We have ,

  • Length of ladder = 10 m
  • Height of wall = 8 m

By using Phythagoras theorem :

Pythagorean theorem, states that the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle :D

  • (Hypotenuse)² = (Base)² + (Height)²

Here,

  • Hypotenuse = 10 m
  • Base = ?
  • Height = 8 m

Now, plug in the given values in above formula of Phythagoras theorem :

→ (10)² = (Base)² + (8)²

→ 100 = (Base)² + 64

→ 100 - 64 = Base²

→ 36 = Base²

→ Base = √36

Base = 6 m

  • Hence ,the distance from foot the ladder to the base of the wall is 6 m.
Answered by EliteZeal
7

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

 \:\:

\large\underline{\green{\bf Given :-}}

 \:\:

  • A ladder 10m long reaches a window 8m above the ground

 \:\:

\large\underline{\red{\bf To \: Find :-}}

 \:\:

  • The distance of the foot of the ladder from base of the wall

 \:\:

\large\underline{\orange{\bf Solution :-}}

 \:\:

  • Let the distance of the foot of the ladder from base of the wall be "x"

 \:\:

The given case will rise a right angled triangle with its height as the height of wall , its hypotenuse would be the length of ladder and its base will work as the distance of the foot of the ladder from base of the wall.

 \:\:

 \underline{\bold{\texttt{By Phythagoras theorem :}}}

 \:\:

 \footnotesize{ \sf(Hypotenuse)^2  = (Base)^2  + (Height)^2 \: \:  ⚊⚊ ⓵}

 \:\:

 \underline{\bold{\texttt{Applying pythagoras theorem in this case :}}}

 \:\:

  • Hypotenuse = 10 m

  • Base = x

  • Height = 8 m

 \:\:

Putting these values in equation ⓵

 \:\:

➜ (Hypotenuse)² = (Base)² + (Height)²

 \:\:

➜ (10)² = (x)² + (8)²

 \:\:

➜ 100 = (x)² + 64

 \:\:

➜ 100 - 64 = x²

 \:\:

➜ x² = 36

 \:\:

 \sf x = \sqrt { 36 }

 \:\:

  • x = 6
  • x = -6

 \:\:

As length can't be negative hence x = 6

 \:\:

∴ The distance of the foot of the ladder from base of the wall is 6 m

 \:\:

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