Question

6.
A ladder of 6 meter length is laid against a wall. The distance
between the foot of the wall and foot of the ledder is 3 meters
Find the height at which the ladder touches the wall.

Answer 1

Required height is 3√3 m

EXPLANATION:

In the given picture, we have considered AB to be the wall and CD to be the ladder. Also BC is the distance between the foot of the wall and the foot of the ladder, and BD be the height at which the ladder touches the wall.

We have to find BD = ?

We see that, BCD is a right-angled triangle, whose base (BC) = 3 m and the hypotenuse (CD) = 6 m

Using Pythogoras theorem, we get

BD² + BC² = CD²

or, BD² + 3² = 6²

or, BD² = 6² - 3² = 36 - 9 = 27

or, BD² = (3√3)²

∴ BD = 3√3 m

∴ at the height of 3√3 m, the ladder touches the wall.

Answer 2

The height at which the ladder touches the wall is $3\sqrt{3}m$

• Let AB be the ladder and BC be the distance between the foot of the wall and foot of the ladder.
• We have to find the height at which the ladder touches the walls i.e, AC
• From figure, ABC is a right angle triangle
• In a right angle triangle

                    $hyp^{2}=opp^{2}+adj^{2}$ [Pythagorean theorem]

                    $AB^{2}=AC^{2}+BC^{2}$

                    AB = 6 m, BC = 3 m, AC = ?

                    $AC=\sqrt{AB^{2}-BC^{2}}$

                    $AC=\sqrt{6^{2}-3^{2}}$

                    $AC=\sqrt{36-9}}$

                    $AC=\sqrt{27}}$

                     $AC=\sqrt{9*3}}$

                     $AC=3\sqrt{3}m$

∴ The height at which the ladder touches the wall is $3\sqrt{3}m$