Question

3. The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground, (a) Find the length of the ladder. (b) If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach

Answer 1

GIVEN:

• The length of AB = 6 cm
• Length of BC = 8 cm

TO FIND:

• What is the length of the ladder and if its foot is 8 m away from the wall, what height does its top reach ?

SOLUTION:

Triangle ABC is a right angled triangle

To find the length AC, we apply Pythagoras theorem:-

❰ (H)² = (P)² + (B)² ❱

According to question:-

➜ (AC)² = (BC)² + (AB)²

➜ x² = (8)² + (6)²

➜ x² = 64 + 36

➜ x² = 100

➜ x = $\sf{\sqrt{100}}$

❛ x = 10 cm ❜

✍ The length of ladder is 10 cm

✎ If the foot of the ladder is 8 m away from the wall, to what height does its top reach

• AC = 10 cm
• AB = 8 cm
• BC = y cm

Again applying Pythagoras theorem:-

➜ (AC)² = (BC)² + (AB)²

➜ (10)² = (y)² + (8)²

➜ 100 = y² + 64

➜ 100 –64 = y²

➜ 36 = y²

➜ $\sf{\sqrt{36}}$ = y

❛ 6 cm = y ❜

❝ Hence, the length of ladder is 10 cm and if the foot of the ladder is 8 m away from the wall, it's height is 6 cm ❞

______________________

Answer 2

$\sf\huge\underline{Given:-}$

• Length of AB= 6 m
• Lengty of BC= 8 m

$\sf\huge\underline{To \: Find:-}$

• The length of the ladder AC.
• The height of the wall if the foot of the ladder is 8 m away from the wall BC.

$\sf\huge\underline{Solution:-}$

a)

✪ABC is a right- angled triangle .

■ Refer to the first picture.

So,we need to apply Pythagoras Theorem.

$\sf{ {Hypotenous}^{2} = {Perpendicular}^{2} + {Base}^{2} }$

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

$\sf{\implies {AC}^{2} = {8}^{2} + {6}^{2} }$

$\sf{\implies {AC}^{2} = 64 + 36}$

$\sf{\implies {AC}^{2} = 100}$

$\sf{\implies AC = \sqrt{100} }$

$\sf{\implies AC = 10 \: m }$

Hence,the length of the ladder is 10 m.

$\rule{300}{2}$

b)

Length of AB= 8 m

Length of AC= 10 m

ABC is a right-angled triangle.

■Refer to the second picture.

Again,we need to apply Pythagoras Theorem,

$\sf{ {Hypotenous}^{2} = {Perpendicular}^{2} + {Base}^{2} }$

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

$\sf{\implies {10}^{2} = {BC}^{2} + {8}^{2} }$

$\sf{\implies {BC}^{2} = {10}^{2} - {8}^{2} }$

$\sf{\implies {BC}^{2} = 100 - 64}$

$\sf{\implies {BC}^{2} = 36}$

$\sf{\implies BC = \sqrt{36} }$

$\sf{\implies BC = 6 \: m}$

The height of the top= 6 m

$\rule{300}{2}$

Therefore the length of the ladder is 10 m and the height of the wall is 6 m if the foot of the wall is 8 m away from the wall.

$\rule{300}{2}$