Question

If a ladder is 10 m long and distance between bottom of ladder and wall is 6 m. What is the maximum size of cube that can be placed between the ladder and wall.
a. 34.28
b. 24.28
c. 21.42
d. 28.56; If a ladder is 10 m long and distance between bottom of ladder and wall is 6 m. What is the maximum size of cube that can be placed between the ladder and wall.; a. 34.28; b. 24.28; c. 21.42; d. 28.56

Answer 1

Answer:

3.428 m

Step-by-step explanation:

Ladder is 10 m long

distance between bottom of ladder and wall = 6m

Height of wall where it touches

$=\sqrt{10^2 - 6^2} \\=\sqrt{100 - 36} \\=\sqrt{64} \\= 8 m$

To have the largest cube placed - Ref Picture attached

let say point D on ladder such that D is equidistant from wall and

bottom DE = Distance from wall & DF = Distance from bottom

Let call Δ ABC  - where A is the bottom of Ladder , B is bottom of wall

& C is top of Ladder & wall

D is the point  on Ladded & DE & DF perpendicular on Wall & Ground

AB = 6 m , AC = 10 m  BC = 8 m

Let say D & E is the point  x m above the ground

DE = DF = BE = BF = x m

CE = BC - BE = 8 - x m

AF = AB - BF = 6 -x  m

AD² = AF² + DF²

AD² = (6-x)² + x²

AD² = 36 + x² - 12x + x²

AD² = 2x² -12x + 36

AD = √(2x² -12x + 36)

Similarly

CD² = DE² + CE²

CD² = x² + (8-x)²

CD² = x² + x² + 64 - 16x

CD² = 2x² -16x + 64

CD = √(2x² -16x + 64)

AD + CD = AC = 10 m

√(2x² -12x + 36) + √(2x² -16x + 64) = 10

Squaring both sides

2x² -12x + 36 + 2x² -16x + 64 + 2√(2x² -12x + 36)√(2x² -16x + 64) = 100

=> 4x²-28x + 100 + 4 √{(x² -6x + 18)(x² -8x + 32)} = 100

=> x² - 7x = - √{(x² -6x + 18)(x² -8x + 32)}

Squaring both sides again

x^4 + 49x² -14x³ = x^4 - 14x^3 + 98x² - 336x + 576

=>  0 = 49x² - 336x + 576

=> 49x² - 336x + 576 = 0

=> (7x - 24)² = 0

=> x = 24/7 m

=> x = 3.428 m

Volume of cube = (24/7)³ = 40.3 m³