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Item 11
A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder is moved so that the base of the ladder travels toward the wall twice the distance that the top of the ladder moves up. How much higher is the top of the ladder now? (Hint: Let 8−2x be the distance from the base of the ladder to the wall.)

Sagot :

oladayoademosu

Answer:

4 ft higher

Step-by-step explanation:

Since the ladder is 10 ft long and its top is 6 feet high(above the ground), we find the distance of its base from the wall since these three (the ladder, wall and ground) form a right angled triangle. Let d be the distance from the wall to the ladder.

So, by Pythagoras' theorem,

10² = 6² + d²  (the length of the ladder is the hypotenuse side)

d² = 10² - 6²

d² = 100 - 36

d² = 64

d = √64

d = 8 ft

Since the ladder is moved so that the base of the ladder travels toward the wall twice the distance that the top of the ladder moves up.

Now, let x be the distance the top of the ladder is moved, the new height of top of the ladder is 6 + x. Since the base moves twice the distance the top of the ladder moves up, the new distance for our base is 8 - 2x(It reduces since it gets closer to the wall).

Now, applying Pythagoras' theorem to the ladder with these new lengths, we have

10² = (6 + x)² + (8 - 2x)²

Expanding the brackets, we have

100 = 36 + 12x + x² + 64 - 32x + 4x²

collecting like terms, we have

100 = 4x² + x² + 12x - 32x + 64 + 36

100 = 5x² - 20x + 100

Subtracting 100 from both sides, we have

100 - 100 = 5x² - 20x + 100 - 100

5x² - 20x = 0

Factorizing, we have

5x(x - 4) = 0

5x = 0 or x - 4 = 0

x = 0 or x = 4

The top of the ladder is thus 4 ft higher

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