If a 650-cm ladder is placed against a building at a certain angle, it just reaches a point on the building th?

Question

If a 650-cm ladder is placed against a building at a certain angle, it just reaches a point on the building that is 520 cm above the ground. if the ladder is moved to reach a point 80 cm higher up, how much closer will the foot of the ladder be to the building?

Answer 1

650^2 = x^2+ (520)^2..........................(i)

422500 = x^2 +270400

x^2 = 422500 - 270400= 152100

x = 390....................................(ii)

650^2 = (x- h)^2 + (600)^2...................(iii)

From (ii) from (i)

650^2= (y^2)^2 +600^2

y = 250................................(iv)

Distance moved = 390-250 = 140 cm................Ans

Answer 2

it is right angle triangle so ACCORDING TO PYTHAGORAS THEOREM

let distance of ladder from the building = x

then x^2 = (650)^2-- ( 520)^2

x = 390 cm

now when ladder is moved 80 cm up ward then height = 520 +80 = 600

& distance of building from ladder = y

then y^2 = (650)^2 -- (600)^2 = 62500

y = 250 cm

SO THE FOOT OF LADDER WILL BE 250 cm FROM THE BUILDING

so 390 -250 = 140 cm the foot of ladder will be 140 cm closer to the building.ans

Answer 3

closer by sqrt [ 650^2 -- 520^2] cm -- [650^2 -- 600^2] cm = 390 cm -- 250 cm = 140 cm

Answer 4

650² = 520² + a²

650² = 600² + b²



b - a = √(650²-520²) - √(650²-600²) = 140 cm

Answer 5

650^2 - 520^2 = 422500 - 270400 = 152100



\/152100 = 390cm



At first the ladder is 390cm from the wall



650^2 - 600^2 = 422500 - 360000= 62500



\/62500 = 250cm



ANSWER = 390cm -250cm = 140cm

Answer 6

pythagorean theorum. (a2+b2=c2)