A ladder 10m long is made to rest against a wall. Its lower end touches the ground at a distance of 6m from the wall.At what height above the ground the upper end of the ladder rests against a wall?
2007-09-06 23:51:48 · 8 answers · asked by AhsanAbdul Rehman R 1 in Science & Mathematics ➔ Mathematics
Let the height be 'x'
From the pythagorean theorem,
(10)^2 = x^2 + (6)^2
or, 100 = x^2 + 36
or, x^2 = 100 - 36
or, x^2 = 64
so, x = ±8
But length/distance cannot be negative.
So, the height is 8 meter. [Answer]
2007-09-10 22:16:07 · answer #1 · answered by defeNder 3 · 0⤊ 0⤋
ladder makes a right angle with wall so
(height)^2 = (ladder)^2 -- (distance)^2
=> h^2 = 10^2 -- 6^2 = 100 -- 36 = 64 = 8^2
=> height h = 8m
2007-09-07 00:08:07 · answer #2 · answered by sv 7 · 0⤊ 0⤋
This is a right triangle problem
c² = a² + c²
10² = a² + 6²
a² = 64
a = 8 meters (height)
.
2007-09-06 23:58:41 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
100-36 =64 , take sq rt = 8m
2007-09-07 00:15:38 · answer #4 · answered by Anonymous · 0⤊ 0⤋
10^2-6^2=b^2
b=8
2007-09-07 00:00:42 · answer #5 · answered by jara 3 · 0⤊ 0⤋
use pythagorean theorom..
height=square root of 10square - 6suare..
= 8...
2007-09-07 01:00:59 · answer #6 · answered by Sunny v 2 · 0⤊ 0⤋
answer is 8m
2007-09-07 00:14:44 · answer #7 · answered by bond_007 2 · 0⤊ 0⤋
4m
2007-09-06 23:58:19 · answer #8 · answered by Anonymous · 0⤊ 1⤋