if an 11 ft ladder is placed on a house, and the ladder is 3 ft away from the house, how high is the ladder?

Question

math homework help!

Answer 1

[10]
The ladder forms a right angled trianle with the wall of the house
Let,the ladder has been placed at a height of h ft on the wall
Therefore,
h^2=11^2-3^2
=121-9
=112
h=sqrt112
=10.583 ft
The ladder has been placed at a height of 10.583 height from the ground level

Answer 2

It's a right triangle. And in a right triangle, A squared + B square = C squared (where C is the long side).

Setting up the equation:
(3 * 3) + (B * B) = (11 * 11)
9 + B sqaured = 121
-9 from each side
B squared = 112

The square root of 112 can be approximated. 10 squared = 100; 11 squared = 121.

Answer 3

x ² = 11 ² - 3 ²
x ² = (11 - 3)(11 + 3)
x ² = 8 x 14
x ² = 112
x = √ 112
x = 4 √7
Height of ladder = 10.6 ft

Answer 4

............11../|
................/ | a
.............../_|
................3

forms a right triangle

pythagorean theorem
let c = 11, the hypotenuse
a = height of the diagonally placed ladder
b = 3, the distance from the house

c^2 = a^2 + b^2
11^2 = a^2 + 3^2
121 = a^2 + 9
121 - 9 = a^2
112 = a^2
sqrt(112) = a
10.5830 ft = a --> the height of the diagonally placed ladder

Answer 5

11^2 plus 3^2

Answer 6

lol, Pythagorean Theorem:
A^2+B^2=C^2 in right triangles where C is the hypotenuse.

therefore, in your problem
(11^2)-(3^2)=x^2
121-9
114=x^2
Sqrt(114)= how far off the ground he tip of the ladder is.

......DUH, do your own work next time

Answer 7

3^2 + x^2 = 11^2

solve

Answer 8

still 11 ft, that was easy

Answer 9

square root of 112

Answer 10

sqrt(11^2-9)=sqrt112=4sqrt7=10.58ft