I don't understand this question?

Question

A wall of a building is to be braced by a beam that must rest on the ground and pass over a vertical wall 10 ft high that is 8 ft from the building. Find the lengh L of the shortest beam that can be used.

Answer 1

I think you might mean this scenario:
(Click to view)
http://www.mudandmuck.com/str2/buildingWall.JPG

If so, take the derivative of L with respect to x and set equal to zero. This will determine x that minimizes L. Then, substitute this value of x into the equation for L to find the answer.
I will leave it to you to derive that:
dL/dx = (x^3 - 800) / [x^2√(x^2+100)]
so, x^3 = 800 or x = 9.283...
and L = 25.4033...

(This is shorter than 2√(10^2+8^2) = 2√164 = 25.612... ,
and also √(10^2+10^2)+√(8^2+8^2) = √200+√128 = 25.455.. , as where some other answers seem to be leading.)

Answer 2

Well, suppose x is the distance from the 8' wall to the bottom of the beam. The angle A from the ground to the top of that wall is:

A = atan(8/1)

This of course will be the same angle from the ground to the wall being supported. The total distance from the supported wall to the beam is x+10. So the length of the beam L, being the hypotenuse of our wall-ground-beam triangle, is given by:

L = (x+10)/cos(A)
or
L = (x+10)/cos(atan(8/x))

Draw a digram and you see the shortest length will be found if x = 10, in other words if the beam is placed the same distance from the 8' wall as that wall is from the building.

Answer 3

Think of it as a right triangle with a rectangle inside. The building is the right side of the square and the triangle. The 10 ft wall is the left side of the square. The beam is the hypotenuse. The bottom of the square = 8 ft. What is the hypotenuse of smallest right triangle that can contain the entire 10 x 8 rectangle? hint: the whole thing can be represented by four 10 x 8 triangles.

Answer 4

There does appear to be 1 piece of information missing. I'm thinking that the shortest hypotenuse will be in a triangle with the other 2 sides equal. If this is correct, then we have 2 triangles, one with 2 sides of 10' and a hypotenuse and the other with 2 8' sides. Work out the hypotenuse for both triangles and add them together. Do you need to make an allowance for the width of the wall? Have you been told whether the brace rests on the wall or clears it by a specified distance?

Good luck.

Answer 5

Its a right triangle with the height of the wall as one leg and the 8 ft distance from the wall is another leg. The brace goes from the top of the wall to the point 8 ft away from the wall. Hope this helps.

Answer 6

Use the Pythagorean Theorum. a^2 + b^2 = c^2
so 10^2 + 8^2 = c^2
100 + 64 = 164, which is c
take the square root of 164 and you get 12.8.

Draw a picture if you need

Answer 7

You don't understand the question because there is not enough information.

Two walls are mentioned.
We don't know how high the building wall is.
Presumably the beam will attach to the top of this wall.

Answer 8

L = sqr[ 10^2 + 8 ^2 ] = sqr(164) = 12.81
. . . . . . . . this bracing is from ground to top of fence
form shortest length use 45 deg triangle . .
brace up to 8 ft. vertical up from ground . 2 sides of triangle = 8
L = sqr( 8^2 + 8^2) = sqr(128) = sqr(64x2) = 8 sqr(2) = 11.31 ft.

Answer 9

Zaphod's answer is correct, also great sketch.
Alternatively, working with the angle, A, that the beam makes with the horizontal, we have:
10/sinA +8/cosA = L
10 cscA + 8 secA = L
Take the derivative and set it equal to zero:
10(-cscA cotA) +8(secA tanA) = 0
-10(1/sinA)(cosA/sinA) +8(1/cosA)(sinA/cosA) = 0
Multiply both sides by (sinA)^2(cosA)^2:
-10 (cosA)^3 + 8 (sinA)^3 = 0
8 (sinA)^3 = 10 (cosA)^3
(sinA)^3/(cosA)^3 = 10/8
(tanA)^3 = 10/8
tanA = (cube root)1.25
A = 47.1289
L = 10/sinA +8/cosA
L = 25.4033

Answer 10

10' squared + 8'squared...add together the square root of answer is the hypoteneuse....or a2 +b2=c2