SOLUTION: A ladder of length 15 ft is placed against a building in such a way that the distance from the top of the ladder to the ground is 12ft. Find the distance from the bottom of the l

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Question 96: A ladder of length 15 ft is placed against a building in such a way that the distance from the top
of the ladder to the ground is 12ft. Find the distance from the bottom of the ladder to the building.

The length of a rectangular frame is 7 cm more than the width. the area inside the frame is 98 square cm.
Find the width of the frame.

PLEASE HELP !!
Found 4 solutions by terrtwo, KMST, ikleyn, MathTherapy:
Answer by terrtwo(10) About Me  (Show Source):
You can put this solution on YOUR website!
+graph%28+600%2C+400%2C+-4%2C+20%2C+-4%2C+20%2C+-12%2F13x+%2B+12%29+
Since the ladder is leaning against a wall and we are given the height of the ladder, as well as how high is sits while leaning on the wall, we are able to form a right triangle. The original height of the ladder, which is now leaning, is the hypothenus. The height upward on the wall is the height from the ground and the vertical side. The remaining side of the triangle, the distance from the foot of the wall to the foot of the ladder is the distance we are seeking.
Using the Pythagorean theory (x^2 + y^2) = z^2, we realize that z is equal to 15, and that either of the remaining sides (I'll choose x) can be 12.
Thus 12%5E2+%2B+y%5E2+=+15%5E2 --> 144+%2B+y%5E2+=+225.
We subtract 144 from both sides --> 144+%2B+y%5E2+-+144+=+225+-+144+ --> y%5E2+=+169.
We take the square root of both sides --> sqrt%28y%5E2%29+=+sqrt%28169%29 --> y = 13

Answer by KMST(5484) About Me  (Show Source):
You can put this solution on YOUR website!
To find the distance x from the bottom of the ladder to the building wall,
we could apply the Pythagorean theorem to the right triangle in the sketch and solve for x :
x%5E2%2B12%5E2=15%5E2 --> x%5E2%2B144=225 --> x%5E2=225-144 --> x%5E2=81 --> highlight%28x=9%29
We could also realize the the triangle is similar (scaled up by a factor of red%283%29)
to the 3-4-5 right triangle corresponding to the simplest Pythagorean triple.
We have a hypotenuse with a length of red%283%29%2A5=15 and a vertical leg with a length of red%283%29%2A4=12 .
The other leg length, must be red%283%29%2A3=9 .
The bottom of the ladder is at highlight%289feet%29 from the building .
L, W, l, w, and x are measures in cm, L=W%2B7 l=L-2x w=W-2x
l=W%2B7-2x l=%28W-2x%29%2B7 l=w%2B7
The area inside the frame, in square cm, is l%2Aw=98 --> %28w%2B7%29w=98 --> w%5E2%2B7w-98=0 --> w=7
The width of the area inside the frame is highlight%287cm%29 .
There is no way to determine x or W=w%2B2x , which could be called the width of the frame.

Answer by ikleyn(54062) About Me  (Show Source):
You can put this solution on YOUR website!
.
(1) A ladder of length 15 ft is placed against a building in such a way that the distance from the top
of the ladder to the ground is 12ft. Find the distance from the bottom of the ladder to the building.

(2) The length of a rectangular frame is 7 cm more than the width. the area inside the frame is 98 square cm.
Find the width of the frame.

PLEASE HELP !!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


For problem (1), the calculations and the answer in @terrtwo's post are totally wrong.
For this problem, use the solution by tutor @KMST.


For problem (2), the problem is posed incorrectly, so it can not be solved
and the problem's question can not be answered, as tutor @KMST showed/explained in her post.



Answer by MathTherapy(10898) About Me  (Show Source):
You can put this solution on YOUR website!
A ladder of length 15 ft is placed against a building in such a way that the distance from the top 
of the ladder to the ground is 12ft. Find the distance from the bottom of the ladder to the building.

The length of a rectangular frame is 7 cm more than the width. the area inside the frame is 98 square cm. 
Find the width of the frame.

PLEASE HELP !!
**************
TYPICAL PYTHAGOREAN TRIPLE of the form: (3-4-5)x3 = 9-12-15, with the LONGEST side being the hypotenuse!
With 2 sides being 15 (hypotenuse), and 12 (longer leg), 3rd side (shorter leg) is 9 ft