This is an open question so there are no answers in the textbook, is there any methodical solution that is not trial and error? Please keep it simple!

Question

An engineer designs a slide for a playground. As shown in the diagram, the slide AB is 20 cm long. Angle ACB is 90°. The gradient of the slide is required to be greater than 1/2 but less than 1. Find two possible sets of measurements for the lengths of AC and BC.
Hint: Gradient of AB = BC/AC

    B/|

    /  |

  /    |

A      C

(I got AC = 16 and BC = 12 cm, but that was from trial and error. I cannot find another set that has whole numbers as an answer. All the other values had too many decimal places.

and I hope the diagram is understandable)

Answer 1

There are many pairs of numbers that you can use.

A starting point is to choose any slope that is between 1/2 and 1.

Suppose we use 2/3, then BC = 2x ,  AC = 3x,  AB = 20

Use the Pythagorean theorem to find x

In this case, 4 x^2 + 9 x^2 = 400

13 x^2 = 400

x^2 = 400/13          x = (20√13)/13

This means that AC = (60√13)/13 =16.64  and BC = (40√13)/13 = 11.09

 

Choosing another fraction between 1/2 and 1, and following the same method as the one I have used above will produce another pair of values for AC and BC.

 

Answer 2

First, I want to say that your diagram is very nicely drawn and accurate.

 

 

So the slope must be between 1/2 and 1.

 

Slope is vertical distance over horizontal distance.  Therefore, BC must be less than AC.

 

One slope can be 3/4.  The other can be 4/5.  These two slopes are one of many possibilities, but we will use these slopes.

 

 

Because we have a right triangle, the three sides must satisfy the Pythagorean theorem.  Since AB (longest side) is 20 cm, the other to legs (BC and AB) must be less than 20 cm.

 

 

Lets pick some values of BC and AB that might work using the slope and know side length.

 

BC: 3 , 6 , 9 , 12

AC: 4 , 8, 12 , 16

 

If we add the squares of each corresponding pair, the sum must be equal to 20² , which is 400.

 

12² + 16² = 400

 

This pair works and we have our first set. I see that you already have that down.  Nice work.

 

 

Now we need to get the second set, using the slope of 4/5.

 

BC: 4 , 8 , 12

AC: 5, 10 , 15

 

 

This slope does not work out because none of the corresponding pairs if you add their squares, gets you a sum of 400.  Lets try slope 4/6.  It is between 1/2 and 1.

 

BC: 4, 8, 12

AC: 6, 12, 18

 

 

Does not work.  Try 5/6.

 

BC: 5, 10,  15

AC: 6, 12, 18

 

 

Does not work.  Try slope 4/7.

 

BC: 4, 8

AC: 7, 14

 

 

This does not work.

 

Okay now that you have the idea of how this works, try these slopes:

 

5/7 , 6/7 , 5/8 , 7/8 , 5/9 , 8/9 , and   7/9

 

 

 

 

 

Answer 3

THX for the picture !  If there isn’t one, draw it yourself. 

It helps to read and re-read the problem, then put it into your own words.  Often, the problem is really a math problem with other things just made up (for example, playground slide) in order to make it interesting.  Here’s my version:

         What Pythagorean triples (as simple as possible) have 20 as the hypotenuse?

 

How did I get that?  Well, it was all that math while studying Engineering.  You will see some math information many, many times (for example, do you use 3.14 or 22/7 for the value of PI?) because we learn by repetition.

 

So, let me clue you in.  Trial-and-error is not bad, it just takes time and is often unorganized.  Another method, one that is great with computers because they are so very fast, is called “exhaustive enumeration.”  That means to systematically try every possibility and then test to see if the conditions are met.

 

So, when learning about the Pythagorean Theorem, you will often see:

             Sides                                 Angles

          1n, 1n, n√2                        45-45-90

          1n, n√3, 2n                           30-60-90

          3n, 4n, 5n                        (learn later)

 

The values 12, 16, 20 (with n=4) are the third triangle just listed.  You found one!  Now, notice that the first triangle listed will have a gradient of 1 (from 1n/1n), so you can’t use it.  But, the second triangle (10, 10√3, 20; with n=10 and gradient 1/√3) works.