My Math Forum Pythagorean Theorem

 Geometry Geometry Math Forum

 March 7th, 2016, 03:32 PM #1 Newbie   Joined: Mar 2016 From: new york Posts: 1 Thanks: 0 Pythagorean Theorem To get from one corner of a rectangular court to the diagonally opposite corner by walking along two sides, a distance of 160 meters must be covered. By going diagonally across the court, 40 meters are saved. Find the dimensions of the court, to the nearest cm. I'm using the pythagorean theorem to solve but my answers don't make sense Last edited by sidra76; March 7th, 2016 at 03:37 PM.
 March 7th, 2016, 03:45 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,834 Thanks: 650 Math Focus: Yet to find out. You might find it much easier to solve if you draw a diagram of the situation! Have you tried doing this? Also, what part of your answer isn't making sense exactly? Thanks from skeeter and topsquark
March 7th, 2016, 04:11 PM   #3
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
 Originally Posted by sidra76 To get from one corner of a rectangular court to the diagonally opposite corner by walking along two sides, a distance of 160 meters must be covered. By going diagonally across the court, 40 meters are saved. Find the dimensions of the court, to the nearest cm. I'm using the pythagorean theorem to solve but my answers don't make sense
Then show us what you did and what answers you got so we can point out any possible errors.

March 7th, 2016, 06:19 PM   #4
Senior Member

Joined: Jul 2014
From: भारत

Posts: 1,178
Thanks: 230

Quote:
 Originally Posted by sidra76 To get from one corner of a rectangular court to the diagonally opposite corner by walking along two sides, a distance of 160 meters must be covered. By going diagonally across the court, 40 meters are saved. Find the dimensions of the court, to the nearest cm. I'm using the pythagorean theorem to solve but my answers don't make sense
We know that diagonal is 120 metres and sum of other two sides is 160 metres.

Let us name the diagonal as AC and other two sides as AB and BC.

For finding the other two sides, we need two numbers whose sum is 160 and the sum of whose squares is equal to 120² (which is equal to 14400).

By Pythagoras theorem, we have: AB² + BC² = 120² (First Equation)

We need: AB + BC = 160 (Second Equation)

Solving this system of equations will give approximate values for AB and BC (in meters).

Note: The question asks you to find the dimensions to the nearest centimetres, so you might prefer to convert all given dimensions to centimetres and then solve the system of equations rather that finding in metres and then forgetting to convert .

Last edited by Prakhar; March 7th, 2016 at 06:25 PM.

March 7th, 2016, 06:30 PM   #5
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1588

Quote:
 Originally Posted by Prakhar Note: The question asks you to find the dimensions to the nearest centimetres, so you might prefer to convert all given dimensions to centimetres and then solve the system of equations rather that finding in metres and then forgetting to convert
Rounding the dimensions in meters to two decimal places would be to the nearest centimeter.

March 7th, 2016, 06:33 PM   #6
Senior Member

Joined: Jul 2014
From: भारत

Posts: 1,178
Thanks: 230

Quote:
 Originally Posted by skeeter Rounding the dimensions in meters to two decimal places would be to the nearest centimeter.
I know, but I said that because many people often forget to convert units later. You said "rounding", but it should be "multiplying".

Last edited by Prakhar; March 7th, 2016 at 06:38 PM.

 March 7th, 2016, 06:38 PM #7 Senior Member   Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus Since it would be a triangular figure, here base + height = 160 and hypotenuse = 120, so b + h = 160 hy = 120 we know that hy^2 = b^2 + h^2 14400 = (h - 160)^2 + h^2 14400 = 2h^2 - 320h + 25600 h^2 - 160h + 5600 = 0 h~109; and h~51 Last edited by skipjack; October 26th, 2016 at 01:46 PM.
 March 8th, 2016, 03:44 AM #8 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 No, rounding the solutions to two decimal places after performing the required calculations will suffice ... $x^2+(160-x)^2=120^2$ solving the quadratic for x yields ... $x=20(4 \pm \sqrt{2})$ choosing $x=20(4+\sqrt{2}) \approx 108.28 \, m$ $160-x=20(4-\sqrt{2}) \approx 51.72 \, m$ both solutions have been rounded to the nearest cm.
 March 9th, 2016, 05:46 PM #9 Senior Member   Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230 OK, the question states "Find the dimensions of the court, to the nearest cm." It doesn't state that the answer has to be given in centimetres rather than metres. This is what you were pointing out.
 March 9th, 2016, 07:16 PM #10 Newbie   Joined: Mar 2016 From: San Carlos City, Negros Occidental, Philippines Posts: 1 Thanks: 0 I suggest that you draw the problem so you can understand and know how you will approach the problem

 Tags pythagorean, theorem