Understanding Geometry | MathHelp.com - By MathHelp.com
| 00:0-1 | to find the value of X . In this example | |
| 00:03 | , our first step is to create a right triangle | |
| 00:07 | by drawing the following perpendicular segment . Now we can | |
| 00:19 | see that X . Is the hype a tennis of | |
| 00:21 | the right triangle that is formed . But we need | |
| 00:25 | to find the lengths of the legs of the right | |
| 00:27 | triangle to find the lengths of the legs notice that | |
| 00:31 | a rectangle is formed on the left side of the | |
| 00:34 | figure , and since opposite sides of a rectangle are | |
| 00:39 | congruent , We know that one of the legs of | |
| 00:42 | our right triangle has a length of 12 to find | |
| 00:47 | the length of the other leg of the right triangle | |
| 00:50 | . Since opposite sides of a rectangle are congruent . | |
| 00:54 | We know that the left part of the segment at | |
| 00:57 | the top of the figure has a length of 14 | |
| 01:01 | . And since the full segment at the top of | |
| 01:04 | the figure has a length of 19 , then the | |
| 01:07 | right part of the segment at the top of the | |
| 01:10 | figure has a length of 19-14 or five . Now | |
| 01:17 | we have the lengths of both legs of the right | |
| 01:20 | triangle , so we can use the Pythagorean theorem to | |
| 01:24 | find the value of X . The pythagorean theorem states | |
| 01:30 | that the sum of the squares of the lengths of | |
| 01:33 | the legs of a right triangle is equal to the | |
| 01:36 | square of the length of the hypotenuse , or a | |
| 01:40 | squared plus B squared equals c squared . So we | |
| 01:48 | have 12 squared plus five squared equals X squared , | |
| 01:58 | solving from here , 12 squared is 144 and five | |
| 02:05 | squared is 25 so we have 144 plus 25 equals | |
| 02:13 | X squared , Or 169 equals x squared . Finally | |
| 02:21 | , we take the square root of both sides . | |
| 02:25 | Yeah , To get 13 equals x right . |
DESCRIPTION:
This lesson covers wind and current problems. Students learn to solve wind and current word problems using a system of linear equations, as demonstrated in the following problem. Into a headwind, the plane flew 2000 miles in 5 hours. With a tailwind, the return trip took 4 hours. Find the speed of the plane in still air and the speed of the wind. The two variables used in this problem are p, the speed of the plane in still air, and w, the speed of the wind. So the speed of the plane with a tailwind can be represented as p + w, and the speed of the plane against a headwind can be represented as p -- w. Note that this problem requires a chart to organize the information. The chart is based on the formula rate times time = distance. The chart is then used to set up the two equations.
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