WEBVTT
00:00:01.780 --> 00:00:17.490
Quadrilateral πππ
π has vertices π at negative one, six; π at five, six; π
at five, three; and π at negative one, three.
00:00:18.550 --> 00:00:20.620
Find the length of ππ.
00:00:21.920 --> 00:00:30.620
ππ cuts our quadrilateral into two triangles, and we can actually use either of those triangles to find the length of ππ.
00:00:32.540 --> 00:00:36.460
Letβs go ahead and choose triangle ππ
π.
00:00:37.600 --> 00:00:41.640
And at angle π
, thatβs a 90-degree angle.
00:00:42.320 --> 00:00:43.210
How do we know that?
00:00:44.660 --> 00:00:50.000
Well, segment ππ
is vertical because our π₯-coordinate doesnβt change.
00:00:50.060 --> 00:00:56.130
So it just goes directly up from three to six and stays at five for π₯.
00:00:57.290 --> 00:01:03.470
ππ
is horizontal because from π to π
, we donβt rise at all.
00:01:04.150 --> 00:01:10.270
We stay at three and we go from negative one to five for our π₯.
00:01:10.670 --> 00:01:11.980
So it runs left and right.
00:01:12.480 --> 00:01:14.160
That means these are perpendicular.
00:01:14.960 --> 00:01:18.150
Therefore, angle π
is a right angle.
00:01:19.040 --> 00:01:28.840
And since we have a right angle at π
, that means triangle ππ
π is a right triangle and we can use the Pythagorean theorem.
00:01:30.200 --> 00:01:35.900
The Pythagorean theorem is useful because it will allow us to find ππ.
00:01:37.290 --> 00:01:50.400
Because the Pythagorean theorem states that π squared plus π squared equals π squared, where π and π are the two shorter sides and π is the longest side β the side across from the 90-degree angle.
00:01:52.270 --> 00:02:00.490
And because we have this diagram, we actually know the side lengths of π and π because we can count them.
00:02:01.990 --> 00:02:05.800
From π
to π, we went up three spaces.
00:02:06.630 --> 00:02:07.740
So π is three.
00:02:08.790 --> 00:02:14.500
You can also look at our points: π
is at five, three and π is at five, six.
00:02:15.370 --> 00:02:18.290
So π₯ didnβt change at all, but our π¦ is changed.
00:02:18.320 --> 00:02:22.340
And it changed from three to six; thatβs a difference of three.
00:02:23.570 --> 00:02:30.620
Now looking at ππ
, from π to π
, we can count the spaces and find that itβs six.
00:02:31.720 --> 00:02:40.470
Or looking at our points, we stayed at three for π¦, but we changed for π₯ β from negative one to five.
00:02:40.690 --> 00:02:43.290
And the difference between these numbers is six.
00:02:44.190 --> 00:02:47.510
So letβs go ahead and fill in this information into our Pythagorean theorem.
00:02:48.890 --> 00:02:56.870
So we have three squared plus six squared equals π squared, and π will be the side ππ that we wanna find.
00:02:57.980 --> 00:03:03.520
So three squared is nine, six squared is 36.
00:03:04.040 --> 00:03:06.220
Bring down your equals π squared.
00:03:07.240 --> 00:03:10.210
Now, we need to add nine plus 36.
00:03:11.570 --> 00:03:14.480
So 45 equals π squared.
00:03:15.830 --> 00:03:18.990
In order to solve for π, we need to get rid of the squared.
00:03:19.740 --> 00:03:24.760
The inverse operation of squaring a number would be to square root a number.
00:03:25.200 --> 00:03:26.940
So letβs square root both sides.
00:03:28.260 --> 00:03:33.260
So π is equal to the square root of 45; however, this can be simplified.
00:03:34.740 --> 00:03:37.520
45 is five times nine.
00:03:38.900 --> 00:03:42.060
The reason weβre doing this is because nine is a perfect square.
00:03:42.850 --> 00:03:47.880
So you can actually take that out of the square root and the square root of nine is three.
00:03:48.980 --> 00:03:51.050
Three times three equals nine.
00:03:51.900 --> 00:04:00.380
So three will come outside of the square root and five will be left inside of the square root, also known as the radical sign.
00:04:01.910 --> 00:04:05.330
So π is equal to three square root five.
00:04:06.420 --> 00:04:11.230
Therefore, the length of ππ is three square root five.
00:04:12.910 --> 00:04:16.830
ππ equals three square root five.