Question 1207590: In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $QR$, and let $Y$ be the foot of the perpendicular from $X$ to side $PR$. If $PQ = 10,$ $QR = 10,$ and $PR = 12,$ then compute the length of $XY$.
Found 3 solutions by Timnewman, Edwin McCravy, greenestamps:
Answer by Timnewman(323)   (Show Source):
You can put this solution on YOUR website! Hello, here is the solution to your mathematics problem. I hope you find this helpful.
==========================
Let's break down the problem step by step:
1. Draw the triangle $PQR$ with the given side lengths: $PQ = 10$, $QR = 10$, and $PR = 12$.
2. Draw the angle bisector of $\angle P$ and label the intersection point with $QR$ as $X$.
3. Draw the perpendicular from $X$ to $PR$ and label the foot of the perpendicular as $Y$.
4. Let's use the angle bisector theorem, which states that the ratio of the lengths of the sides is equal to the ratio of the distances from the angle bisector to the sides. In this case:
$$\frac{PQ}{QR} = \frac{PX}{XQ}$$
Since $PQ = QR = 10$, we get:
$$\frac{10}{10} = \frac{PX}{XQ}$$
$$1 = \frac{PX}{XQ}$$
$$PX = XQ$$
This means that $X$ is the midpoint of $QR$.
1. Since $QR = 10$, we have:
$$XQ = \frac{1}{2}QR = \frac{1}{2}(10) = 5$$
1. Now, let's use the Pythagorean theorem in triangle $PXQ$:
$$PX^2 + XQ^2 = PQ^2$$
$$PX^2 + 5^2 = 10^2$$
$$PX^2 + 25 = 100$$
$$PX^2 = 75$$
$$PX = \sqrt{75} = 5\sqrt{3}$$
1. Now, let's use the Pythagorean theorem in triangle $XYR$:
$$XY^2 + YR^2 = XR^2$$
Since $XR = PX = 5\sqrt{3}$, we get:
$$XY^2 + YR^2 = (5\sqrt{3})^2$$
$$XY^2 + YR^2 = 75$$
1. We can also use the Pythagorean theorem in triangle $PYR$:
$$PY^2 + YR^2 = PR^2$$
$$PY^2 + YR^2 = 12^2$$
$$PY^2 + YR^2 = 144$$
1. Subtract the two equations:
$$XY^2 - PY^2 = -69$$
$$XY^2 = PY^2 - 69$$
1. We know that $PY = PX = 5\sqrt{3}$, so:
$$XY^2 = (5\sqrt{3})^2 - 69$$
$$XY^2 = 75 - 69$$
$$XY^2 = 6$$
$$XY = \sqrt{6}$$
Therefore, the length of $XY$ is $\sqrt{6}$.
Note: This problem is a classic example of using the angle bisector theorem and the Pythagorean theorem to solve a triangle problem.
==========================
If you're satisfied with my solution and interested in further learning, I'm available for a one-on-one online session. I'd be happy to share my knowledge with you and provide personalized guidance.
Answer by Edwin McCravy(20056)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
One response to your question has the right answer by a difficult path. The other has the wrong solution because it applies the angle bisector theorem incorrectly.
Here is a simpler correct solution.
PQ=QR=10, so triangle PQR is isosceles with vertex Q.
Let Z be the foot of the perpendicular from Q to side PR. Since PQR is isosceles, Z is the midpoint of PR, making ZR=6.
Triangle QZR is a right triangle with leg ZR=6 and hypotenuse QR=10. By the Pythagorean Theorem, that makes leg QZ=8.
Triangles QZR and XYR are now similar. (Both are right triangles that have angle R in common.)
PX is the bisector of angle P; by the angle bisector theorem, QX/RX = PQ/PR = 10/12 = 5/6. That makes RX/RQ = 6/11.
RX and RQ are the hypotenuses of similar triangles XYR and QZR. Then, by corresponding parts of similar triangles, XY/QZ = 6/11.
But QZ=8, making XY (6/11)*8 = 48/11.
ANSWER: XZ = 48/11
|
|
|
