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Geometry problem
- Thread starter maine
- Start date
D
Deleted member 4993
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Hello folks, I have serious problems with this task. Anyone have time to help?
The rectangle ABCD has side lengths AB = 8 og BC = 6.
The inscribed circles in ΔABD and ΔBCD have centers K and L.
Determine the length of the segment KL.
View attachment 1613
The problem can be solved easily using in-center properties discussed in:
http://mathworld.wolfram.com/Incenter.html
Hello, maine!
Consider the right triangle \(\displaystyle ABD.\)
The area of the triangle is: .\(\displaystyle \frac{1}{2}bh \,=\,\frac{1}{2}(6)(8) = 24\)
Using Pythagorus, we find the hypotenuse \(\displaystyle BD = 10.\)
The perimeter of the triangle is: .\(\displaystyle p \:=\:6+8+10 \:=\:24\)
Let \(\displaystyle r\) = radius of the inscribed circle.
The area of the triangle is given by: .\(\displaystyle \text{Area }\:=\: \frac{1}{2}rp\)
Hence, we have: .\(\displaystyle 24 \:=\:\frac{1}{2}r(24) \quad\Rightarrow\quad r = 2\)
Place the rectangle on a coordinate system with vertex \(\displaystyle D\) at the origin.
Then \(\displaystyle K\) is 2 unit to the right and 2 units down from vertex \(\displaystyle A(0,6).\)
. . We have: .\(\displaystyle K(2,4)\)
Similarly, \(\displaystyle L\) is 2 units to the left and 2 units up from vertex \(\displaystyle C(8,0).\)
. . We have: .\(\displaystyle L(6,2)\)
Use the Distance Formula on point \(\displaystyle K\) and \(\displaystyle L.\)
The rectangle ABCD has side lengths AB = 8 and BC = 6.
The inscribed circles in ΔABD and ΔBCD have centers K and L.
Determine the length of the segment KL.
View attachment 1613
Consider the right triangle \(\displaystyle ABD.\)
The area of the triangle is: .\(\displaystyle \frac{1}{2}bh \,=\,\frac{1}{2}(6)(8) = 24\)
Using Pythagorus, we find the hypotenuse \(\displaystyle BD = 10.\)
The perimeter of the triangle is: .\(\displaystyle p \:=\:6+8+10 \:=\:24\)
Let \(\displaystyle r\) = radius of the inscribed circle.
The area of the triangle is given by: .\(\displaystyle \text{Area }\:=\: \frac{1}{2}rp\)
Hence, we have: .\(\displaystyle 24 \:=\:\frac{1}{2}r(24) \quad\Rightarrow\quad r = 2\)
Place the rectangle on a coordinate system with vertex \(\displaystyle D\) at the origin.
Then \(\displaystyle K\) is 2 unit to the right and 2 units down from vertex \(\displaystyle A(0,6).\)
. . We have: .\(\displaystyle K(2,4)\)
Similarly, \(\displaystyle L\) is 2 units to the left and 2 units up from vertex \(\displaystyle C(8,0).\)
. . We have: .\(\displaystyle L(6,2)\)
Use the Distance Formula on point \(\displaystyle K\) and \(\displaystyle L.\)