Segments BF, CD, and AE are medians for triangle ABC. Point K is the intersection of all three medians. The measure of segment FK is 3.40 cm and segment EK = 1.63 cm. Find the sum of the lengths of segments AK and BK.
Answers
The point of concurrency of the three median is the centroid of the triangle ABC.
The sum of the lengths of AK and BK is 10.06Reasons:
The given parameters are;
The medians of the triangle ABC = Segments BF, CD, and AE
The length of segment FK = 3.40
Length of segment EK = 1.63
Required:
The sum of the lengths of segment AK and BK
Solution:
The ratios of the lengths of AK to KE = 2:1
Therefore, we have;
[tex]\dfrac{AK}{KE} = \dfrac{2}{1} = \mathbf{\dfrac{AK}{1.63}}[/tex]
1.63 × 2 = 1 × AK
AK = 3.26
Similarly, we have;
[tex]\dfrac{BK}{KF} = \dfrac{2}{1} = \mathbf{\dfrac{BK}{3.40}}[/tex]
3.40 × 2 = BK
6.80 = BK
Which gives;
AK + BK = 3.26 + 6.80 = 10.06
The sum of the lengths of AK and BK = 10.06
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